\input{preamble} % OK, start here. % \begin{document} \title{Exercises} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Algebra} \label{section-algebra} \noindent This first section just contains some assorted questions. \begin{exercise} \label{exercise-isomorphism-localizations} Let $A$ be a ring, and ${\mathfrak m}$ a maximal ideal. In $A[X]$ let $\tilde {\mathfrak m}_1 = ({\mathfrak m}, X)$ and $\tilde {\mathfrak m}_2 = ({\mathfrak m}, X-1)$. Show that $$A[X]_{\tilde {\mathfrak m}_1} \cong A[X]_{\tilde {\mathfrak m}_2}.$$ \end{exercise} \begin{exercise} \label{exercise-coherent} Find an example of a non Noetherian ring $R$ such that every finitely generated ideal of $R$ is finitely presented as an $R$-module. (A ring is said to be {\it coherent} if the last property holds.) \end{exercise} \begin{exercise} \label{exercise-flat-ideals-pid} Suppose that $(A, {\mathfrak m}, k)$ is a Noetherian local ring. For any finite $A$-module $M$ define $r(M)$ to be the minimum number of generators of $M$ as an $A$-module. This number equals $\dim_k M/{\mathfrak m} M = \dim_k M \otimes_A k$ by NAK. \begin{enumerate} \item Show that $r(M \otimes_A N) = r(M)r(N)$. \item Let $I\subset A$ be an ideal with $r(I) > 1$. Show that $r(I^2) < r(I)^2$. \item Conclude that if every ideal in $A$ is a flat module, then $A$ is a PID (or a field). \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-not-isomorphic} Let $k$ be a field. Show that the following pairs of $k$-algebras are not isomorphic: \begin{enumerate} \item $k[x_1, \ldots, x_n]$ and $k[x_1, \ldots, x_{n + 1}]$ for any $n\geq 1$. \item $k[a, b, c, d, e, f]/(ab + cd + ef)$ and $k[x_1, \ldots, x_n]$ for $n = 5$. \item $k[a, b, c, d, e, f]/(ab + cd + ef)$ and $k[x_1, \ldots, x_n]$ for $n = 6$. \end{enumerate} \end{exercise} \begin{remark} \label{remark-simple-geometric} Of course the idea of this exercise is to find a simple argument in each case rather than applying a big'' theorem. Nonetheless it is good to be guided by general principles. \end{remark} \begin{exercise} \label{exercise-silly} Algebra. (Silly and should be easy.) \begin{enumerate} \item Give an example of a ring $A$ and a nonsplit short exact sequence of $A$-modules $$0 \to M_1 \to M_2 \to M_3 \to 0.$$ \item Give an example of a nonsplit sequence of $A$-modules as above and a faithfully flat $A \to B$ such that $$0 \to M_1\otimes_A B \to M_2\otimes_A B \to M_3\otimes_A B \to 0.$$ is split as a sequence of $B$-modules. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-field-kummer} Suppose that $k$ is a field having a primitive $n$th root of unity $\zeta$. This means that $\zeta^n = 1$, but $\zeta^m\not = 1$ for $0 < m < n$. \begin{enumerate} \item Show that the characteristic of $k$ is prime to $n$. \item Suppose that $a \in k$ is an element of $k$ which is not an $d$th power in $k$ for any divisor $d$ of $n$ for $n \geq d > 1$. Show that $k[x]/(x^n-a)$ is a field. (Hint: Consider a splitting field for $x^n-a$ and use Galois theory.) \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-valuation} Let $\nu : k[x]\setminus \{0\} \to {\mathbf Z}$ be a map with the following properties: $\nu(fg) = \nu(f) + \nu(g)$ whenever $f$, $g$ not zero, and $\nu(f + g) \geq min(\nu(f), \nu(g))$ whenever $f$, $g$, $f + g$ are not zero, and $\nu(c) = 0$ for all $c\in k^*$. \begin{enumerate} \item Show that if $f$, $g$, and $f + g$ are nonzero and $\nu(f) \not = \nu(g)$ then we have equality $\nu(f + g) = min(\nu(f), \nu(g))$. \item Show that if $f = \sum a_i x^i$, $f\not = 0$, then $\nu(f) \geq min(\{i\nu(x)\}_{a_i\not = 0})$. When does equality hold? \item Show that if $\nu$ attains a negative value then $\nu(f) = -n \deg(f)$ for some $n\in {\mathbf N}$. \item Suppose $\nu(x) \geq 0$. Show that $\{f \mid f = 0, \ or\ \nu(f) > 0\}$ is a prime ideal of $k[x]$. \item Describe all possible $\nu$. \end{enumerate} \end{exercise} \noindent Let $A$ be a ring. An {\it idempotent} is an element $e \in A$ such that $e^2 = e$. The elements $1$ and $0$ are always idempotent. A {\it nontrivial idempotent} is an idempotent which is not equal to zero. Two idempotents $e, e' \in A$ are called {\it orthogonal} if $ee' = 0$. \begin{exercise} \label{exercise-product} Let $A$ be a ring. Show that $A$ is a product of two nonzero rings if and only if $A$ has a nontrivial idempotent. \end{exercise} \begin{exercise} \label{exercise-lift-idempotents} Let $A$ be a ring and let $I \subset A$ be a locally nilpotent ideal. Show that the map $A \to A/I$ induces a bijection on idempotents. (Hint: It may be easier to prove this when $I$ is nilpotent. Do this first. Then use absolute Noetherian reduction'' to reduce to the nilpotent case.) \end{exercise} \section{Colimits} \label{section-colimits} \begin{definition} \label{definition-directed-poset} A {\it directed set} is a nonempty set $I$ endowed with a preorder $\leq$ such that given any pair $i, j \in I$ there exists a $k \in I$ such that $i \leq k$ and $j \leq k$. A {\it system of rings} over $I$ is given by a ring $A_i$ for each $i \in I$ and a map of rings $\varphi_{ij} : A_i \to A_j$ whenever $i \leq j$ such that the composition $A_i \to A_j \to A_k$ is equal to $A_i \to A_k$ whenever $i \leq j \leq k$. \end{definition} \noindent One similarly defines systems of groups, modules over a fixed ring, vector spaces over a field, etc. \begin{exercise} \label{exercise-directed-colimit} Let $I$ be a directed set and let $(A_i, \varphi_{ij})$ be a system of rings over $I$. Show that there exists a ring $A$ and maps $\varphi_i : A_i \to A$ such that $\varphi_j \circ \varphi_{ij} = \varphi_i$ for all $i \leq j$ with the following universal property: Given any ring $B$ and maps $\psi_i : A_i \to B$ such that $\psi_j \circ \varphi_{ij} = \psi_i$ for all $i \leq j$, then there exists a unique ring map $\psi : A \to B$ such that $\psi_i = \psi \circ \varphi_i$. \end{exercise} \begin{definition} \label{definition-colimit} The ring $A$ constructed in Exercise \ref{exercise-directed-colimit} is called the {\it colimit} of the system. Notation $\colim A_i$. \end{definition} \begin{exercise} \label{exercise-prime-in-colimit} Let $(I, \geq)$ be a directed set and let $(A_i, \varphi_{ij})$ be a system of rings over $I$ with colimit $A$. Prove that there is a bijection $$\Spec(A) = \{(\mathfrak p_i)_{i \in I} \mid \mathfrak p_i \subset A_i \text{ and } \mathfrak p_i = \varphi_{ij}^{-1}(\mathfrak p_j)\ \forall i \leq j\} \subset \prod\nolimits_{i \in I} \Spec(A_i)$$ The set on the right hand side of the equality is the limit of the sets $\Spec(A_i)$. Notation $\lim \Spec(A_i)$. \end{exercise} \begin{exercise} \label{exercise-colimit-surjective} Let $(I, \geq)$ be a directed set and let $(A_i, \varphi_{ij})$ be a system of rings over $I$ with colimit $A$. Suppose that $\Spec(A_j) \to \Spec(A_i)$ is surjective for all $i \leq j$. Show that $\Spec(A) \to \Spec(A_i)$ is surjective for all $i$. (Hint: You can try to use Tychonoff, but there is also a basically trivial direct algebraic proof based on Algebra, Lemma \ref{algebra-lemma-in-image}.) \end{exercise} \begin{exercise} \label{exercise-integral-colimit-finite} Let $A \subset B$ be an integral ring extension. Prove that $\Spec(B) \to \Spec(A)$ is surjective. Use the exercises above, the fact that this holds for a finite ring extension (proved in the lectures), and by proving that $B = \colim B_i$ is a directed colimit of finite extensions $A \subset B_i$. \end{exercise} \begin{exercise} \label{exercise-colimit-tensor} Let $(I, \geq)$ be a directed set. Let $A$ be a ring and let $(N_i, \varphi_{i, i'})$ be a directed system of $A$-modules indexed by $I$. Suppose that $M$ is another $A$-module. Prove that $$\colim_{i\in I} M \otimes_A N_i\cong M \otimes_A \Big( \colim_{i\in I} N_i\Big).$$ \end{exercise} \begin{definition} \label{definition-finite-presentation} A module $M$ over $R$ is said to be of {\it finite presentation} over $R$ if it is isomorphic to the cokernel of a map of finite free modules $R^{\oplus n} \to R^{\oplus m}$. \end{definition} \begin{exercise} \label{exercise-colimit-modules} Prove that any module over any ring is \begin{enumerate} \item the colimit of its finitely generated submodules, and \item in some way a colimit of finitely presented modules. \end{enumerate} \end{exercise} \section{Additive and abelian categories} \label{section-additive} \begin{exercise} \label{exercise-filtered-vector-spaces} Let $k$ be a field. Let $\mathcal{C}$ be the category of filtered vector spaces over $k$, see Homology, Definition \ref{homology-definition-filtered} for the definition of a filtered object of any category. \begin{enumerate} \item Show that this is an additive category (explain carefuly what the direct sum of two objects is). \item Let $f : (V, F) \to (W, F)$ be a morphism of $\mathcal{C}$. Show that $f$ has a kernel and cokernel (explain precisely what the kernel and cokernel of $f$ are). \item Give an example of a map of $\mathcal{C}$ such that the canonical map $\Coim(f) \to \Im(f)$ is not an isomorphism. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-torsion-free} Let $R$ be a Noetherian domain. Let $\mathcal{C}$ be the category of finitely generated torsion free $R$-modules. \begin{enumerate} \item Show that this is an additive category. \item Let $f : N \to M$ be a morphism of $\mathcal{C}$. Show that $f$ has a kernel and cokernel (make sure you define precisely what the kernel and cokernel of $f$ are). \item Give an example of a Noetherian domain $R$ and a map of $\mathcal{C}$ such that the canonical map $\Coim(f) \to \Im(f)$ is not an isomorphism. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-other} Give an example of a category which is additive and has kernels and cokernels but which is not as in Exercises \ref{exercise-filtered-vector-spaces} and \ref{exercise-torsion-free}. \end{exercise} \section{Tensor product} \label{section-tensor-product} \noindent Tensor products are introduced in Algebra, Section \ref{algebra-section-tensor-product}. Let $R$ be a ring. Let $\text{Mod}_R$ be the category of $R$-modules. We will say that a functor $F : \text{Mod}_R \to \text{Mod}_R$ \begin{enumerate} \item is additive if $F : \Hom_R(M, N) \to \Hom_R(F(M), F(N))$ is a homomorphism of abelian groups for any $R$-modules $M, N$, see Homology, Definition \ref{homology-definition-preadditive}. \item $R$-linear if $F : \Hom_R(M, N) \to \Hom_R(F(M), F(N))$ is $R$-linear for any $R$-modules $M, N$, \item right exact if for any short exact sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ the sequence $F(M_1) \to F(M_2) \to F(M_3) \to 0$ is exact, \item left exact if for any short exact sequence $0 \to M_1 \to M_2 \to M_3 \to 0$ the sequence $0 \to F(M_1) \to F(M_2) \to F(M_3)$ is exact, \item commutes with direct sums, if given a set $I$ and $R$-modules $M_i$ the maps $F(M_i) \to F(\bigoplus M_i)$ induce an isomorphism $\bigoplus F(M_i) = F(\bigoplus M_i)$. \end{enumerate} \begin{exercise} \label{exercise-characterize-tensor-functor} Let $R$ be a ring. With notation as above. \begin{enumerate} \item Give an example of a ring $R$ and an additive functor $F : \text{Mod}_R \to \text{Mod}_R$ which is not $R$-linear. \item Let $N$ be an $R$-module. Show that the functor $F(M) = M \otimes_R N$ is $R$-linear, right exact, and commutes with direct sums, \item Conversely, show that any functor $F : \text{Mod}_R \to \text{Mod}_R$ which is $R$-linear, right exact, and commutes with direct sums is of the form $F(M) = M \otimes_R N$ for some $R$-module $N$. \item Show that if in (3) we drop the assumption that $F$ commutes with direct sums, then the conclusion no longer holds. \end{enumerate} \end{exercise} \section{Flat ring maps} \label{section-flat} \begin{exercise} \label{exercise-localization-flat} Let $S$ be a multiplicative subset of the ring $A$. \begin{enumerate} \item For an $A$-module $M$ show that $S^{-1}M = S^{-1}A \otimes_A M$. \item Show that $S^{-1}A$ is flat over $A$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-examples-not-flat} Find an injection $M_1 \to M_2$ of $A$-modules such that $M_1\otimes N \to M_2 \otimes N$ is not injective in the following cases: \begin{enumerate} \item $A = k[x, y]$ and $N = (x, y) \subset A$. (Here and below $k$ is a field.) \item $A = k[x, y]$ and $N = A/(x, y)$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-flat-not-projective} Give an example of a ring $A$ and a finite $A$-module $M$ which is a flat but not a projective $A$-module. \end{exercise} \begin{remark} \label{remark-flat-not-projective} If $M$ is of finite presentation and flat over $A$, then $M$ is projective over $A$. Thus your example will have to involve a ring $A$ which is not Noetherian. I know of an example where $A$ is the ring of ${\mathcal C}^\infty$-functions on ${\mathbf R}$. \end{remark} \begin{exercise} \label{exercise-flat-not-free-dvr} Find a flat but not free module over ${\mathbf Z}_{(2)}$. \end{exercise} \begin{exercise} \label{exercise-flat-deformations} Flat deformations. \begin{enumerate} \item Suppose that $k$ is a field and $k[\epsilon]$ is the ring of dual numbers $k[\epsilon] = k[x]/(x^2)$ and $\epsilon = \bar x$. Show that for any $k$-algebra $A$ there is a flat $k[\epsilon]$-algebra $B$ such that $A$ is isomorphic to $B/\epsilon B$. \item Suppose that $k = {\mathbf F}_p = {\mathbf Z}/p{\mathbf Z}$ and $$A = k[x_1, x_2, x_3, x_4, x_5, x_6]/ (x_1^p, x_2^p, x_3^p, x_4^p, x_5^p, x_6^p).$$ Show that there exists a flat ${\mathbf Z}/p^2{\mathbf Z}$-algebra $B$ such that $B/pB$ is isomorphic to $A$. (So here $p$ plays the role of $\epsilon$.) \item Now let $p = 2$ and consider the same question for $k = {\mathbf F}_2 = {\mathbf Z}/2{\mathbf Z}$ and $$A = k[x_1, x_2, x_3, x_4, x_5, x_6]/ (x_1^2, x_2^2, x_3^2, x_4^2, x_5^2, x_6^2, x_1x_2 + x_3x_4 + x_5x_6).$$ However, in this case show that there does {\it not} exist a flat ${\mathbf Z}/4{\mathbf Z}$-algebra $B$ such that $B/2B$ is isomorphic to $A$. (Find the trick! The same example works in arbitrary characteristic $p > 0$, except that the computation is more difficult.) \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-flat-given-residue-field-extension} Let $(A, {\mathfrak m}, k)$ be a local ring and let $k'/k$ be a finite field extension. Show there exists a flat, local map of local rings $A \to B$ such that ${\mathfrak m}_B = {\mathfrak m} B$ and $B/{\mathfrak m} B$ is isomorphic to $k'$ as $k$-algebra. (Hint: first do the case where $k \subset k'$ is generated by a single element.) \end{exercise} \begin{remark} \label{remark-flat-given-residue-field-extension-general} The same result holds for arbitrary field extensions $K/k$. \end{remark} \section{The Spectrum of a ring} \label{section-spectrum-ring} \begin{exercise} \label{exercise-spec-Z} Compute $\Spec(\mathbf{Z})$ as a set and describe its topology. \end{exercise} \begin{exercise} \label{exercise-basis-opens-standard} Let $A$ be any ring. For $f\in A$ we define $D(f):= \{\mathfrak p \subset A \mid f \not \in \mathfrak p\}$. Prove that the open subsets $D(f)$ form a basis of the topology of $\Spec(A)$. \end{exercise} \begin{exercise} \label{exercise-radical-ideals-closed} Prove that the map $I\mapsto V(I)$ defines a natural bijection $$\{I\subset A\text{ with }I = \sqrt{I}\} \longrightarrow \{T\subset \Spec(A)\text{ closed}\}$$ \end{exercise} \begin{definition} \label{definition-quasi-compact} A topological space $X$ is called {\it quasi-compact} if for any open covering $X = \bigcup_{i\in I} U_i$ there is a finite subset $\{i_1, \ldots, i_n\}\subset I$ such that $X = U_{i_1}\cup\ldots U_{i_n}$. \end{definition} \begin{exercise} \label{exercise-spec-quasi-compact} Prove that $\Spec(A)$ is quasi-compact for any ring $A$. \end{exercise} \begin{definition} \label{definition-Hausdorff} A topological space $X$ is said to verify the separation axiom $T_0$ if for any pair of points $x, y\in X$, $x\not = y$ there is an open subset of $X$ containing one but not the other. We say that $X$ is {\it Hausdorff} if for any pair $x, y\in X$, $x\not = y$ there are disjoint open subsets $U, V$ such that $x\in U$ and $y\in V$. \end{definition} \begin{exercise} \label{exercise-not-hausdorff} Show that $\Spec(A)$ is {\bf not} Hausdorff in general. Prove that $\Spec(A)$ is $T_0$. Give an example of a topological space $X$ that is not $T_0$. \end{exercise} \begin{remark} \label{remark-not-hausdorff} Usually the word compact is reserved for quasi-compact and Hausdorff spaces. \end{remark} \begin{definition} \label{definition-irreducible} A topological space $X$ is called {\it irreducible} if $X$ is not empty and if $X = Z_1\cup Z_2$ with $Z_1, Z_2\subset X$ closed, then either $Z_1 = X$ or $Z_2 = X$. A subset $T\subset X$ of a topological space is called {\it irreducible} if it is an irreducible topological space with the topology induced from $X$. This definition implies $T$ is irreducible if and only if the closure $\bar T$ of $T$ in $X$ is irreducible. \end{definition} \begin{exercise} \label{exercise-irreducible-spec} Prove that $\Spec(A)$ is irreducible if and only if $Nil(A)$ is a prime ideal and that in this case it is the unique minimal prime ideal of $A$. \end{exercise} \begin{exercise} \label{exercise-irreducible-prime} Prove that a closed subset $T\subset \Spec(A)$ is irreducible if and only if it is of the form $T = V({\mathfrak p})$ for some prime ideal ${\mathfrak p}\subset A$. \end{exercise} \begin{definition} \label{definition-generic-point} A point $x$ of an irreducible topological space $X$ is called a {\it generic point} of $X$ if $X$ is equal to the closure of the subset $\{x\}$. \end{definition} \begin{exercise} \label{exercise-irreducible-T0-at-most-one-generic} Show that in a $T_0$ space $X$ every irreducible closed subset has at most one generic point. \end{exercise} \begin{exercise} \label{exercise-spec-sober} Prove that in $\Spec(A)$ every irreducible closed subset {\it does} have a generic point. In fact show that the map ${\mathfrak p} \mapsto \overline{\{{\mathfrak p}\}}$ is a bijection of $\Spec(A)$ with the set of irreducible closed subsets of $X$. \end{exercise} \begin{exercise} \label{exercise-irreducible-subset-not-generic} Give an example to show that an irreducible subset of $\Spec(\mathbf{Z})$ does not necessarily have a generic point. \end{exercise} \begin{definition} \label{definition-Noetherian-space} A topological space $X$ is called {\it Noetherian} if any decreasing sequence $Z_1\supset Z_2 \supset Z_3\supset \ldots$ of closed subsets of $X$ stabilizes. (It is called {\it Artinian} if any increasing sequence of closed subsets stabilizes.) \end{definition} \begin{exercise} \label{exercise-Noetherian-spec} Show that if the ring $A$ is Noetherian then the topological space $\Spec(A)$ is Noetherian. Give an example to show that the converse is false. (The same for Artinian if you like.) \end{exercise} \begin{definition} \label{definition-irreducible-component} A maximal irreducible subset $T\subset X$ is called an {\it irreducible component} of the space $X$. Such an irreducible component of $X$ is automatically a closed subset of $X$. \end{definition} \begin{exercise} \label{exercise-irreducible-in-irreducible} Prove that any irreducible subset of $X$ is contained in an irreducible component of $X$. \end{exercise} \begin{exercise} \label{exercise-Noetherian-finite-nr-irreducible} Prove that a Noetherian topological space $X$ has only finitely many irreducible components, say $X_1, \ldots, X_n$, and that $X = X_1\cup X_2\cup\ldots\cup X_n$. (Note that any $X$ is always the union of its irreducible components, but that if $X = {\mathbf R}$ with its usual topology for instance then the irreducible components of $X$ are the one point subsets. This is not terribly interesting.) \end{exercise} \begin{exercise} \label{exercise-irreducible-components-minimal-primes} Show that irreducible components of $\Spec(A)$ correspond to minimal primes of $A$. \end{exercise} \begin{definition} \label{definition-closed} A point $x\in X$ is called {\it closed} if $\overline{\{x\}} = \{ x\}$. Let $x, y$ be points of $X$. We say that $x$ is a {\it specialization} of $y$, or that $y$ is a {\it generalization} of $x$ if $x\in \overline{\{y\}}$. \end{definition} \begin{exercise} \label{exercise-closed-maximal} Show that closed points of $\Spec(A)$ correspond to maximal ideals of $A$. \end{exercise} \begin{exercise} \label{exercise-generalization} Show that ${\mathfrak p}$ is a generalization of ${\mathfrak q}$ in $\Spec(A)$ if and only if ${\mathfrak p}\subset {\mathfrak q}$. Characterize closed points, maximal ideals, generic points and minimal prime ideals in terms of generalization and specialization. (Here we use the terminology that a point of a possibly reducible topological space $X$ is called a generic point if it is a generic points of one of the irreducible components of $X$.) \end{exercise} \begin{exercise} \label{exercise-disjoint-closed-spec} Let $I$ and $J$ be ideals of $A$. What is the condition for $V(I)$ and $V(J)$ to be disjoint? \end{exercise} \begin{definition} \label{definition-connected-component} A topological space $X$ is called {\it connected} if it is nonempty and not the union of two nonempty disjoint open subsets. A {\it connected component} of $X$ is a maximal connected subset. Any point of $X$ is contained in a connected component of $X$ and any connected component of $X$ is closed in $X$. (But in general a connected component need not be open in $X$.) \end{definition} \begin{exercise} \label{exercise-disconnected-spec} Let $A$ be a nonzero ring. Show that $\Spec(A)$ is disconnected iff $A\cong B \times C$ for certain nonzero rings $B, C$. \end{exercise} \begin{exercise} \label{exercise-connected-component-stable-generalization} Let $T$ be a connected component of $\Spec(A)$. Prove that $T$ is stable under generalization. Prove that $T$ is an open subset of $\Spec(A)$ if $A$ is Noetherian. (Remark: This is wrong when $A$ is an infinite product of copies of ${\mathbf F}_2$ for example. The spectrum of this ring consists of infinitely many closed points.) \end{exercise} \begin{exercise} \label{exercise-primes-kx} Compute $\Spec(k[x])$, i.e., describe the prime ideals in this ring, describe the possible specializations, and describe the topology. (Work this out when $k$ is algebraically closed but also when $k$ is not.) \end{exercise} \begin{exercise} \label{exercise-primes-kxy} Compute $\Spec(k[x, y])$, where $k$ is algebraically closed. [Hint: use the morphism $\varphi : \Spec(k[x, y]) \to \Spec(k[x])$; if $\varphi({\mathfrak p}) = (0)$ then localize with respect to $S = \{f\in k[x] \mid f \not = 0\}$ and use result of lecture on localization and $\Spec$.] (Why do you think algebraic geometers call this affine 2-space?) \end{exercise} \begin{exercise} \label{exercise-primes-Zy} Compute $\Spec(\mathbf{Z}[y])$. [Hint: as above.] (Affine 1-space over $\mathbf{Z}$.) \end{exercise} \section{Localization} \label{section-localization} \begin{exercise} \label{exercise-submodule-localization} Let $A$ be a ring. Let $S \subset A$ be a multiplicative subset. Let $M$ be an $A$-module. Let $N \subset S^{-1}M$ be an $S^{-1}A$-submodule. Show that there exists an $A$-submodule $N' \subset M$ such that $N = S^{-1}N'$. (This useful result applies in particular to ideals of $S^{-1}A$.) \end{exercise} \begin{exercise} \label{exercise-localize-zero} Let $A$ be a ring. Let $M$ be an $A$-module. Let $m \in M$. \begin{enumerate} \item Show that $I = \{a \in A \mid am = 0\}$ is an ideal of $A$. \item For a prime $\mathfrak p$ of $A$ show that the image of $m$ in $M_\mathfrak p$ is zero if and only if $I \not \subset \mathfrak p$. \item Show that $m$ is zero if and only if the image of $m$ is zero in $M_\mathfrak p$ for all primes $\mathfrak p$ of $A$. \item Show that $m$ is zero if and only if the image of $m$ is zero in $M_\mathfrak m$ for all maximal ideals $\mathfrak m$ of $A$. \item Show that $M = 0$ if and only if $M_{\mathfrak m}$ is zero for all maximal ideals $\mathfrak m$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-localization-is-field} Find a pair $(A, f)$ where $A$ is a domain with three or more pairwise distinct primes and $f \in A$ is an element such that the principal localization $A_f = \{1, f, f^2, \ldots \}^{-1}A$ is a field. \end{exercise} \begin{exercise} \label{exercise-localize-finite-module-zero} Let $A$ be a ring. Let $M$ be a finite $A$-module. Let $S \subset A$ be a multiplicative set. Assume that $S^{-1}M = 0$. Show that there exists an $f \in S$ such that the principal localization $M_f = \{1, f, f^2, \ldots \}^{-1}M$ is zero. \end{exercise} \begin{exercise} \label{exercise-localization-is-quotient} Give an example of a triple $(A, I, S)$ where $A$ is a ring, $0 \not = I \not = A$ is a proper nonzero ideal, and $S \subset A$ is a multiplicative subset such that $A/I \cong S^{-1}A$ as $A$-algebras. \end{exercise} \section{Nakayama's Lemma} \label{section-nakayama} \begin{exercise} \label{exercise-nakayama} Let $A$ be a ring. Let $I$ be an ideal of $A$. Let $M$ be an $A$-module. Let $x_1, \ldots, x_n \in M$. Assume that \begin{enumerate} \item $M/IM$ is generated by $x_1, \ldots, x_n$, \item $M$ is a finite $A$-module, \item $I$ is contained in every maximal ideal of $A$. \end{enumerate} Show that $x_1, \ldots, x_n$ generate $M$. (Suggested solution: Reduce to a localization at a maximal ideal of $A$ using Exercise \ref{exercise-localize-zero} and exactness of localization. Then reduce to the statement of Nakayama's lemma in the lectures by looking at the quotient of $M$ by the submodule generated by $x_1, \ldots, x_n$.) \end{exercise} \section{Length} \label{section-length} \begin{definition} \label{definition-length} Let $A$ be a ring. Let $M$ be an $A$-module. The {\it length} of $M$ as an $R$-module is $$\text{length}_A(M) = \sup \{ n \mid \exists\ 0 = M_0 \subset M_1 \subset \ldots \subset M_n = M, \text{ }M_i \not = M_{i + 1} \}.$$ In other words, the supremum of the lengths of chains of submodules. \end{definition} \begin{exercise} \label{exercise-length-is-one} Show that a module $M$ over a ring $A$ has length $1$ if and only if it is isomorphic to $A/\mathfrak m$ for some maximal ideal $\mathfrak m$ in $A$. \end{exercise} \begin{exercise} \label{exercise-length-easy} Compute the length of the following modules over the following rings. Briefly(!) explain your answer. (Please feel free to use additivity of the length function in short exact sequences, see Algebra, Lemma \ref{algebra-lemma-length-additive}). \begin{enumerate} \item The length of $\mathbf{Z}/120\mathbf{Z}$ over $\mathbf{Z}$. \item The length of $\mathbf{C}[x]/(x^{100} + x + 1)$ over $\mathbf{C}[x]$. \item The length of $\mathbf{R}[x]/(x^4 + 2x^2 + 1)$ over $\mathbf{R}[x]$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-compute-length} Let $A = k[x, y]_{(x, y)}$ be the local ring of the affine plane at the origin. Make any assumption you like about the field $k$. Suppose that $f = x^3 + x^2y^2 + y^{100}$ and $g = y^3 - x^{999}$. What is the length of $A/(f, g)$ as an $A$-module? (Possible way to proceed: think about the ideal that $f$ and $g$ generate in quotients of the form $A/{\mathfrak m}_A^n= k[x, y]/(x, y)^n$ for varying $n$. Try to find $n$ such that $A/(f, g)+{\mathfrak m}_A^n \cong A/(f, g)+{\mathfrak m}_A^{n + 1}$ and use NAK.) \end{exercise} \section{Associated primes} \label{section-ass} \noindent Associated primes are discussed in Algebra, Section \ref{algebra-section-ass} \begin{exercise} \label{exercise-compute-ass} Compute the set of associated primes for each of the following modules. \begin{enumerate} \item $R = k[x, y]$ and $M = R/(xy(x + y))$, \item $R = \mathbf{Z}[x]$ and $M = R/(300x + 75)$, and \item $R = k[x, y, z]$ and $M = R/(x^3, x^2y, xz)$. \end{enumerate} Here as usual $k$ is a field. \end{exercise} \begin{exercise} \label{exercise-prime-power-not-primary} Give an example of a Noetherian ring $R$ and a prime ideal $\mathfrak p$ such that $\mathfrak p$ is not the only associated prime of $R/\mathfrak p^2$. \end{exercise} \begin{exercise} \label{exercise-product-primes-not-only-associated} Let $R$ be a Noetherian ring with incomparable prime ideals $\mathfrak p$, $\mathfrak q$, i.e., $\mathfrak p \not \subset \mathfrak q$ and $\mathfrak q \not \subset \mathfrak p$. \begin{enumerate} \item Show that for $N = R/(\mathfrak p \cap \mathfrak q)$ we have $\text{Ass}(N) = \{\mathfrak p, \mathfrak q\}$. \item Show by an example that the module $M = R/\mathfrak p \mathfrak q$ can have an associated prime not equal to $\mathfrak p$ or $\mathfrak q$. \end{enumerate} \end{exercise} \section{Ext groups} \label{section-ext} \noindent Ext groups are defined in Algebra, Section \ref{algebra-section-ext}. \begin{exercise} \label{exercise-compute-ext-abelian-groups} Compute all the Ext groups $\Ext^i(M, N)$ of the given modules in the category of $\mathbf{Z}$-modules (also known as the category of abelian groups). \begin{enumerate} \item $M = \mathbf{Z}$ and $N = \mathbf{Z}$, \item $M = \mathbf{Z}/4\mathbf{Z}$ and $N = \mathbf{Z}/8\mathbf{Z}$, \item $M = \mathbf{Q}$ and $N = \mathbf{Z}/2\mathbf{Z}$, and \item $M = \mathbf{Z}/2\mathbf{Z}$ and $N = \mathbf{Q}/\mathbf{Z}$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-compute-in-regular} Let $R = k[x, y]$ where $k$ is a field. \begin{enumerate} \item Show by hand that the Koszul complex $$0 \to R \xrightarrow{ \left( \begin{matrix} y \\ -x \end{matrix} \right) } R^{\oplus 2} \xrightarrow{(x, y)} R \xrightarrow{f \mapsto f(0, 0)} k \to 0$$ is exact. \item Compute $\Ext^i_R(k, k)$ where $k = R/(x, y)$ as an $R$-module. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-infinitely-many-nonzero-ext} Give an example of a Noetherian ring $R$ and finite modules $M$, $N$ such that $\Ext^i_R(M, N)$ is nonzero for all $i \geq 0$. \end{exercise} \begin{exercise} \label{exercise-infinite-ext} Give an example of a ring $R$ and ideal $I$ such that $\Ext^1_R(R/I, R/I)$ is not a finite $R$-module. (We know this cannot happen if $R$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-ext-noetherian}.) \end{exercise} \section{Depth} \label{section-depth} \noindent Depth is defined in Algebra, Section \ref{algebra-section-depth} and further studied in Dualizing Complexes, Section \ref{dualizing-section-depth}. \begin{exercise} \label{exercise-compute-depth} Let $R$ be a ring, $I \subset R$ an ideal, and $M$ an $R$-module. Compute $\text{depth}_I(M)$ in the following cases. \begin{enumerate} \item $R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}$, \item $R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}/(300)$, \item $R = \mathbf{Z}$, $I = (30)$, $M = \mathbf{Z}/(7)$, \item $R = k[x, y, z]/(x^2 + y^2 + z^2)$, $I = (x, y, z)$, $M = R$, \item $R = k[x, y, z, w]/(xz, xw, yz, yw)$, $I = (x, y, z, w)$, $M = R$. \end{enumerate} Here $k$ is a field. In the last two cases feel free to localize at the maximal ideal $I$. \end{exercise} \begin{exercise} \label{exercise-depth-not-inherited-localization} Give an example of a Noetherian local ring $(R, \mathfrak m, \kappa)$ of depth $\geq 1$ and a prime ideal $\mathfrak p$ such that \begin{enumerate} \item $\text{depth}_\mathfrak m(R) \geq 1$, \item $\text{depth}_\mathfrak p(R_\mathfrak p) = 0$, and \item $\dim(R_\mathfrak p) \geq 1$. \end{enumerate} If we don't ask for (3) then the exercise is too easy. Why? \end{exercise} \begin{exercise} \label{exercise-depth-torsion-free} Let $(R, \mathfrak m)$ be a local Noetherian domain. Let $M$ be a finite $R$-module. \begin{enumerate} \item If $M$ is torsion free, show that $M$ has depth at least $1$ over $R$. \item Give an example with depth equal to $1$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-depth-examples} For every $m \geq n \geq 0$ give an example of a Noetherian local ring $R$ with $\dim(R) = m$ and $\text{depth}(R) = n$. \end{exercise} \begin{exercise} \label{exercise-make-depth-1} Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $R$-module. Show that there exists a canonical short exact sequence $$0 \to K \to M \to Q \to 0$$ such that the following are true \begin{enumerate} \item $\text{depth}(Q) \geq 1$, \item $K$ is zero or $\text{Supp}(K) = \{\mathfrak m\}$, and \item $\text{length}_R(K) < \infty$. \end{enumerate} Hint: using the Noetherian property show that there exists a maximal submodule $K$ as in (2) and then show that $Q = M/K$ satisfies (1) and $K$ satisfies (3). \end{exercise} \begin{exercise} \label{exercise-make-depth-2} Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a finite $R$-module of depth $\geq 2$. Let $N \subset M$ be a nonzero submodule. \begin{enumerate} \item Show that $\text{depth}(N) \geq 1$. \item Show that $\text{depth}(N) = 1$ if and only if the quotient module $M/N$ has $\text{depth}(M/N) = 0$. \item Show there exists a submodule $N' \subset M$ with $N \subset N'$ of finite colength, i.e., $\text{length}_R(N'/N) < \infty$, such that $N'$ has depth $\geq 2$. Hint: Apply Exercise \ref{exercise-make-depth-1} to $M/N$ and choose $N'$ to be the inverse image of $K$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-Hartshorne-reduced} Let $(R, \mathfrak m)$ be a Noetherian local ring. Assume that $R$ is reduced, i.e., $R$ has no nonzero nilpotent elements. Assume moreover that $R$ has two distinct minimal primes $\mathfrak p$ and $\mathfrak q$. \begin{enumerate} \item Show that the sequence of $R$-modules $$0 \to R \to R/\mathfrak p \oplus R/\mathfrak q \to R/\mathfrak p + \mathfrak q \to 0$$ is exact (check at all the spots). The maps are $x \mapsto (x \bmod \mathfrak p, x \bmod \mathfrak q)$ and $(y \bmod \mathfrak p, z \bmod \mathfrak q) \mapsto (y - z \bmod \mathfrak p + \mathfrak q)$. \item Show that if $\text{depth}(R) \geq 2$, then $\dim(R/\mathfrak p + \mathfrak q) \geq 1$. \item Show that if $\text{depth}(R) \geq 2$, then $U = \Spec(R) \setminus \{\mathfrak m\}$ is a connected topological space. \end{enumerate} This proves a very special case of Hartshorne's connectedness theorem which says that the punctured spectrum $U$ of a local Noetherian ring of $\text{depth} \geq 2$ is connected. \end{exercise} \begin{exercise} \label{exercise-depth-2} Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $x, y \in \mathfrak m$ be a regular sequence of length $2$. For any $n \geq 2$ show that there do not exist $a, b \in R$ with $$x^{n - 1}y^{n - 1} = a x^n + b y^n$$ Suggestion: First try for $n = 2$ to see how to argue. Remark: There is a vast generalization of this result called the monomial conjecture. \end{exercise} \section{Cohen-Macaulay modules and rings} \label{section-CM} \noindent Cohen-Macaulay modules are studied in Algebra, Section \ref{algebra-section-CM} and Cohen-Macaulay rings are studied in Algebra, Section \ref{algebra-section-CM-ring}. \begin{exercise} \label{exercise-examples-CM} In the following cases, please answer yes or no. No explanation or proof necessary. \begin{enumerate} \item Let $p$ be a prime number. Is the local ring $\mathbf{Z}_{(p)}$ a Cohen-Macaulay local ring? \item Let $p$ be a prime number. Is the local ring $\mathbf{Z}_{(p)}$ a regular local ring? \item Let $k$ be a field. Is the local ring $k[x]_{(x)}$ a Cohen-Macaulay local ring? \item Let $k$ be a field. Is the local ring $k[x]_{(x)}$ a regular local ring? \item Let $k$ be a field. Is the local ring $(k[x, y]/(y^2 - x^3))_{(x, y)} = k[x, y]_{(x, y)}/(y^2 - x^3)$ a Cohen-Macaulay local ring? \item Let $k$ be a field. Is the local ring $(k[x, y]/(y^2, xy))_{(x, y)} = k[x, y]_{(x, y)}/(y^2, xy)$ a Cohen-Macaulay local ring? \end{enumerate} \end{exercise} \section{Singularities} \label{section-singularities} \begin{exercise} \label{exercise-singularities} Let $k$ be any field. Suppose that $A = k[[x, y]]/(f)$ and $B = k[[u, v]]/(g)$, where $f = xy$ and $g = uv + \delta$ with $\delta \in (u, v)^3$. Show that $A$ and $B$ are isomorphic rings. \end{exercise} \begin{remark} \label{remark-singularities} A singularity on a curve over a field $k$ is called an ordinary double point if the complete local ring of the curve at the point is of the form $k'[[x, y]]/(f)$, where (a) $k'$ is a finite separable extension of $k$, (b) the initial term of $f$ has degree two, i.e., it looks like $q = ax^2 + bxy + cy^2$ for some $a, b, c\in k'$ not all zero, and (c) $q$ is a nondegenerate quadratic form over $k'$ (in char 2 this means that $b$ is not zero). In general there is one isomorphism class of such rings for each isomorphism class of pairs $(k', q)$. \end{remark} \begin{exercise} \label{exercise-periodic-resolution} Let $R$ be a ring. Let $n \geq 1$. Let $A$, $B$ be $n \times n$ matrices with coefficients in $R$ such that $AB = f 1_{n \times n}$ for some nonzerodivisor $f$ in $R$. Set $S = R/(f)$. Show that $$\ldots \to S^{\oplus n} \xrightarrow{B} S^{\oplus n} \xrightarrow{A} S^{\oplus n} \xrightarrow{B} S^{\oplus n} \to \ldots$$ is exact. \end{exercise} \section{Constructible sets} \label{section-constructible} \noindent Let $k$ be an algebraically closed field, for example the field $\mathbf{C}$ of complex numbers. Let $n \geq 0$. A polynomial $f \in k[x_1, \ldots, x_n]$ gives a function $f : k^n \to k$ by evaluation. A subset $Z \subset k^n$ is called an {\it algebraic set} if it is the common vanishing set of a collection of polynomials. \begin{exercise} \label{exercise-finite-nr-equations} Prove that an algebraic set can always be written as the zero locus of finitely many polynomials. \end{exercise} \noindent With notation as above a subset $E \subset k^n$ is called {\it constructible} if it is a finite union of sets of the form $Z \cap \{f \not = 0\}$ where $f$ is a polynomial. \begin{exercise} \label{exercise-constructible-classical} Show the following \begin{enumerate} \item the complement of a constructible set is a constructible set, \item a finite union of constructible sets is a constructible set, \item a finite intersection of constructible sets is a constructible set, and \item any constructible set $E$ can be written as a finite disjoint union $E = \coprod E_i$ with each $E_i$ of the form $Z \cap \{f \not = 0\}$ where $Z$ is an algebraic set and $f$ is a polynomial. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-division-with-remainder} Let $R$ be a ring. Let $f = a_d x^d + a_{d - 1} x^{d - 1} + \ldots + a_0 \in R[x]$. (As usual this notation means $a_0, \ldots, a_d \in R$.) Let $g \in R[x]$. Prove that we can find $N \geq 0$ and $r, q \in R[x]$ such that $$a_d^N g = q f + r$$ with $\deg(r) < d$, i.e., for some $c_i \in R$ we have $r = c_0 + c_1 x + \ldots + c_{d - 1}x^{d - 1}$. \end{exercise} \section{Hilbert Nullstellensatz} \label{section-Hilbert-Nullstellensatz} \begin{exercise} \label{exercise-uncountable} {\it A silly argument using the complex numbers!} Let ${\mathbf C}$ be the complex number field. Let $V$ be a vector space over ${\mathbf C}$. The spectrum of a linear operator $T : V \to V$ is the set of complex numbers $\lambda \in {\mathbf C}$ such that the operator $T - \lambda \text{id}_V$ is not invertible. \begin{enumerate} \item Show that $\mathbf{C}(X)$ has uncountable dimension over ${\mathbf C}$. \item Show that any linear operator on $V$ has a nonempty spectrum if the dimension of $V$ is finite or countable. \item Show that if a finitely generated ${\mathbf C}$-algebra $R$ is a field, then the map ${\mathbf C}\to R$ is an isomorphism. \item Show that any maximal ideal ${\mathfrak m}$ of ${\mathbf C}[x_1, \ldots, x_n]$ is of the form $(x_1-\alpha_1, \ldots, x_n-\alpha_n)$ for some $\alpha_i \in {\mathbf C}$. \end{enumerate} \end{exercise} \begin{remark} \label{remark-HNSS} Let $k$ be a field. Then for every integer $n\in {\mathbf N}$ and every maximal ideal ${\mathfrak m} \subset k[x_1, \ldots, x_n]$ the quotient $k[x_1, \ldots, x_n]/{\mathfrak m}$ is a finite field extension of $k$. This will be shown later in the course. Of course (please check this) it implies a similar statement for maximal ideals of finitely generated $k$-algebras. The exercise above proves it in the case $k = {\mathbf C}$. \end{remark} \begin{exercise} \label{exercise-Hilbert-Nullstellensatz} Let $k$ be a field. Please use Remark \ref{remark-HNSS}. \begin{enumerate} \item Let $R$ be a $k$-algebra. Suppose that $\dim_k R < \infty$ and that $R$ is a domain. Show that $R$ is a field. \item Suppose that $R$ is a finitely generated $k$-algebra, and $f\in R$ not nilpotent. Show that there exists a maximal ideal ${\mathfrak m} \subset R$ with $f\not\in {\mathfrak m}$. \item Show by an example that this statement fails when $R$ is not of finite type over a field. \item Show that any radical ideal $I \subset {\mathbf C}[x_1, \ldots, x_n]$ is the intersection of the maximal ideals containing it. \end{enumerate} \end{exercise} \begin{remark} \label{remark-Hilbert-Nullstellensatz} This is the Hilbert Nullstellensatz. Namely it says that the closed subsets of $\Spec(k[x_1, \ldots, x_n])$ (which correspond to radical ideals by a previous exercise) are determined by the closed points contained in them. \end{remark} \begin{exercise} \label{exercise-product-matrices-ring} Let $A = {\mathbf C}[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}]$. Let $I$ be the ideal of $A$ generated by the entries of the matrix $XY$, with $$X = \left( \begin{matrix} x_{11} & x_{12}\\ x_{21} & x_{22} \end{matrix} \right) \quad\text{and}\quad Y = \left( \begin{matrix} y_{11} & y_{12}\\ y_{21} & y_{22} \end{matrix} \right).$$ Find the irreducible components of the closed subset $V(I)$ of $\Spec(A)$. (I mean describe them and give equations for each of them. You do not have to prove that the equations you write down define prime ideals.) Hints: \begin{enumerate} \item You may use the Hilbert Nullstellensatz, and it suffices to find irreducible locally closed subsets which cover the set of closed points of $V(I)$. \item There are two easy components. \item An image of an irreducible set under a continuous map is irreducible. \end{enumerate} \end{exercise} \section{Dimension} \label{section-dimension} \begin{exercise} \label{exercise-dimension-bigger-one-finite-nr-primes} Construct a ring $A$ with finitely many prime ideals having dimension $> 1$. \end{exercise} \begin{exercise} \label{exercise-hypersurface-in-A2-dimension-one} Let $f \in \mathbf{C}[x, y]$ be a nonconstant polynomial. Show that $\mathbf{C}[x, y]/(f)$ has dimension $1$. \end{exercise} \begin{exercise} \label{exercise-dimension-polynomial-ring} Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $n \geq 1$. Let $\mathfrak m' = (\mathfrak m, x_1, \ldots, x_n)$ in the polynomial ring $R[x_1, \ldots, x_n]$. Show that $$\dim(R[x_1, \ldots, x_n]_{\mathfrak m'}) = \dim(R) + n.$$ \end{exercise} \section{Catenary rings} \label{section-catenary} \begin{definition} \label{definition-catenary} A Noetherian ring $A$ is said to be {\it catenary} if for any triple of prime ideals ${\mathfrak p}_1 \subset {\mathfrak p}_2 \subset {\mathfrak p}_3$ we have $$ht({\mathfrak p}_3 / {\mathfrak p}_1) = ht({\mathfrak p}_3/{\mathfrak p}_2) + ht({\mathfrak p}_2/{\mathfrak p}_1).$$ Here $ht(\mathfrak p/\mathfrak q)$ means the height of $\mathfrak p/\mathfrak q$ in the ring $A/\mathfrak q$. In a formula $$ht(\mathfrak p/\mathfrak q) = \dim(A_\mathfrak p/\mathfrak qA_\mathfrak p) = \dim((A/\mathfrak q)_\mathfrak p) = \dim((A/\mathfrak q)_{\mathfrak p/\mathfrak q})$$ A topological space $X$ is {\it catenary}, if given $T \subset T' \subset X$ with $T$ and $T'$ closed and irreducible, then there exists a maximal chain of irreducible closed subsets $$T = T_0 \subset T_1 \subset \ldots \subset T_n = T'$$ and every such chain has the same (finite) length. \end{definition} \begin{exercise} \label{exercise-catenary-the-same} Show that the notion of catenary defined in Algebra, Definition \ref{algebra-definition-catenary} agrees with the notion of Definition \ref{definition-catenary} for Noetherian rings. \end{exercise} \begin{exercise} \label{exercise-Noetherian-local-domain-dim-2-catenary} Show that a Noetherian local domain of dimension $2$ is catenary. \end{exercise} \begin{exercise} \label{exercise-finite-type-over-field-catenary} Let $k$ be a field. Show that a finite type $k$-algebra is catenary. \end{exercise} \begin{exercise} \label{exercise-example-no-dim-function} Give an example of a finite, sober, catenary topological space $X$ which does not have a dimension function $\delta : X \to \mathbf{Z}$. Here $\delta : X \to \mathbf{Z}$ is a dimension function if for $x, y \in X$ we have \begin{enumerate} \item $x \leadsto y$ and $x \not = y$ implies $\delta(x) > \delta(y)$, \item $x \leadsto y$ and $\delta(x) \geq \delta(y) + 2$ implies there exists a $z \in X$ with $x \leadsto z \leadsto y$ and $\delta(x) > \delta(z) > \delta(y)$. \end{enumerate} Describe your space clearly and succintly explain why there cannot be a dimension function. \end{exercise} \section{Fraction fields} \label{section-fraction-fields} \begin{exercise} \label{exercise-find-fraction-field} Consider the domain $${\mathbf Q}[r, s, t]/(s^2-(r-1)(r-2)(r-3), t^2-(r + 1)(r + 2)(r + 3)).$$ Find a domain of the form ${\mathbf Q}[x, y]/(f)$ with isomorphic field of fractions. \end{exercise} \section{Transcendence degree} \label{section-transcendence} \begin{exercise} \label{exercise-algebraic-extension} Let $K'/K/k$ be field extensions with $K'$ algebraic over $K$. Prove that $\text{trdeg}_k(K) = \text{trdeg}_k(K')$. (Hint: Show that if $x_1, \ldots, x_d \in K$ are algebraically independent over $k$ and $d < \text{trdeg}_k(K')$ then $k(x_1, \ldots, x_d) \subset K$ cannot be algebraic.) \end{exercise} \begin{exercise} \label{exercise-growth-powers-subvector-space} Let $k$ be a field. Let $K/k$ be a finitely generated extension of transcendence degree $d$. If $V, W \subset K$ are finite dimensional $k$-subvector spaces denote $$VW = \{f \in K \mid f = \sum\nolimits_{i = 1, \ldots, n} v_i w_i \text{ for some }n\text{ and }v_i \in V, w_i \in W\}$$ This is a finite dimensional $k$-subvector space. Set $V^2 = VV$, $V^3 = V V^2$, etc. \begin{enumerate} \item Show you can find $V \subset K$ and $\epsilon > 0$ such that $\dim V^n \geq \epsilon n^d$ for all $n \geq 1$. \item Conversely, show that for every finite dimensional $V \subset K$ there exists a $C > 0$ such that $\dim V^n \leq C n^d$ for all $n \geq 1$. (One possible way to proceed: First do this for subvector spaces of $k[x_1, \ldots, x_d]$. Then do this for subvector spaces of $k(x_1, \ldots, x_d)$. Finally, if $K/k(x_1, \ldots, x_d)$ is a finite extension choose a basis of $K$ over $k(x_1, \ldots, x_d)$ and argue using expansion in terms of this basis.) \item Conclude that you can redefine the transcendence degree in terms of growth of powers of finite dimensional subvector spaces of $K$. \end{enumerate} This is related to Gelfand-Kirillov dimension of (noncommutative) algebras over $k$. \end{exercise} \section{Dimension of fibres} \label{section-dimension-fibres} \noindent Some questions related to the dimension formula, see Algebra, Section \ref{algebra-section-dimension-formula}. \begin{exercise} \label{exercise-nr-components-fibre} Let $k$ be your favorite algebraically closed field. Below $k[x]$ and $k[x, y]$ denote the polynomial rings. \begin{enumerate} \item For every integer $n \geq 0$ find a finite type extension $k[x] \subset A$ of domains such that the spectrum of $A/xA$ has exactly $n$ irreducible components. \item Make an example of a finite type extension $k[x] \subset A$ of domains such that the spectrum of $A/(x - \alpha)A$ is nonempty and reducible for every $\alpha \in k$. \item Make an example of a finite type extension $k[x, y] \subset A$ of domains such that the spectrum of $A/(x - \alpha, y - \beta)A$ is irreducible\footnote{Recall that irreducible implies nonempty.} for all $(\alpha, \beta) \in k^2 \setminus \{(0, 0)\}$ and the spectrum of $A/(x, y)A$ is nonempty and reducible. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-codim-1} Let $k$ be your favorite algebraically closed field. Let $n \geq 1$. Let $k[x_1, \ldots, x_n]$ be the polynomial ring. Set $\mathfrak m = (x_1, \ldots, x_n)$. Let $k[x_1, \ldots, x_n] \subset A$ be a finite type extension of domains. Set $d = \dim(A)$. \begin{enumerate} \item Show that $d - 1 \geq \dim(A/\mathfrak m A) \geq d - n$ if $A/\mathfrak mA \not = 0$. \item Show by example that every value can occur. \item Show by example that $\Spec(A/\mathfrak m A)$ can have irreducible components of different dimensions. \end{enumerate} \end{exercise} \section{Finite locally free modules} \label{section-finite-locally-free} \begin{definition} \label{definition-finite-locally-free} Let $A$ be a ring. Recall that a {\it finite locally free} $A$-module $M$ is a module such that for every ${\mathfrak p} \in \Spec(A)$ there exists an $f\in A$, $f \not \in {\mathfrak p}$ such that $M_f$ is a finite free $A_f$-module. We say $M$ is an {\it invertible module} if $M$ is finite locally free of rank $1$, i.e., for every ${\mathfrak p} \in \Spec(A)$ there exists an $f\in A$, $f \not \in \mathfrak p$ such that $M_f \cong A_f$ as an $A_f$-module. \end{definition} \begin{exercise} \label{exercise-tensor-finite-locally-free} Prove that the tensor product of finite locally free modules is finite locally free. Prove that the tensor product of two invertible modules is invertible. \end{exercise} \begin{definition} \label{definition-class-group} Let $A$ be a ring. The {\it class group of $A$}, sometimes called the {\it Picard group of $A$} is the set $\Pic(A)$ of isomorphism classes of invertible $A$-modules endowed with a group operation defined by tensor product (see Exercise \ref{exercise-tensor-finite-locally-free}). \end{definition} \noindent Note that the class group of $A$ is trivial exactly when every invertible module is isomorphic to a free module of rank 1. \begin{exercise} \label{exercise-class-group-trivial} Show that the class groups of the following rings are trivial \begin{enumerate} \item a polynomial ring $A = k[x]$ where $k$ is a field, \item the integers $A = \mathbf{Z}$, \item a polynomial ring $A = k[x, y]$ where $k$ is a field, and \item the quotient $k[x, y]/(xy)$ where $k$ is a field. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-class-group-not-trivial} Show that the class group of the ring $A = k[x, y]/(y^2 - f(x))$ where $k$ is a field of characteristic not $2$ and where $f(x) = (x - t_1) \ldots (x - t_n)$ with $t_1, \ldots, t_n \in k$ distinct and $n \geq 3$ an odd integer is not trivial. (Hint: Show that the ideal $(y, x - t_1)$ defines a nontrivial element of $\Pic(A)$.) \end{exercise} \begin{exercise} \label{exercise-trace-det} Let $A$ be a ring. \begin{enumerate} \item Suppose that $M$ is a finite locally free $A$-module, and suppose that $\varphi : M \to M$ is an endomorphism. Define/construct the {\it trace} and {\it determinant} of $\varphi$ and prove that your construction is functorial in the triple $(A, M, \varphi)$''. \item Show that if $M, N$ are finite locally free $A$-modules, and if $\varphi : M \to N$ and $\psi : N \to M$ then $\text{Trace}(\varphi \circ \psi) = \text{Trace}(\psi \circ \varphi)$ and $\det(\varphi \circ \psi) = \det(\psi \circ \varphi)$. \item In case $M$ is finite locally free show that $\text{Trace}$ defines an $A$-linear map $\text{End}_A(M) \to A$ and $\det$ defines a multiplicative map $\text{End}_A(M) \to A$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-trace-det-rings} Now suppose that $B$ is an $A$-algebra which is finite locally free as an $A$-module, in other words $B$ is a finite locally free $A$-algebra. \begin{enumerate} \item Define $\text{Trace}_{B/A}$ and $\text{Norm}_{B/A}$ using $\text{Trace}$ and $\det$ from Exercise \ref{exercise-trace-det}. \item Let $b\in B$ and let $\pi : \Spec(B) \to \Spec(A)$ be the induced morphism. Show that $\pi(V(b)) = V(\text{Norm}_{B/A}(b))$. (Recall that $V(f) = \{ {\mathfrak p} \mid f \in {\mathfrak p}\}$.) \item (Base change.) Suppose that $i : A \to A'$ is a ring map. Set $B' = B \otimes_A A'$. Indicate why $i(\text{Norm}_{B/A}(b))$ equals $\text{Norm}_{B'/A'}(b \otimes 1)$. \item Compute $\text{Norm}_{B/A}(b)$ when $B = A \times A \times A \times \ldots \times A$ and $b = (a_1, \ldots, a_n)$. \item Compute the norm of $y-y^3$ under the finite flat map ${\mathbf Q}[x] \to {\mathbf Q}[y]$, $x \to y^n$. (Hint: use the base change'' $A = {\mathbf Q}[x] \subset A' = {\mathbf Q}(\zeta_n)(x^{1/n})$.) \end{enumerate} \end{exercise} \section{Glueing} \label{section-glueing} \begin{exercise} \label{exercise-cover} Suppose that $A$ is a ring and $M$ is an $A$-module. Let $f_i$, $i \in I$ be a collection of elements of $A$ such that $$\Spec(A) = \bigcup D(f_i).$$ \begin{enumerate} \item Show that if $M_{f_i}$ is a finite $A_{f_i}$-module, then $M$ is a finite $A$-module. \item Show that if $M_{f_i}$ is a flat $A_{f_i}$-module, then $M$ is a flat $A$-module. (This is kind of silly if you think about it right.) \end{enumerate} \end{exercise} \begin{remark} \label{remark-cover} In algebraic geometric language this means that the property of being finitely generated'' or being flat'' is local for the Zariski topology (in a suitable sense). You can also show this for the property being of finite presentation''. \end{remark} \begin{exercise} \label{exercise-cover-ring-map} Suppose that $A \to B$ is a ring map. Let $f_i \in A$, $i \in I$ and $g_j \in B$, $j \in J$ be collections of elements such that $$\Spec(A) = \bigcup D(f_i) \quad\text{and}\quad \Spec(B) = \bigcup D(g_j).$$ Show that if $A_{f_i} \to B_{f_ig_j}$ is of finite type for all $i, j$ then $A \to B$ is of finite type. \end{exercise} \section{Going up and going down} \label{section-going-up} \begin{definition} \label{definition-GU-GD} Let $\phi : A \to B$ be a homomorphism of rings. We say that the {\it going-up theorem} holds for $\phi$ if the following condition is satisfied: \begin{itemize} \item[(GU)] for any ${\mathfrak p}, {\mathfrak p}' \in \Spec(A)$ such that ${\mathfrak p} \subset {\mathfrak p}'$, and for any $P \in \Spec(B)$ lying over ${\mathfrak p}$, there exists $P'\in \Spec(B)$ lying over ${\mathfrak p}'$ such that $P \subset P'$. \end{itemize} Similarly, we say that the {\it going-down theorem} holds for $\phi$ if the following condition is satisfied: \begin{itemize} \item[(GD)] for any ${\mathfrak p}, {\mathfrak p}' \in \Spec(A)$ such that ${\mathfrak p} \subset {\mathfrak p}'$, and for any $P' \in \Spec(B)$ lying over ${\mathfrak p}'$, there exists $P\in \Spec(B)$ lying over ${\mathfrak p}$ such that $P \subset P'$. \end{itemize} \end{definition} \begin{exercise} \label{exercise-GU-GD} In each of the following cases determine whether (GU), (GD) holds, and explain why. (Use any Prop/Thm/Lemma you can find, but check the hypotheses in each case.) \begin{enumerate} \item $k$ is a field, $A = k$, $B = k[x]$. \item $k$ is a field, $A = k[x]$, $B = k[x, y]$. \item $A = {\mathbf Z}$, $B = {\mathbf Z}[1/11]$. \item $k$ is an algebraically closed field, $A = k[x, y]$, $B = k[x, y, z]/(x^2-y, z^2-x)$. \item $A = {\mathbf Z}$, $B = {\mathbf Z}[i, 1/(2 + i)]$. \item $A = {\mathbf Z}$, $B = {\mathbf Z}[i, 1/(14 + 7i)]$. \item $k$ is an algebraically closed field, $A = k[x]$, $B = k[x, y, 1/(xy-1)]/(y^2-y)$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-image} Let $A$ be a ring. Let $B = A[x]$ be the polynomial algebra in one variable over $A$. Let $f = a_0 + a_1 x + \ldots + a_r x^r \in B = A[x]$. Prove carefully that the image of $D(f)$ in $\Spec(A)$ is equal to $D(a_0) \cup \ldots \cup D(a_r)$. \end{exercise} \begin{exercise} \label{exercise-images} Let $k$ be an algebraically closed field. Compute the image in $\Spec(k[x, y])$ of the following maps: \begin{enumerate} \item $\Spec(k[x, yx^{-1}]) \to \Spec(k[x, y])$, where $k[x, y] \subset k[x, yx^{-1}] \subset k[x, y, x^{-1}]$. (Hint: To avoid confusion, give the element $yx^{-1}$ another name.) \item $\Spec(k[x, y, a, b]/(ax-by-1))\to \Spec(k[x, y])$. \item $\Spec(k[t, 1/(t-1)]) \to \Spec(k[x, y])$, induced by $x \mapsto t^2$, and $y \mapsto t^3$. \item $k = {\mathbf C}$ (complex numbers), $\Spec(k[s, t]/(s^3 + t^3-1)) \to \Spec(k[x, y])$, where $x\mapsto s^2$, $y \mapsto t^2$. \end{enumerate} \end{exercise} \begin{remark} \label{remark-elimination-theory} Finding the image as above usually is done by using elimination theory. \end{remark} \section{Fitting ideals} \label{section-fitting-ideals} \begin{exercise} \label{exercise-fitting} Let $R$ be a ring and let $M$ be a finite $R$-module. Choose a presentation $$\bigoplus\nolimits_{j \in J} R \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0.$$ of $M$. Note that the map $R^{\oplus n} \to M$ is given by a sequence of elements $x_1, \ldots, x_n$ of $M$. The elements $x_i$ are {\it generators} of $M$. The map $\bigoplus_{j \in J} R \to R^{\oplus n}$ is given by a $n \times J$ matrix $A$ with coefficients in $R$. In other words, $A = (a_{ij})_{i = 1, \ldots, n, j \in J}$. The columns $(a_{1j}, \ldots, a_{nj})$, $j \in J$ of $A$ are said to be the {\it relations}. Any vector $(r_i) \in R^{\oplus n}$ such that $\sum r_i x_i = 0$ is a linear combination of the columns of $A$. Of course any finite $R$-module has a lot of different presentations. \begin{enumerate} \item Show that the ideal generated by the $(n - k) \times (n - k)$ minors of $A$ is independent of the choice of the presentation. This ideal is the {\it $k$th Fitting ideal of $M$}. Notation $Fit_k(M)$. \item Show that $Fit_0(M) \subset Fit_1(M) \subset Fit_2(M) \subset \ldots$. (Hint: Use that a determinant can be computed by expanding along a column.) \item Show that the following are equivalent: \begin{enumerate} \item $Fit_{r - 1}(M) = (0)$ and $Fit_r(M) = R$, and \item $M$ is locally free of rank $r$. \end{enumerate} \end{enumerate} \end{exercise} \section{Hilbert functions} \label{section-hilbert} \begin{definition} \label{definition-numerical-polynomial} A {\it numerical polynomial} is a polynomial $f(x) \in {\mathbf Q}[x]$ such that $f(n) \in {\mathbf Z}$ for every integer $n$. \end{definition} \begin{definition} \label{definition-graded-module} A {\it graded module} $M$ over a ring $A$ is an $A$-module $M$ endowed with a direct sum decomposition $\bigoplus\nolimits_{n \in {\mathbf Z}} M_n$ into $A$-submodules. We will say that $M$ is {\it locally finite} if all of the $M_n$ are finite $A$-modules. Suppose that $A$ is a Noetherian ring and that $\varphi$ is a {\it Euler-Poincar\'e function} on finite $A$-modules. This means that for every finitely generated $A$-module $M$ we are given an integer $\varphi(M) \in {\mathbf Z}$ and for every short exact sequence $$0 \longrightarrow M' \longrightarrow M \longrightarrow M'' \longrightarrow 0$$ we have $\varphi(M) = \varphi(M') + \varphi(M'')$. The {\it Hilbert function} of a locally finite graded module $M$ (with respect to $\varphi$) is the function $\chi_\varphi(M, n) = \varphi(M_n)$. We say that $M$ has a {\it Hilbert polynomial} if there is some numerical polynomial $P_\varphi$ such that $\chi_\varphi(M, n) = P_\varphi(n)$ for all sufficiently large integers $n$. \end{definition} \begin{definition} \label{definition-graded-algebra} A {\it graded $A$-algebra} is a graded $A$-module $B = \bigoplus_{n \geq 0} B_n$ together with an $A$-bilinear map $$B \times B \longrightarrow B, \ (b, b') \longmapsto bb'$$ that turns $B$ into an $A$-algebra so that $B_n \cdot B_m \subset B_{n + m}$. Finally, a {\it graded module $M$ over a graded $A$-algebra $B$} is given by a graded $A$-module $M$ together with a (compatible) $B$-module structure such that $B_n \cdot M_d \subset M_{n + d}$. Now you can define {\it homomorphisms of graded modules/rings}, {\it graded submodules}, {\it graded ideals}, {\it exact sequences of graded modules}, etc, etc. \end{definition} \begin{exercise} \label{exercise-Euler-Poincare-field} Let $A = k$ a field. What are all possible Euler-Poincar\'e functions on finite $A$-modules in this case? \end{exercise} \begin{exercise} \label{exercise-Euler-Poincare-Z} Let $A ={\mathbf Z}$. What are all possible Euler-Poincar\'e functions on finite $A$-modules in this case? \end{exercise} \begin{exercise} \label{exercise-Euler-Poincare-node} Let $A = k[x, y]/(xy)$ with $k$ algebraically closed. What are all possible Euler-Poincar\'e functions on finite $A$-modules in this case? \end{exercise} \begin{exercise} \label{exercise-kernel-locally-finite} Suppose that $A$ is Noetherian. Show that the kernel of a map of locally finite graded $A$-modules is locally finite. \end{exercise} \begin{exercise} \label{exercise-no-hilbert} Let $k$ be a field and let $A = k$ and $B = k[x, y]$ with grading determined by $\deg(x) = 2$ and $\deg(y) = 3$. Let $\varphi(M) = \dim_k(M)$. Compute the Hilbert function of $B$ as a graded $k$-module. Is there a Hilbert polynomial in this case? \end{exercise} \begin{exercise} \label{exercise-no-hilbert-or-is-there} Let $k$ be a field and let $A = k$ and $B = k[x, y]/(x^2, xy)$ with grading determined by $\deg(x) = 2$ and $\deg(y) = 3$. Let $\varphi(M) = \dim_k(M)$. Compute the Hilbert function of $B$ as a graded $k$-module. Is there a Hilbert polynomial in this case? \end{exercise} \begin{exercise} \label{exercise-hilbert-to-compute} Let $k$ be a field and let $A = k$. Let $\varphi(M) = \dim_k(M)$. Fix $d\in {\mathbf N}$. Consider the graded $A$-algebra $B = k[x, y, z]/(x^d + y^d + z^d)$, where $x, y, z$ each have degree $1$. Compute the Hilbert function of $B$. Is there a Hilbert polynomial in this case? \end{exercise} \section{Proj of a ring} \label{section-proj-ring} \begin{definition} \label{definition-homogeneous-ideal} Let $R$ be a graded ring. A {\it homogeneous} ideal is simply an ideal $I \subset R$ which is also a graded submodule of $R$. Equivalently, it is an ideal generated by homogeneous elements. Equivalently, if $f \in I$ and $$f = f_0 + f_1 + \ldots + f_n$$ is the decomposition of $f$ into homogeneous pieces in $R$ then $f_i \in I$ for each $i$. \end{definition} \begin{definition} \label{definition-Proj-R} We define the {\it homogeneous spectrum $\text{Proj}(R)$} of the graded ring $R$ to be the set of homogeneous, prime ideals ${\mathfrak p}$ of $R$ such that $R_{+} \not \subset {\mathfrak p}$. Note that $\text{Proj}(R)$ is a subset of $\Spec(R)$ and hence has a natural induced topology. \end{definition} \begin{definition} \label{definition-Dplus-Vplus} Let $R = \oplus_{d \geq 0} R_d$ be a graded ring, let $f\in R_d$ and assume that $d \geq 1$. We define {\it $R_{(f)}$} to be the subring of $R_f$ consisting of elements of the form $r/f^n$ with $r$ homogeneous and $\deg(r) = nd$. Furthermore, we define $$D_{+}(f) = \{ {\mathfrak p} \in \text{Proj}(R) | f \not\in {\mathfrak p} \}.$$ Finally, for a homogeneous ideal $I \subset R$ we define $V_{+}(I) = V(I) \cap \text{Proj}(R)$. \end{definition} \begin{exercise} \label{exercise-topology-proj} On the topology on $\text{Proj}(R)$. With definitions and notation as above prove the following statements. \begin{enumerate} \item Show that $D_{+}(f)$ is open in $\text{Proj}(R)$. \item Show that $D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$. \item Let $g = g_0 + \ldots + g_m$ be an element of $R$ with $g_i \in R_i$. Express $D(g) \cap \text{Proj}(R)$ in terms of $D_{+}(g_i)$, $i \geq 1$ and $D(g_0) \cap \text{Proj}(R)$. No proof necessary. \item Let $g\in R_0$ be a homogeneous element of degree $0$. Express $D(g) \cap \text{Proj}(R)$ in terms of $D_{+}(f_\alpha)$ for a suitable family $f_\alpha \in R$ of homogeneous elements of positive degree. \item Show that the collection $\{D_{+}(f)\}$ of opens forms a basis for the topology of $\text{Proj}(R)$. \item \label{item-bijection} Show that there is a canonical bijection $D_{+}(f) \to \Spec(R_{(f)})$. (Hint: Imitate the proof for $\Spec$ but at some point thrown in the radical of an ideal.) \item Show that the map from (\ref{item-bijection}) is a homeomorphism. \item Give an example of an $R$ such that $\text{Proj}(R)$ is not quasi-compact. No proof necessary. \item Show that any closed subset $T \subset \text{Proj}(R)$ is of the form $V_{+}(I)$ for some homogeneous ideal $I \subset R$. \end{enumerate} \end{exercise} \begin{remark} \label{remark-continuous-proj-spec} There is a continuous map $\text{Proj}(R) \longrightarrow \Spec(R_0)$. \end{remark} \begin{exercise} \label{exercise-iso-polynomial-ring-one-variable} If $R = A[X]$ with $\deg(X) = 1$, show that the natural map $\text{Proj}(R) \to \Spec(A)$ is a bijection and in fact a homeomorphism. \end{exercise} \begin{exercise} \label{exercise-blowing-up-I} Blowing up: part I. In this exercise $R = Bl_I(A) = A \oplus I \oplus I^2 \oplus \ldots$. Consider the natural map $b : \text{Proj}(R) \to \Spec(A)$. Set $U = \Spec(A) - V(I)$. Show that $$b : b^{-1}(U) \longrightarrow U$$ is a homeomorphism. Thus we may think of $U$ as an open subset of $\text{Proj}(R)$. Let $Z \subset \Spec(A)$ be an irreducible closed subscheme with generic point $\xi \in Z$. Assume that $\xi \not\in V(I)$, in other words $Z \not\subset V(I)$, in other words $\xi \in U$, in other words $Z\cap U \not = \emptyset$. We define the {\it strict transform} $Z'$ of $Z$ to be the closure of the unique point $\xi'$ lying above $\xi$. Another way to say this is that $Z'$ is the closure in $\text{Proj}(R)$ of the locally closed subset $Z\cap U \subset U \subset \text{Proj}(R)$. \end{exercise} \begin{exercise} \label{exercise-blowing-up-II} Blowing up: Part II. Let $A = k[x, y]$ where $k$ is a field, and let $I = (x, y)$. Let $R$ be the blowup algebra for $A$ and $I$. \begin{enumerate} \item Show that the strict transforms of $Z_1 = V(\{x\})$ and $Z_2 = V(\{y\})$ are disjoint. \item Show that the strict transforms of $Z_1 = V(\{x\})$ and $Z_2 = V(\{x-y^2\})$ are not disjoint. \item Find an ideal $J \subset A$ such that $V(J) = V(I)$ and such that the strict transforms of $Z_1 = V(\{x\})$ and $Z_2 = V(\{x-y^2\})$ in the blowup along $J$ are disjoint. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-proj-when-empty} Let $R$ be a graded ring. \begin{enumerate} \item Show that $\text{Proj}(R)$ is empty if $R_n = (0)$ for all $n >> 0$. \item Show that $\text{Proj}(R)$ is an irreducible topological space if $R$ is a domain and $R_{+}$ is not zero. (Recall that the empty topological space is not irreducible.) \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-blowing-up-III} Blowing up: Part III. Consider $A$, $I$ and $U$, $Z$ as in the definition of strict transform. Let $Z = V({\mathfrak p})$ for some prime ideal ${\mathfrak p}$. Let $\bar A = A/{\mathfrak p}$ and let $\bar I$ be the image of $I$ in $\bar A$. \begin{enumerate} \item Show that there exists a surjective ring map $R: = Bl_I(A) \to \bar R: = Bl_{\bar I}(\bar A)$. \item Show that the ring map above induces a bijective map from $\text{Proj}(\bar R)$ onto the strict transform $Z'$ of $Z$. (This is not so easy. Hint: Use 5(b) above.) \item Conclude that the strict transform $Z' = V_{+}(P)$ where $P \subset R$ is the homogeneous ideal defined by $P_d = I^d \cap {\mathfrak p}$. \item Suppose that $Z_1 = V({\mathfrak p})$ and $Z_2 = V({\mathfrak q})$ are irreducible closed subsets defined by prime ideals such that $Z_1 \not \subset Z_2$, and $Z_2 \not \subset Z_1$. Show that blowing up the ideal $I = {\mathfrak p} + {\mathfrak q}$ separates the strict transforms of $Z_1$ and $Z_2$, i.e., $Z_1' \cap Z_2' = \emptyset$. (Hint: Consider the homogeneous ideal $P$ and $Q$ from part (c) and consider $V(P + Q)$.) \end{enumerate} \end{exercise} \section{Cohen-Macaulay rings of dimension 1} \label{section-CM-dim-1} \begin{definition} \label{definition-CM} A Noetherian local ring $A$ is said to be {\it Cohen-Macaulay} of dimension $d$ if it has dimension $d$ and there exists a system of parameters $x_1, \ldots, x_d$ for $A$ such that $x_i$ is a nonzerodivisor in $A/(x_1, \ldots, x_{i-1})$ for $i = 1, \ldots, d$. \end{definition} \begin{exercise} \label{exercise-CM-dim-1-I} Cohen-Macaulay rings of dimension 1. Part I: Theory. \begin{enumerate} \item Let $(A, {\mathfrak m})$ be a local Noetherian with $\dim A = 1$. Show that if $x\in {\mathfrak m}$ is not a zerodivisor then \begin{enumerate} \item $\dim A/xA = 0$, in other words $A/xA$ is Artinian, in other words $\{x\}$ is a system of parameters for $A$. \item $A$ is has no embedded prime. \end{enumerate} \item Conversely, let $(A, {\mathfrak m})$ be a local Noetherian ring of dimension $1$. Show that if $A$ has no embedded prime then there exists a nonzerodivisor in ${\mathfrak m}$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-CM-dim-1-II} Cohen-Macaulay rings of dimension 1. Part II: Examples. \begin{enumerate} \item Let $A$ be the local ring at $(x, y)$ of $k[x, y]/(x^2, xy)$. \begin{enumerate} \item Show that $A$ has dimension 1. \item Prove that every element of ${\mathfrak m}\subset A$ is a zerodivisor. \item Find $z\in {\mathfrak m}$ such that $\dim A/zA = 0$ (no proof required). \end{enumerate} \item Let $A$ be the local ring at $(x, y)$ of $k[x, y]/(x^2)$. Find a nonzerodivisor in ${\mathfrak m}$ (no proof required). \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-embedding-dim-1} Local rings of embedding dimension $1$. Suppose that $(A, {\mathfrak m}, k)$ is a Noetherian local ring of embedding dimension $1$, i.e., $$\dim_k {\mathfrak m}/{\mathfrak m}^2 = 1.$$ Show that the function $f(n) = \dim_k {\mathfrak m}^n/{\mathfrak m}^{n + 1}$ is either constant with value $1$, or its values are $$1, 1, \ldots, 1, 0, 0, 0, 0, 0, \ldots$$ \end{exercise} \begin{exercise} \label{exercise-regular-local-dim-1} Regular local rings of dimension $1$. Suppose that $(A, {\mathfrak m}, k)$ is a regular Noetherian local ring of dimension $1$. Recall that this means that $A$ has dimension $1$ and embedding dimension $1$, i.e., $$\dim_k {\mathfrak m}/{\mathfrak m}^2 = 1.$$ Let $x\in{\mathfrak m}$ be any element whose class in ${\mathfrak m}/{\mathfrak m}^2$ is not zero. \begin{enumerate} \item Show that for every element $y$ of ${\mathfrak m}$ there exists an integer $n$ such that $y$ can be written as $y = ux^n$ with $u\in A^\ast$ a unit. \item Show that $x$ is a nonzerodivisor in $A$. \item Conclude that $A$ is a domain. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-nonzerodivisor-graded} Let $(A, {\mathfrak m}, k)$ be a Noetherian local ring with associated graded $Gr_{\mathfrak m}(A)$. \begin{enumerate} \item Suppose that $x\in {\mathfrak m}^d$ maps to a nonzerodivisor $\bar x \in {\mathfrak m}^d/{\mathfrak m}^{d + 1}$ in degree $d$ of $Gr_{\mathfrak m}(A)$. Show that $x$ is a nonzerodivisor. \item Suppose the depth of $A$ is at least $1$. Namely, suppose that there exists a nonzerodivisor $y \in {\mathfrak m}$. In this case we can do better: assume just that $x\in {\mathfrak m}^d$ maps to the element $\bar x \in {\mathfrak m}^d/{\mathfrak m}^{d + 1}$ in degree $d$ of $Gr_{\mathfrak m}(A)$ which is a nonzerodivisor on sufficiently high degrees: $\exists N$ such that for all $n \geq N$ the map of multiplication by $\bar x$ $${\mathfrak m}^n/{\mathfrak m}^{n + 1} \longrightarrow {\mathfrak m}^{n + d}/{\mathfrak m}^{n + d + 1}$$ is injective. Then show that $x$ is a nonzerodivisor. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-embedding-2-dim-1} Suppose that $(A, {\mathfrak m}, k)$ is a Noetherian local ring of dimension $1$. Assume also that the embedding dimension of $A$ is $2$, i.e., assume that $$\dim_k {\mathfrak m}/{\mathfrak m}^2 = 2.$$ Notation: $f(n) = \dim_k {\mathfrak m}^n/{\mathfrak m}^{n + 1}$. Pick generators $x, y \in {\mathfrak m}$ and write $Gr_{\mathfrak m}(A) = k[\bar x, \bar y]/I$ for some homogeneous ideal $I$. \begin{enumerate} \item Show that there exists a homogeneous element $F\in k[\bar x, \bar y]$ such that $I \subset (F)$ with equality in all sufficiently high degrees. \item Show that $f(n) \leq n + 1$. \item Show that if $f(n) < n + 1$ then $n \geq \deg(F)$. \item Show that if $f(n) < n + 1$, then $f(n + 1) \leq f(n)$. \item Show that $f(n) = \deg(F)$ for all $n >> 0$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-CM-dim-1-embedding-dim-2} Cohen-Macaulay rings of dimension 1 and embedding dimension 2. Suppose that $(A, {\mathfrak m}, k)$ is a Noetherian local ring which is Cohen-Macaulay of dimension $1$. Assume also that the embedding dimension of $A$ is $2$, i.e., assume that $$\dim_k {\mathfrak m}/{\mathfrak m}^2 = 2.$$ Notations: $f$, $F$, $x, y\in {\mathfrak m}$, $I$ as in Ex.\ 6 above. Please use any results from the problems above. \begin{enumerate} \item Suppose that $z\in {\mathfrak m}$ is an element whose class in ${\mathfrak m}/{\mathfrak m}^2$ is a linear form $\alpha \bar x + \beta \bar y \in k[\bar x, \bar y]$ which is coprime with $f$. \begin{enumerate} \item Show that $z$ is a nonzerodivisor on $A$. \item Let $d = \deg(F)$. Show that ${\mathfrak m}^n = z^{n + 1-d}{\mathfrak m}^{d-1}$ for all sufficiently large $n$. (Hint: First show $z^{n + 1-d}{\mathfrak m}^{d-1} \to {\mathfrak m}^n/{\mathfrak m}^{n + 1}$ is surjective by what you know about $Gr_{\mathfrak m}(A)$. Then use NAK.) \end{enumerate} \item What condition on $k$ guarantees the existence of such a $z$? (No proof required; it's too easy.) \noindent Now we are going to assume there exists a $z$ as above. This turns out to be a harmless assumption (in the sense that you can reduce to the situation where it holds in order to obtain the results in parts (d) and (e) below). \item Now show that ${\mathfrak m}^\ell = z^{\ell - d + 1} {\mathfrak m}^{d-1}$ for all $\ell \geq d$. \item Conclude that $I = (F)$. \item Conclude that the function $f$ has values $$2, 3, 4, \ldots, d-1, d, d, d, d, d, d, d, \ldots$$ \end{enumerate} \end{exercise} \begin{remark} \label{remark-CM-dim-1-embedding-dim-2} This suggests that a local Noetherian Cohen-Macaulay ring of dimension 1 and embedding dimension 2 is of the form $B/FB$, where $B$ is a 2-dimensional regular local ring. This is more or less true (under suitable niceness'' properties of the ring). \end{remark} \section{Infinitely many primes} \label{section-many-primes} \noindent A section with a collection of strange questions on rings where infinitely many primes are not invertible. \begin{exercise} \label{exercise-not-in-Q} Give an example of a finite type ${\mathbf Z}$-algebra $R$ with the following two properties: \begin{enumerate} \item There is no ring map $R \to {\mathbf Q}$. \item For every prime $p$ there exists a maximal ideal ${\mathfrak m} \subset R$ such that $R/{\mathfrak m} \cong {\mathbf F}_p$. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-strange-fp-1} For $f \in {\mathbf Z}[x, u]$ we define $f_p(x) = f(x, x^p) \bmod p \in {\mathbf F}_p[x]$. Give an example of an $f \in {\mathbf Z}[x, u]$ such that the following two properties hold: \begin{enumerate} \item There exist infinitely many $p$ such that $f_p$ does not have a zero in ${\mathbf F}_p$. \item For all $p >> 0$ the polynomial $f_p$ either has a linear or a quadratic factor. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-strange-fp-2} For $f \in {\mathbf Z}[x, y, u, v]$ we define $f_p(x, y) = f(x, y, x^p, y^p) \bmod p \in {\mathbf F}_p[x, y]$. Give an interesting'' example of an $f$ such that $f_p$ is reducible for all $p >> 0$. For example, $f = xv-yu$ with $f_p = xy^p-x^py = xy(x^{p-1}-y^{p-1})$ is uninteresting''; any $f$ depending only on $x, u$ is uninteresting'', etc. \end{exercise} \begin{remark} \label{remark-strange-fp} Let $h \in {\mathbf Z}[y]$ be a monic polynomial of degree $d$. Then: \begin{enumerate} \item The map $A = {\mathbf Z}[x] \to B ={\mathbf Z}[y]$, $x \mapsto h$ is finite locally free of rank $d$. \item For all primes $p$ the map $A_p = {\mathbf F}_p[x]\to B_p = {\mathbf F}_p[y]$, $y \mapsto h(y) \bmod p$ is finite locally free of rank $d$. \end{enumerate} \end{remark} \begin{exercise} \label{exercise-strange-fp-3} Let $h, A, B, A_p, B_p$ be as in the remark. For $f \in {\mathbf Z}[x, u]$ we define $f_p(x) = f(x, x^p) \bmod p \in {\mathbf F}_p[x]$. For $g \in {\mathbf Z}[y, v]$ we define $g_p(y) = g(y, y^p) \bmod p \in {\mathbf F}_p[y]$. \begin{enumerate} \item Give an example of a $h$ and $g$ such that there does not exist a $f$ with the property $$f_p = Norm_{B_p/A_p}(g_p).$$ \item Show that for any choice of $h$ and $g$ as above there exists a nonzero $f$ such that for all $p$ we have $$Norm_{B_p/A_p}(g_p)\quad\text{divides}\quad f_p .$$ If you want you can restrict to the case $h = y^n$, even with $n = 2$, but it is true in general. \item Discuss the relevance of this to Exercises 6 and 7 of the previous set. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-strange-fp-unsolved} Unsolved problems. They may be really hard or they may be easy. I don't know. \begin{enumerate} \item Is there any $f \in {\mathbf Z}[x, u]$ such that $f_p$ is irreducible for an infinite number of $p$? (Hint: Yes, this happens for $f(x, u) = u - x - 1$ and also for $f(x, u) = u^2 - x^2 + 1$.) \item Let $f \in {\mathbf Z}[x, u]$ nonzero, and suppose $\deg_x(f_p) = dp + d'$ for all large $p$. (In other words $\deg_u(f) = d$ and the coefficient $c$ of $u^d$ in $f$ has $\deg_x(c) = d'$.) Suppose we can write $d = d_1 + d_2$ and $d' = d'_1 + d'_2$ with $d_1, d_2 > 0$ and $d'_1, d'_2 \geq 0$ such that for all sufficiently large $p$ there exists a factorization $$f_p = f_{1, p} f_{2, p}$$ with $\deg_x(f_{1, p}) = d_1p + d'_1$. Is it true that $f$ comes about via a norm construction as in Exercise 4? (More precisely, are there a $h$ and $g$ such that $Norm_{B_p/A_p}(g_p)$ divides $f_p$ for all $p >> 0$.) \item Analogous question to the one in (b) but now with $f \in {\mathbf Z}[x_1, x_2, u_1, u_2]$ irreducible and just assuming that $f_p(x_1, x_2) = f(x_1, x_2, x_1^p, x_2^p) \bmod p$ factors for all $p >> 0$. \end{enumerate} \end{exercise} \section{Filtered derived category} \label{section-filtered-derived} \noindent In order to do the exercises in this section, please read the material in Homology, Section \ref{homology-section-filtrations}. We will say $A$ is a filtered object of $\mathcal{A}$, to mean that $A$ comes endowed with a filtration $F$ which we omit from the notation. \begin{exercise} \label{exercise-split-injective} Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume that the filtration on $I$ is finite and that each $\text{gr}^p(I)$ is an injective object of $\mathcal{A}$. Show that there exists an isomorphism $I \cong \bigoplus \text{gr}^p(I)$ with filtration $F^p(I)$ corresponding to $\bigoplus_{p' \geq p} \text{gr}^p(I)$. \end{exercise} \begin{exercise} \label{exercise-filtered-injective} Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume that the filtration on $I$ is finite. Show the following are equivalent: \begin{enumerate} \item For any solid diagram $$\xymatrix{ A \ar[r]_\alpha \ar[d] & B \ar@{-->}[ld] \\ I & }$$ of filtered objects with (\romannumeral1) the filtrations on $A$ and $B$ are finite, and (\romannumeral2) $\text{gr}(\alpha)$ injective the dotted arrow exists making the diagram commute. \item Each $\text{gr}^p I$ is injective. \end{enumerate} \end{exercise} \noindent Note that given a morphism $\alpha : A \to B$ of filtered objects with finite filtrations to say that $\text{gr}(\alpha)$ injective is the same thing as saying that $\alpha$ is a {\it strict monomorphism} in the category $\text{Fil}(\mathcal{A})$. Namely, being a monomorphism means $\Ker(\alpha) = 0$ and strict means that this also implies $\Ker(\text{gr}(\alpha)) = 0$. See Homology, Lemma \ref{homology-lemma-characterize-strict}. (We only use the term injective'' for a morphism in an abelian category, although it makes sense in any additive category having kernels.) The exercises above justifies the following definition. \begin{definition} \label{definition-injective-filtered} Let $\mathcal{A}$ be an abelian category. Let $I$ be a filtered object of $\mathcal{A}$. Assume the filtration on $I$ is finite. We say $I$ is {\it filtered injective} if each $\text{gr}^p(I)$ is an injective object of $\mathcal{A}$. \end{definition} \noindent We make the following definition to avoid having to keep saying with a finite filtration'' everywhere. \begin{definition} \label{definition-finite-filtration-category} Let $\mathcal{A}$ be an abelian category. We denote {\it $\text{Fil}^f(\mathcal{A})$} the full subcategory of $\text{Fil}(\mathcal{A})$ whose objects consist of those $A \in \Ob(\text{Fil}(\mathcal{A}))$ whose filtration is finite. \end{definition} \begin{exercise} \label{exercise-inject-into-injective} Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $A$ be an object of $\text{Fil}^f(\mathcal{A})$. Show that there exists a strict monomorphism $\alpha : A \to I$ of $A$ into a filtered injective object $I$ of $\text{Fil}^f(\mathcal{A})$. \end{exercise} \begin{definition} \label{definition-filtered-quasi-isomorphism} Let $\mathcal{A}$ be an abelian category. Let $\alpha : K^\bullet \to L^\bullet$ be a morphism of complexes of $\text{Fil}(\mathcal{A})$. We say that $\alpha$ is a {\it filtered quasi-isomorphism} if for each $p \in \mathbf{Z}$ the morphism $\text{gr}^p(K^\bullet) \to \text{gr}^p(L^\bullet)$ is a quasi-isomorphism. \end{definition} \begin{definition} \label{definition-filtered-acyclic} Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex of $\text{Fil}^f(\mathcal{A})$. We say that $K^\bullet$ is {\it filtered acyclic} if for each $p \in \mathbf{Z}$ the complex $\text{gr}^p(K^\bullet)$ is acyclic. \end{definition} \begin{exercise} \label{exercise-filtered-quasi-isomorphism} Let $\mathcal{A}$ be an abelian category. Let $\alpha : K^\bullet \to L^\bullet$ be a morphism of bounded below complexes of $\text{Fil}^f(\mathcal{A})$. (Note the superscript $f$.) Show that the following are equivalent: \begin{enumerate} \item $\alpha$ is a filtered quasi-isomorphism, \item for each $p \in \mathbf{Z}$ the map $\alpha : F^pK^\bullet \to F^pL^\bullet$ is a quasi-isomorphism, \item for each $p \in \mathbf{Z}$ the map $\alpha : K^\bullet/F^pK^\bullet \to L^\bullet/F^pL^\bullet$ is a quasi-isomorphism, and \item the cone of $\alpha$ (see Derived Categories, Definition \ref{derived-definition-cone}) is a filtered acyclic complex. \end{enumerate} Moreover, show that if $\alpha$ is a filtered quasi-isomorphism then $\alpha$ is also a usual quasi-isomorphism. \end{exercise} \begin{exercise} \label{exercise-injective-resolution} Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $A$ be an object of $\text{Fil}^f(\mathcal{A})$. Show there exists a complex $I^\bullet$ of $\text{Fil}^f(\mathcal{A})$, and a morphism $A[0] \to I^\bullet$ such that \begin{enumerate} \item each $I^p$ is filtered injective, \item $I^p = 0$ for $p < 0$, and \item $A[0] \to I^\bullet$ is a filtered quasi-isomorphism. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-injective-resolution-complex} Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough injectives. Let $K^\bullet$ be a bounded below complex of objects of $\text{Fil}^f(\mathcal{A})$. Show there exists a filtered quasi-isomorphism $\alpha : K^\bullet \to I^\bullet$ with $I^\bullet$ a complex of $\text{Fil}^f(\mathcal{A})$ having filtered injective terms $I^n$, and bounded below. In fact, we may choose $\alpha$ such that each $\alpha^n$ is a strict monomorphism. \end{exercise} \begin{exercise} \label{exercise-morphisms-lift} Let $\mathcal{A}$ be an abelian category. Consider a solid diagram $$\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^\beta \\ I^\bullet }$$ of complexes of $\text{Fil}^f(\mathcal{A})$. Assume $K^\bullet$, $L^\bullet$ and $I^\bullet$ are bounded below and assume each $I^n$ is a filtered injective object. Also assume that $\alpha$ is a filtered quasi-isomorphism. \begin{enumerate} \item There exists a map of complexes $\beta$ making the diagram commute up to homotopy. \item If $\alpha$ is a strict monomorphism in every degree then we can find a $\beta$ which makes the diagram commute. \end{enumerate} \end{exercise} \begin{exercise} \label{exercise-acyclic-is-zero} Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$, $K^\bullet$ be complexes of $\text{Fil}^f(\mathcal{A})$. Assume \begin{enumerate} \item $K^\bullet$ bounded below and filtered acyclic, and \item $I^\bullet$ bounded below and consisting of filtered injective objects. \end{enumerate} Then any morphism $K^\bullet \to I^\bullet$ is homotopic to zero. \end{exercise} \begin{exercise} \label{exercise-morphisms-equal-up-to-homotopy} Let $\mathcal{A}$ be an abelian category. Consider a solid diagram $$\xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_\gamma & L^\bullet \ar@{-->}[dl]^{\beta_i} \\ I^\bullet }$$ of complexes of $\text{Fil}^f(\mathcal{A})$. Assume $K^\bullet$, $L^\bullet$ and $I^\bullet$ bounded below and each $I^n$ a filtered injective object. Also assume $\alpha$ a filtered quasi-isomorphism. Any two morphisms $\beta_1, \beta_2$ making the diagram commute up to homotopy are homotopic. \end{exercise} \section{Regular functions} \label{section-regular-functions} \begin{exercise} \label{exercise-extra-function} Consider the affine curve $X$ given by the equation $t^2 = s^5 + 8$ in $\mathbf{C}^2$ with coordinates $s, t$. Let $x \in X$ be the point with coordinates $(1, 3)$. Let $U = X \setminus \{x\}$. Prove that there is a regular function on $U$ which is not the restriction of a regular function on $\mathbf{C}^2$, i.e., is not the restriction of a polynomial in $s$ and $t$ to $U$. \end{exercise} \begin{exercise} \label{exercise-no-extra-function} Let $n \geq 2$. Let $E \subset \mathbf{C}^n$ be a finite subset. Show that any regular function on $\mathbf{C}^n \setminus E$ is a polynomial. \end{exercise} \begin{exercise} \label{exercise-cone} Let $X \subset \mathbf{C}^n$ be an affine variety. Let us say $X$ is a {\it cone} if $x = (a_1, \ldots, a_n) \in X$ and $\lambda \in \mathbf{C}$ implies $(\lambda a_1, \ldots, \lambda a_n) \in X$. Of course, if $\mathfrak p \subset \mathbf{C}[x_1, \ldots, x_n]$ is a prime ideal generated by homogeneous polynomials in $x_1, \ldots, x_n$, then the affine variety $X = V(\mathfrak p) \subset \mathbf{C}^n$ is a cone. Show that conversely the prime ideal \$\mathfrak p \subset