Spell checking algebra.tex

algebra.tex

@@ -2900,7 +2900,8 @@ \section{The spectrum of a ring}
 \in I$ for some $n$. Hence if $I \subset \mathfrak{p}$, then $a^n \in
 \mathfrak{p}$. But since $\mathfrak{p}$ is prime, we have $a \in \mathfrak{p}$.
 Thus the equality is shown.
-\item $I$ is not the unit ideal iff $I$ is contained in some maximal ideal (to
+\item $I$ is not the unit ideal if and only if $I$
+is contained in some maximal ideal (to
 see this, apply (2) to the ring $R/I$) which is therefore prime.
 \item If $\mathfrak{p} \in V(I) \cup V(J)$, then $I \subset \mathfrak{p}$ or $J
 \subset \mathfrak{p}$ which means that $I \cap J \subset \mathfrak{p}$. Now if
@@ -8426,7 +8427,7 @@ \section{Flat modules and flat ring maps}
 \begin{lemma}
 \label{lemma-ff}
 \begin{slogan}
-A flat module is faithfully flat iff it has nonzero fibers.
+A flat module is faithfully flat if and only if it has nonzero fibers.
 \end{slogan}
 Let $M$ be a flat $R$-module.
 The following are equivalent:
@@ -9336,7 +9337,7 @@ \section{Geometrically reduced algebras}
 
 \begin{lemma}
 \label{lemma-subalgebra-separable}
-Elementary properties of geometrically reducedness.
+Elementary properties of geometrically reduced algebras.
 Let $k$ be a field. Let $S$ be a $k$-algebra.
 \begin{enumerate}
 \item If $S$ is geometrically reduced over $k$ so is every
@@ -9418,7 +9419,9 @@ \section{Geometrically reduced algebras}
 \ref{lemma-Noetherian-irreducible-components}, and
 \ref{lemma-minimal-prime-reduced-ring}).
 Hence we may assume $R$ is a finite product of
-fields. In this case the reducedness follows from the definition.
+fields. In this case it follows from
+Definition \ref{definition-geometrically-reduced}
+that $R \otimes_k S$ is reduced.
 \end{proof}
 
 \begin{lemma}
@@ -11426,7 +11429,7 @@ \section{Artinian rings}
 $\mathfrak m^j/\mathfrak m^{j + 1}$ is the dimension of this
 as a vector space over $\kappa(\mathfrak m)$. This has to be
 finite since otherwise we would have an infinite descending
-chain of subvector spaces which would correspond to an
+chain of sub vector spaces which would correspond to an
 infinite descending chain of ideals in $R$.
 \end{proof}
 
@@ -15313,7 +15316,7 @@ \section{Quasi-regular sequences}
 
 \begin{proof}
 Set $J = (f_1, \ldots, f_r)$, $J' = JR'$ and $M' = M \otimes_R R'$.
-We have to show the canonincal map
+We have to show the canonical map
 $\mu : R'/J'[X_1, \ldots X_n] \otimes_{R'/J'} M'/J'M' \to
 \bigoplus (J')^nM'/(J')^{n + 1}M'$ is an isomorphism.
 Because $R \to R'$ is flat the sequences
@@ -15578,7 +15581,7 @@ \section{Ext groups and depth}
 \begin{proof}
 Proof of (1). Because $F_0$ is free we can find a map $F_0 \to N_0$
 lifting the map $F_0 \to M \to N$. We obtain an induced
-map $F_1 \to F_0 \to N_0$ wich ends up in the image
+map $F_1 \to F_0 \to N_0$ which ends up in the image
 of $N_1 \to N_0$. Since $F_1$ is free we may lift this
 to a map $F_1 \to N_1$. This in turn induces a map
 $F_2 \to F_1 \to N_1$ which maps to zero into
@@ -15658,7 +15661,7 @@ \section{Ext groups and depth}
 be a map inducing $\psi :
 M_2 = \Coker(d_{G, 1}) \to M_1 = \Coker(d_{F, 1})$,
 see Lemma \ref{lemma-compare-resolutions}.
-Ok, and now consider the map
+OK, and now consider the map
 $H^i(\alpha) \circ H^i(\beta) =
 H^i(\alpha \circ \beta)$. By the above the
 map $H^i(\alpha \circ \beta)$ is the {\it same}
@@ -15744,7 +15747,7 @@ \section{Ext groups and depth}
 $$
 By the snake lemma we see that the sequence of kernels
 $0 \to K' \to K \to K'' \to 0$ is short exact sequence of $R$-modules.
-Hence we can continue this process indefinitively. In other words
+Hence we can continue this process indefinitely. In other words
 we obtain a short exact sequence of resolutions fitting into the diagram
 $$
 \begin{matrix}
@@ -19608,7 +19611,7 @@ \section{Countably generated Mittag-Leffler modules}
 \Im(M_{k(i)}^* \to M_i^*)
 $$
 Choose $i \in I$ such that $\Im(P \to M) \subset \Im(M_i \to M)$.
-This is possible as $P$ is finitely generated. Set $k = k(i)$ for this i.
+This is possible as $P$ is finitely generated. Set $k = k(i)$ for this $i$.
 Let
 $x = (0, \ldots, 0, \text{id}_{M_k}, 0, \ldots, 0) \in M_k^*$ and
 note that this maps to $y = (0, \ldots, 0, f_{ik}, 0, \ldots, 0) \in M_i^*$.
@@ -22018,7 +22021,7 @@ \section{What makes a complex exact?}
 Note that the ring $R_{\mathfrak q}$ has depth $0$.
 We apply Lemmas \ref{lemma-exact-depth-zero-local} and
 \ref{lemma-trivial-case-exact} to the localized complex
-over $R_{\mathfrak q}$. all of the ideals
+over $R_{\mathfrak q}$. All of the ideals
 $I(\varphi_i)_{\mathfrak q}$, $e \geq i \geq 1$
 are equal to $R_{\mathfrak q}$. Thus none of the ideals
 $I(\varphi_i)$ is contained in $\mathfrak q$.
@@ -22344,7 +22347,7 @@ \section{Cohen-Macaulay modules}
 \ref{lemma-support-closed}). Hence the dimension is $1$.
 To finish the proof it suffices to find a single
 $Q$-regular element $f \in \mathfrak m$. Since $\mathfrak m$ is
-a maxmimal ideal, the field extension
+a maximal ideal, the field extension
 $\kappa(\mathfrak p) \subset \kappa(\mathfrak m)$
 is finite (Theorem \ref{theorem-nullstellensatz}).
 Hence we can find $f \in \mathfrak m$ which
@@ -23720,7 +23723,7 @@ \section{Regular rings and global dimension}
 Conversely, suppose that all the $R_{\mathfrak m}$
 have global dimension $n$. Let $M$ be a finite
 $R$-module. Let
-$0 \to K_n \to F_{n-1} \to \ldots \to F_0 \to M \to 0$.
+$0 \to K_n \to F_{n-1} \to \ldots \to F_0 \to M \to 0$
 be a resolution with $F_i$ finite free.
 Then $K_n$ is a finite $R$-module.
 According to
@@ -24755,7 +24758,7 @@ \section{Noether normalization}
 f(y_1 + x_n^{e_1}, \ldots, y_{n-1} + x_n^{e_{n-1}}, x_n) = 0.
 $$
 Hence we are done if we can show there exists integers $e_i$ such
-that the leading coefficient w.r.t.\ $x_n$ of the equation
+that the leading coefficient with respect to $x_n$ of the equation
 above is a nonzero element of $k$. This can be achieved for example
 by choosing $e_1 \gg e_2 \gg \ldots \gg e_{n-1}$, see
 Lemma \ref{lemma-helper-polynomial}.
@@ -25946,7 +25949,7 @@ \section{Around Krull-Akizuki}
 is Artinian (Lemma \ref{lemma-finite-dimensional-algebra}). The
 structural results on Artinian rings implies parts (1) and (2), see
 for example Lemma \ref{lemma-artinian-finite-length}.
-Part (3) is implied by the finiteness esthablished above.
+Part (3) is implied by the finiteness established above.
 \end{proof}
 
 \begin{lemma}
@@ -26097,7 +26100,7 @@ \section{Factorization}
 x_1 | x_2 | x_3 | \ldots
 $$
 of elements of $R$ with $x_n/x_{n + 1}$ not a unit.
-This gives a strictly increasin sequence of principal ideals
+This gives a strictly increasing sequence of principal ideals
 $(x_1) \subset (x_2) \subset (x_3) \subset \ldots$ thereby finishing the proof.
 \end{proof}
 
@@ -26336,7 +26339,7 @@ \section{Factorization}
 
 \medskip\noindent
 Writing $(x) = \mathfrak p_1 \ldots \mathfrak p_r$ anew with
-$\mathfrak p_1 \subset \mathfrak p$ we conlude that
+$\mathfrak p_1 \subset \mathfrak p$ we conclude that
 $\mathfrak p_1 R_\mathfrak p = \mathfrak p R_\mathfrak p$, i.e.,
 $\mathfrak p_1 = \mathfrak p$. Moreover, $\mathfrak p_1 = \mathfrak p$ is
 a finitely generated ideal of $R$ by
@@ -28370,7 +28373,7 @@ \section{Colimits and maps of finite presentation}
 \noindent
 In this section we prove some preliminary lemmas
 which will eventually help us prove result using
-absolute Noetherian reduction. we begin discussing
+absolute Noetherian reduction. We begin discussing
 how we will think about colimits in this section.
 
 \medskip\noindent
@@ -28669,7 +28672,7 @@ \section{Colimits and maps of finite presentation}
 To prove (4) assume $u \otimes 1$ is an isomorphism, that
 $B$ is a finite type $A$-algebra, and that $C$ is a finitely presented
 $A$-algebra. Let $v : B \otimes_A R \to C \otimes_A R$ be an inverse to
-$u \otimes 1$. let $v_i : C \otimes_A R_i \to B \otimes_A R_i$ be as
+$u \otimes 1$. Let $v_i : C \otimes_A R_i \to B \otimes_A R_i$ be as
 in part (3). Apply part (1) to see that, after increasing $i$ we have
 $v_i \circ (u \otimes 1) = \text{id}_{B \otimes_R R_i}$ and
 $(u \otimes 1) \circ v_i = \text{id}_{C \otimes_R R_i}$.
@@ -34293,7 +34296,7 @@ \section{Smooth algebras over fields}
 \bigoplus S \cdot \text{d}x_i
 $$
 see Lemma \ref{lemma-relative-global-complete-intersection-conormal}.
-By Lemma \ref{lemma-relative-global-complete-intersection-smooth}.
+By Lemma \ref{lemma-relative-global-complete-intersection-smooth}
 in order to show that $S$ is smooth at
 $\mathfrak m$ we have to show that one of the $c \times c$
 minors $g_I$ of the matrix ``$A$'' giving the map above
@@ -35940,7 +35943,7 @@ \section{\'Etale ring maps}
 have isomorphic open neighbourhoods in $\Spec(F)$
 and $\Spec(F')$. We conclude the set
 $\{\overline{\mathfrak q}'\} \subset \Spec(F')$ is
-open. Combined with closedness shown above
+open. Combined with $\mathfrak q'$ being closed (shown above)
 we conclude that $\overline{\mathfrak q}'$ defines
 an isolated closed point of $\Spec(F')$ as well.
 
@@ -39997,7 +40000,7 @@ \section{Nagata and Japanese rings}
 
 \begin{definition}
 \label{definition-N}
-Let $R$ be a a domain with field of fractions $K$.
+Let $R$ be a domain with field of fractions $K$.
 \begin{enumerate}
 \item We say $R$ is {\it N-1} if the integral closure of $R$ in $K$
 is a finite $R$-module.
@@ -40621,7 +40624,7 @@ \section{Nagata and Japanese rings}
 For each $i$, let $\mathfrak q_{i1}, \ldots, \mathfrak q_{is_i}$
 be the associated primes of the $R^\wedge$-module
 $R^\wedge/\mathfrak p_iR^\wedge$.
-By Lemma \ref{lemma-codimension-1-analytically-unramified}.
+By Lemma \ref{lemma-codimension-1-analytically-unramified}
 we see that $(R^\wedge)_{\mathfrak q_{ij}}$ is regular.
 By Lemma \ref{lemma-bourbaki} we see that
 $$
@@ -41218,7 +41221,7 @@ \section{Ascending properties}
 First assume $R$ is Noetherian.
 In this case being reduced is the same as having properties
 $(S_1)$ and $(R_0)$, see Lemma \ref{lemma-criterion-reduced}.
-Note that $S$ is noetherian, and
+Note that $S$ is Noetherian, and
 $R \to S$ is flat with regular fibres (see the list of
 results on smooth ring maps in Section \ref{section-smooth-overview}).
 Hence we may apply Lemmas \ref{lemma-Sk-goes-up} and \ref{lemma-Rk-goes-up}
@@ -41252,7 +41255,7 @@ \section{Ascending properties}
 First assume $R$ is Noetherian.
 In this case being reduced is the same as having properties
 $(S_2)$ and $(R_1)$, see Lemma \ref{lemma-criterion-normal}.
-Note that $S$ is noetherian, and
+Note that $S$ is Noetherian, and
 $R \to S$ is flat with regular fibres (see the list of
 results on smooth ring maps in Section \ref{section-smooth-overview}).
 Hence we may apply Lemmas \ref{lemma-Sk-goes-up} and \ref{lemma-Rk-goes-up}
@@ -41838,7 +41841,7 @@ \section{Geometrically regular algebras}
 \begin{lemma}
 \label{lemma-geometrically-regular-over-separable-algebraic}
 Let $k \subset k'$ be a separable algebraic field extension.
-Let $A$ be a an algebra over $k'$. Then $A$ is geometrically
+Let $A$ be an algebra over $k'$. Then $A$ is geometrically
 regular over $k$ if and only if it is geometrically regular over $k'$.
 \end{lemma}