# stacks/stacks-project

Be a bit more precise with ranks of modules

aisejohan committed Sep 7, 2019
1 parent 120c1fe commit 0b2b5ddd0e9963f766a9f75e6901435f0634492d
Showing with 110 additions and 125 deletions.
1. +108 −81 algebra.tex
2. +1 −43 examples.tex
3. +1 −1 tags/tags
 @@ -2654,6 +2654,62 @@ \section{Miscellany} in the statement of the lemma. \end{proof} \begin{lemma} \label{lemma-map-cannot-be-injective} \begin{slogan} A map of finite free modules cannot be injective if the source has rank bigger than the target. \end{slogan} Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a nonzero kernel. \end{lemma} \begin{proof} Choose a surjection $R^{\oplus n - 1} \to M$. We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ (Lemma \ref{lemma-lift-map}). It suffices to prove $f'$ has a nonzero kernel. The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say $a = a_{ij}$ is not, then we can replace $R$ by the localization $R_a$ and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Proposition \ref{proposition-localization-exact}. In this case we can do a base change on both $R^{\oplus n}$ and $R^{\oplus n - 1}$ and reduce to the case where $$A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right)$$ Hence in this case we win by induction on $n$. If not then each $a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that $I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer such that $I^m \not = 0$. Then we see that $(I^m)^{\oplus n}$ is contained in the kernel of the map and we win. \end{proof} \begin{lemma} \label{lemma-rank} \begin{slogan} The rank of a finite free module is well defined. \end{slogan} Let $R$ be a nonzero ring. Let $n, m \geq 0$ be integers. If $R^{\oplus n}$ is isomorphic to $R^{\oplus m}$ as $R$-modules, then $n = m$. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-map-cannot-be-injective}. \end{proof} @@ -12501,32 +12557,40 @@ \section{K-groups} \end{example} \begin{example} \label{example-K0-polynomial-ring} Let $k$ be a field. Then $K_0(k[x]) = K'_0(k[x]) = \mathbf{Z}$. \medskip\noindent Since $R = k[x]$ is a principal ideal domain, any finite projective $R$-module is free. In a short exact sequence of modules \label{example-K0-PID} Let $R$ be a PID. We claim $K_0(R) = K'_0(R) = \mathbf{Z}$. Namely, any finite projective $R$-module is finite free. A finite free module has a well defined rank by Lemma \ref{lemma-rank}. Given a short exact sequence of finite free modules $$0 \to M' \to M \to M'' \to 0$$ we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$, which gives $K_0(k[x]) = \mathbf{Z}$. we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$ because we have $M \cong M' \oplus M'$ in this case (for example we have a splitting by Lemma \ref{lemma-lift-map}). We conclude $K_0(R) = \mathbf{Z}$. \medskip\noindent As for $K'_0$, the structure theorem for modules of a PID says that The structure theorem for modules of a PID says that any finitely generated $R$-module is of the form $M = R^r \times R/(d_1) \times \ldots \times R/(d_k)$. $M = R^{\oplus r} \oplus R/(d_1) \oplus \ldots \oplus R/(d_k)$. Consider the short exact sequence $$0 \to (d_i) \to R \to R/(d_i) \to 0$$ Since the ideal $(d_i)$ is isomorphic to $R$ as a module (it is free with generator $d_i$), in $K'_0(R)$ we have $[(d_i)] = [R]$. Then $[R/(d_i)] = [(d_i)]-[R] = 0$. From this it follows that any torsion part disappears'' in $K'_0$. Again the rank of the free part determines that $K'_0(k[x]) = \mathbf{Z}$, and the canonical homomorphism from $K_0$ to $K'_0$ is an isomorphism. follows that a torsion module has zero class in $K'_0(R)$. Using the rank of the free part gives an identification $K'_0(R) = \mathbf{Z}$ and the canonical homomorphism from $K_0(R) \to K'_0(R)$ is an isomorphism. \end{example} \begin{example} \label{example-K0-polynomial-ring} Let $k$ be a field. Then $K_0(k[x]) = K'_0(k[x]) = \mathbf{Z}$. This follows from Example \ref{example-K0-PID} as $R = k[x]$ is a PID. \end{example} \begin{example} @@ -12560,56 +12624,30 @@ \section{K-groups} \begin{lemma} \label{lemma-K0-local} Let $R$ be a local ring. Every finite projective $R$-module Let $(R, \mathfrak m)$ be a local ring. Every finite projective $R$-module is finite free. The map $\text{rank}_R : K_0(R) \to \mathbf{Z}$ defined by $[M] \to \text{rank}_R(M)$ is well defined and an isomorphism. \end{lemma} \begin{proof} Let $P$ be a finite projective $R$-module. The $n$ generators of $P$ give a surjection $R^n \to P$, and since $P$ is projective it follows that $R^n \cong P \oplus Q$ for some projective module $Q$. \medskip\noindent If $\mathfrak m \subset R$ is the maximal ideal, then $P/\mathfrak m$ and $Q/\mathfrak m$ are $R/\mathfrak m$-vector spaces, with $P/\mathfrak m \oplus Q/\mathfrak m \cong (R/\mathfrak m)^n$. Say that $\dim P = p$, $\dim Q = q$, so $p + q = n$. \medskip\noindent Choose elements $a_1, \ldots, a_p$ in $P$ and $b_1, \ldots, b_q$ in $Q$ lying above bases for $P/\mathfrak m$ and $Q/\mathfrak m$. The homomorphism $R^n \to P \oplus Q \cong R^n$ given by $(r_1, \ldots, r_n) \mapsto r_1a_1 + \ldots + r_pa_p + r_{p + 1} b_1 + \ldots + r_nb_q$ is a matrix $A$ which is invertible over $R/\mathfrak m$. Let $B$ be a matrix over $R$ lying over the inverse of $A$ in $R/\mathfrak m$. $AB = I + M$, where $M$ is a matrix whose entries all lie in $\mathfrak m$. Thus $\det AB = 1 + x$, for $x \in \mathfrak m$, so $AB$ is invertible, so $A$ is invertible. \medskip\noindent The homomorphism $R^p \to P$ given by $(r_1, \ldots, r_p) \mapsto r_1a_1 + \ldots + r_pa_p$ inherits injectivity and surjectivity from A. Hence, $P \cong R^p$. \medskip\noindent Next we show that the rank of a finite projective module over $R$ is well defined: if $P \cong R^\alpha \cong R^\beta$, then $\alpha = \beta$. This is immediate in the vector space case, and so it is true in the general module case as well, by dividing out the maximal ideal on both sides. If $0 \to R^\alpha \to R^\beta \to R^\gamma \to 0$ is exact, the sequence splits, so $R^\beta \cong R^\alpha \oplus R^\gamma$, so $\beta = \alpha + \gamma$. \medskip\noindent So far we have seen that the map $\text{rank}_R : K_0(R) \to \mathbf{Z}$ is a well-defined homomorphism. It is surjective because $\text{rank}_R[R] = 1$. It is injective because the element of $K_0(R)$ with rank $\pm\alpha$ is uniquely $\pm [R^\alpha]$. Let $P$ be a finite projective $R$-module. Choose elements $x_1, \ldots, x_n \in P$ which map to a basis of $P/\mathfrak m P$. By Nakayama's Lemma \ref{lemma-NAK} these elements generate $P$. The corresponding surjection $u : R^{\oplus n} \to P$ has a splitting as $P$ is projective. Hence $R^{\oplus n} = P \oplus Q$ with $Q = \Ker(u)$. It follows that $Q/\mathfrak m Q = 0$, hence $Q$ is zero by Nakayama's lemma. In this way we see that every finite projective $R$-module is finite free. A finite free module has a well defined rank by Lemma \ref{lemma-rank}. Given a short exact sequence of finite free $R$-modules $$0 \to M' \to M \to M'' \to 0$$ we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$ because we have $M \cong M' \oplus M'$ in this case (for example we have a splitting by Lemma \ref{lemma-lift-map}). We conclude $K_0(R) = \mathbf{Z}$. \end{proof} \begin{lemma} @@ -12629,27 +12667,13 @@ \section{K-groups} \end{lemma} \begin{proof} By induction on the rank of $M$. Suppose $\left[M\right] \in K_0(R)$. Then $M$ is a finite projective $R$-module over a local ring, so M is free; $M \cong R^n$ for some $n$. The claim is that $\text{rank} (M) \text{length}_R (R) = \text{length}_R(M)$, or equivalently that $n\text{length}_R(R) = \text{length}_R (R^n)$ for all $n \geq 1$. When $n = 1$, this is clearly true. Suppose that $(n-1) \text{length}_R(R) =\text{ length}_R(R^{n-1})$. Then since there is a split short exact sequence $$0 \to R \to R^n \to R^{n-1} \to 0$$ by Lemma \ref{lemma-length-additive} we have \begin{eqnarray*} \text{length}_R(R^n) & = & \text{length}_R(R) + \text{length}_R(R^{n-1}) \\ & = & \text{length}_R(R) + (n-1) \text{length}_R(R) \\ & = & n\text{length}_R(R) \end{eqnarray*} as desired. Let $P$ be a finite projective $R$-module. We have to show that $\text{length}_R(P) = \text{rank}_R(P) \text{length}_R(R)$. By Lemma \ref{lemma-K0-local} the module $P$ is finite free. So $P \cong R^{\oplus n}$ for some $n \geq 0$. Then $\text{rank}_R(P) = n$ and $\text{length}_R(R^{\oplus n}) = n \text{length}_R(R)$ by additivity of lenghts (Lemma \ref{lemma-length-additive}). Thus the result holds. \end{proof} \end{definition} \noindent Note that a finite locally free $R$-module is automatically finitely presented by Lemma \ref{lemma-cover}. Note that a finite locally free $R$-module is automatically finitely presented by Lemma \ref{lemma-cover}. Moreover, if $M$ is a finite locally free module of rank $r$ over a ring $R$ and if $R$ is nonzero, then $r$ is uniquely determined by Lemma \ref{lemma-rank} (because at least one of the localizations $R_{f_i}$ is a nonzero ring). \begin{lemma} \label{lemma-finite-projective}
 @@ -2050,55 +2050,13 @@ \section{A projective module which is not locally free} $R$-module. \end{proof} \begin{lemma} \label{lemma-map-cannot-be-injective} \begin{slogan} A map of finite free modules cannot be injective if the source has rank bigger than the target. \end{slogan} Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a nonzero kernel. \end{lemma} \begin{proof} Choose a surjection $R^{\oplus n - 1} \to M$. We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$. It suffices to prove $f'$ has a nonzero kernel. The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say $a = a_{ij}$ is not, then we can replace $A$ by the localization $A_a$ and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Algebra, Proposition \ref{algebra-proposition-localization-exact}. In this case we can do a base change on both $R^{\oplus n}$ and $R^{\oplus n - 1}$ and reduce to the case where $$A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right)$$ Hence in this case we win by induction on $n$. If not then each $a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that $I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer such that $I^m \not = 0$. Then we see that $(I^m)^{\oplus n}$ is contained in the kernel of the map and we win. \end{proof} \noindent Suppose that $P \subset Q$ is an inclusion of $R$-modules with $Q$ a finite $R$-module and $P$ locally free, see Algebra, Definition \ref{algebra-definition-locally-free}. Suppose that $Q$ can be generated by $N$ elements as an $R$-module. Then it follows from Lemma \ref{lemma-map-cannot-be-injective} Algebra, Lemma \ref{algebra-lemma-map-cannot-be-injective} that $P$ is finite locally free (with the free parts having rank at most $N$). And in this case $P$ is a finite $R$-module, see Algebra, Lemma \ref{algebra-lemma-finite-projective}.
 @@ -7583,7 +7583,7 @@ 05WF,stacks-properties-lemma-zariski-open-cover-stack-is-scheme 05WG,examples-section-projective-not-locally-free 05WH,examples-lemma-ideal-generated-by-idempotents-projective 05WI,examples-lemma-map-cannot-be-injective 05WI,algebra-lemma-map-cannot-be-injective 05WJ,examples-lemma-ideal-projective-not-locally-free 05WK,examples-lemma-chow-group-product 05WL,examples-lemma-projective-not-locally-free