Minor corrections on algebra matters.

algebra.tex:355: The "this" in the "Informally this means..." paragraph
refers to the notion of "finitely presented", and so belongs inside part (2) of the
definition, rather than after it. (Otherwise it should be more explicit
about what it means by "this".)

algebra.tex:718: I have never seen transitive relations referred to as
"associative", so I take this for a typo.

algebra.tex:1863: There are various R-modules around, so I figured a bit more
explicitness wouldn't hurt.

algebra.tex:2021: It is certainly not the only iso $f : S^{-1}R \otimes_R M
\to S^{-1}M$.

algebra.tex:8067: The "(faithfully)" alternatives were added to the second
sentence to help prove the first sentence. It wouldn't hurt splitting the
lemma in two, though, as it now makes four different claims.

more-algebra.tex:4556: Fixed to the best of my knowledge. I am not 100% sure
of this since I don't know your notations.

algebra.tex

@@ -337,7 +337,7 @@ \section{Finite modules and finitely presented modules}
 
 \begin{definition}
 \label{definition-module-finite-type}
-Let $R$ be a ring. Let $M$ be an $R$-module
+Let $R$ be a ring. Let $M$ be an $R$-module.
 \begin{enumerate}
 \item We say $M$ is a {\it finite $R$-module}, or a {\it finitely generated
 $R$-module} if there exist $n \in \mathbf{N}$ and $x_1, \ldots, x_n \in M$
@@ -350,14 +350,14 @@ \section{Finite modules and finitely presented modules}
 $$
 R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0
 $$
-\end{enumerate}
-\end{definition}
 
 \noindent
 Informally this means that $M$ is finitely generated and that the module
 of relations among these generators is finitely generated as well. A choice
 of an exact sequence as in the definition is called a {\it presentation}
 of $M$.
+\end{enumerate}
+\end{definition}
 
 \begin{lemma}
 \label{lemma-lift-map}
@@ -715,7 +715,7 @@ \section{Colimits}
 \begin{definition}
 \label{definition-directed-set}
 A {\it partially ordered set} is a set $I$ together with a relation
-$\leq$ which is associative (if $i \leq j$ and $j \leq k$
+$\leq$ which is transitive (if $i \leq j$ and $j \leq k$
 then $i \leq k$) and reflexive ($i \leq i$ for all $i \in I$).
 A {\it directed set} $(I, \leq)$
 is a partially ordered set $(I, \leq)$ such that
@@ -829,8 +829,8 @@ \section{Colimits}
 \begin{lemma}
 \label{lemma-zero-directed-limit}
 Let $(M_i, \mu_{ij})$ be a directed system.
-Let $M = \colim M_i$ with $\mu_i : M_i \to M$,
-then $\mu_i(x_i) = 0$ for $x_i \in M_i$ if and only if
+Let $M = \colim M_i$ with $\mu_i : M_i \to M$.
+Then, $\mu_i(x_i) = 0$ for $x_i \in M_i$ if and only if
 there exists $j \geq i$ such that $\mu_{ij}(x_i) = 0$.
 \end{lemma}
 
@@ -918,7 +918,7 @@ \section{Colimits}
 =
 \phi_j(x_j) - \phi_k(\mu_{jk}(x_j))
 =
-\phi_j(x_j) - \nu_{jk}(\phi_i(x_j))
+\phi_j(x_j) - \nu_{jk}(\phi_j(x_j))
 $$
 which is in the kernel of the lower horizontal arrow as required.
 \end{proof}
@@ -1545,7 +1545,7 @@ \section{Internal Hom}
 is exact for all $R$-modules $N$.
 \item  Let $0 \to M_1 \to M_2 \to M_3$ be a complex of $R$-modules.
 Then $0 \to M_1 \to M_2 \to M_3$ is exact if and only if
-$0 \to \Hom_R(N, M_1) \to \Hom_R(N, M_2) \to \Hom_R(N, M_1)$
+$0 \to \Hom_R(N, M_1) \to \Hom_R(N, M_2) \to \Hom_R(N, M_3)$
 is exact for all $R$-modules $N$.
 \end{enumerate}
 \end{lemma}
@@ -1775,6 +1775,7 @@ \section{Tensor products}
 $z\in P$, then the mapping $(x, y)\mapsto x \otimes y \otimes z$,
 $x\in M, y\in N$, is $R$-bilinear in $x$ and $y$,
 and hence induces homomorphism $f_z : M \otimes N \to M \otimes N \otimes P$
+which sends
 $f_z(x \otimes y) = x \otimes y \otimes z$.
 Then consider $(M \otimes N)\times P \to M \otimes N \otimes P$ given by
 $(w, z)\mapsto f_z(w)$. The map is
@@ -1853,11 +1854,13 @@ \section{Tensor products}
 f(ax + y, z) = af(x, z)+f(y, z) =
 (a\phi_f(x)+\phi_f(y))(z),
 $$
-for all $y \in N$ and for all $a \in R$, $x, z \in M$. Conversely, any
+for all $a \in R$, $x \in M$, $y \in M$ and
+$z \in N$. Conversely, any
 $f \in \Hom_R(M, \Hom_R(N, P))$ defines an $R$-bilinear
 map $M \times N \to P$, namely $(x, y)\mapsto f(x)(y)$.
 So this is a natural one-to-one correspondence between the
-two modules.
+two modules
+$\Hom_R(M \otimes_R N, P)$ and $\Hom_R(M, \Hom_R(N, P))$.
 \end{proof}
 
 \begin{lemma}[Tensor products commute with colimits]
@@ -2013,9 +2016,9 @@ \section{Tensor products}
 
 \begin{lemma}
 \label{lemma-tensor-localization}
-Let $M$ be an $R$-module. Then the $S^{-1}R$ modules $S^{-1}M$
+Let $M$ be an $R$-module. Then the $S^{-1}R$-modules $S^{-1}M$
 and $S^{-1}R \otimes_R M$ are canonically isomorphic, and the
-unique isomorphism $f : S^{-1}R \otimes_R M \to S^{-1}M$
+canonical isomorphism $f : S^{-1}R \otimes_R M \to S^{-1}M$
 is given by
 $$
 f((a/s) \otimes m) = am/s, \forall a \in R, m \in M, s \in S
@@ -2024,7 +2027,7 @@ \section{Tensor products}
 
 \begin{proof}
 Obviously, the map
-$f' : S^{-1}R \times M \to S^{-1}M$ given by $f((am, s)) = am/s$ is
+$f' : S^{-1}R \times M \to S^{-1}M$ given by $f((a/s, m)) = am/s$ is
 bilinear, and thus by the
 universal property, this map induces a unique $S^{-1}R$-module homomorphism
 $f : S^{-1}R \otimes_R M \to S^{-1}M$ as in the statement of the lemma.
@@ -2123,8 +2126,10 @@ \section{Tensor algebra}
 in $\wedge^n(M)$ is denoted $x_1 \wedge \ldots \wedge x_n$.
 These elements generate $\wedge^n(M)$, they are $R$-linear
 in each $x_i$ and they are zero when two of the $x_i$ are equal
-(i.e., alternating). The multiplication on $\wedge(M)$ is
-graded commutative, i.e., $x \wedge y = - y \wedge x$.
+(i.e., they are alternating as functions of
+$x_1, x_2, \ldots, x_n$). The multiplication on $\wedge(M)$ is
+graded commutative, i.e., every $x \in M$ and $y \in M$
+satisfy $x \wedge y = - y \wedge x$.
 
 \medskip\noindent
 An example of this is when $M = Rx_1 \oplus \ldots \oplus Rx_n$
@@ -2700,7 +2705,7 @@ \section{Cayley-Hamilton}
 $(1 + a_1 + \ldots + a_n)\text{id}_M = 0$ with $a_j \in I$.
 Write $a_j = b_j(x)x$ for some $b_j(x) \in R[x]$.
 Translating back into $\varphi$ we see that
-$\text{id}_M = -(\sum_{j = 1, \ldots, n} b_j(\varphi) \varphi$ and hence
+$\text{id}_M = -(\sum_{j = 1, \ldots, n} b_j(\varphi)) \varphi$, and hence
 $\varphi$ is invertible.
 \end{proof}
 
@@ -8058,8 +8063,10 @@ \section{Flat modules and flat ring maps}
 \begin{lemma}
 \label{lemma-composition-flat}
 A composition of (faithfully) flat ring maps is
-(faithfully) flat. If $R \to R'$ is flat, and $M'$ is a flat
-$R'$-module, then $M'$ is a flat $R$-module.
+(faithfully) flat.
+If $R \to R'$ is (faithfully) flat, and $M'$ is a
+(faithfully) flat $R'$-module, then $M'$ is a
+(faithfully) flat $R$-module.
 \end{lemma}
 
 \begin{proof}
@@ -8083,7 +8090,10 @@ \section{Flat modules and flat ring maps}
 to the complex
 $M' \otimes_R N_1 \to M' \otimes_R N_2 \to M' \otimes_R N_3$,
 which is therefore also exact. This shows that $M'$ is a flat
-$R$-module.
+$R$-module. Tracing this argument backwards, we can show
+that if $R \to R'$ is faithfully flat, and if $M'$ is
+faithfully flat as an $R'$-module, then $M'$ is faithfully
+flat as an $R$-module.
 \end{proof}
 
 \begin{lemma}
@@ -8128,13 +8138,15 @@ \section{Flat modules and flat ring maps}
 injective $R$-module maps into injective $R$-module maps.
 
 \medskip\noindent
-Assume $K \to M$ is an injective $R$-module map and
+Assume $K \to N$ is an injective $R$-module map and
 let $x \in \Ker(K \otimes_R M \to N \otimes_R M)$.
 We have to show that $x$ is zero. It is clear that
 the module $N = \bigcup N_i$ is the union of its
-finitely generated $R$-submodules $N_i$. Set $K_i = K \cap N_i$.
+finitely generated $R$-submodules $N_i$. Set $K_i = K \cap N_i$
+(where we view $K$ as an $R$-submodule of $N$ via the
+injective $R$-module map $K \to N$).
 For some $i$ our $x$ comes from an element $x_i \in K_i \otimes_R M$.
-After increasing $i$ the element $x_i$ maps to zero in $M_i \otimes_R M$
+After increasing $i$ the element $x_i$ maps to zero in $N_i \otimes_R M$
 (because tensor product commutes with colimits).
 Thus we may assume $N$ is a finite $R$-module.
 
@@ -8316,7 +8328,9 @@ \section{Flat modules and flat ring maps}
 
 \begin{proof}
 Assume $M$ is flat and let $\sum f_i x_i = 0$ be a relation.
-Let $I = (f_1, \ldots, f_n)$, and let $K = \Ker(R^n \to I)$.
+Let $I = (f_1, \ldots, f_n)$, and let $K = \Ker(R^n \to I)$,
+where the map $R^n \to I$ is given by sending
+$\left(z_1, z_2, \ldots, z_n\right)$ to $\sum_i f_i z_i$.
 So we have the short exact sequence
 $0 \to K \to R^n \to I \to 0$. Then $\sum f_i \otimes x_i$
 is an element of $I \otimes_R M$ which maps

dga.tex

@@ -139,7 +139,7 @@ \section{Differential graded algebras}
 $$
 endowed with differential $\text{d}$ defined by the rule
 $\text{d}(a \otimes b) = \text{d}(a) \otimes b + (-1)^m a \otimes \text{d}(b)$
-where $m = \deg(b)$.
+where $m = \deg(a)$.
 \end{definition}
 
 \begin{lemma}

dpa.tex

@@ -488,12 +488,12 @@ \section{Extending divided powers}
 \end{lemma}
 
 \begin{proof}
-Any element of $IB$ can be written as a finite sum $\sum b_ix_i$ with
+Any element of $IB$ can be written as a finite sum $\sum\nolimits_{i=1}^t b_ix_i$ with
 $b_i \in B$ and $x_i \in I$. If $\gamma$ extends to $\bar\gamma$ on $IB$
 then $\bar\gamma_n(x_i) = \gamma_n(x_i)$.
-Thus conditions (3) and (4) imply that
+Thus, conditions (3) and (4) in Definition \ref{definition-divided-powers} imply that
 $$
-\bar\gamma_n(\sum b_ix_i) =
+\bar\gamma_n(\sum\nolimits_{i=1}^t b_ix_i) =
 \sum\nolimits_{n_1 + \ldots + n_t = n}
 \prod\nolimits_{i = 1}^t b_i^{n_i}\gamma_{n_i}(x_i)
 $$

more-algebra.tex

@@ -4528,7 +4528,7 @@ \section{The Koszul complex}
 
 \begin{lemma}
 \label{lemma-functorial}
-Let $\varphi : E \to R$ and $\varphi : E' \to R$ be an $R$-module maps.
+Let $\varphi : E \to R$ and $\varphi' : E' \to R$ be $R$-module maps.
 Let $\psi : E \to E'$ be an $R$-module map such that
 $\varphi' \circ \psi = \varphi$. Then $\psi$ induces a
 homomorphism of differential graded algebras
@@ -4546,15 +4546,15 @@ \section{The Koszul complex}
 coefficients in $R$. Then the complexes
 $K_\bullet(f_\bullet)$ and
 $$
-K_\bullet(\sum x_{1j}f_j, \sum x_{2j}f_j, \ldots, \sum x_{cj}f_j)
+K_\bullet(\sum x_{1j}f_j, \sum x_{2j}f_j, \ldots, \sum x_{rj}f_j)
 $$
 are isomorphic.
 \end{lemma}
 
 \begin{proof}
 Set $g_i = \sum x_{ij}f_j$.
-The matrix $(x_{ij})$ gives an isomorphism $x : R^{\oplus r} \to R^{\oplus r}$
-such that $(g_1, \ldots, g_r) \circ x = (f_1, \ldots, f_r)$.
+The matrix $(x_{ji})$ gives an isomorphism $x : R^{\oplus r} \to R^{\oplus r}$
+such that $(g_1, \ldots, g_r) = (f_1, \ldots, f_r) \circ x$.
 Hence this follows from the functoriality of the Koszul complex
 described in
 Lemma \ref{lemma-functorial}.