Typos pointed out by Anfang Zhou

http://stacks.math.columbia.edu/tag/05NE#comment-779
http://stacks.math.columbia.edu/tag/05NH#comment-780
http://stacks.math.columbia.edu/tag/00HL#comment-782
http://stacks.math.columbia.edu/tag/01ED#comment-783
http://stacks.math.columbia.edu/tag/01UP#comment-789

algebra.tex

@@ -8168,7 +8168,7 @@ \section{Flat modules and flat ring maps}
 \begin{proof}
 Let $R^{(I)} \to N$ be a surjection from a free module
 onto $N$ with kernel $K$. The result follows
-by a simple diagram chase from the following diagram
+from the snake lemma applied to the following diagram
 $$
 \begin{matrix}
  & & 0 & & 0 & & 0 & & \\

cohomology.tex

@@ -925,7 +925,7 @@ \section{The {\v C}ech complex and {\v C}ech cohomology}
 This is an abelian group. For
 $s \in \check{\mathcal{C}}^p(\mathcal{U}, \mathcal{F})$ we denote
 $s_{i_0\ldots i_p}$ its value in $\mathcal{F}(U_{i_0\ldots i_p})$.
-Note that if $s \in \check{\mathcal{C}}^2(\mathcal{U}, \mathcal{F})$
+Note that if $s \in \check{\mathcal{C}}^1(\mathcal{U}, \mathcal{F})$
 and $i, j \in I$ then $s_{ij}$ and $s_{ji}$ are both elements
 of $\mathcal{F}(U_i \cap U_j)$ but there is no imposed
 relation between $s_{ij}$ and $s_{ji}$. In other words, we are {\it not}

modules.tex

@@ -2434,7 +2434,7 @@ \section{Flat modules}
 $\mathcal{O}_{X, x}$-module. Note that
 $$
 M \otimes_{\mathcal{O}_{X, x}} \mathcal{F}_x =
-\left(i_{x, *} M \otimes_{\mathcal{O}_X} \mathcal{F})\right)_x
+\left(i_{x, *} M \otimes_{\mathcal{O}_X} \mathcal{F}\right)_x
 $$
 again by
 Lemma \ref{lemma-stalk-tensor-product}.
@@ -2481,7 +2481,7 @@ \section{Flat modules}
 \end{lemma}
 
 \begin{proof}
-The stalks of $j_!\mathcal{O}_U$ are either zero of equal
+The stalks of $j_{U!}\mathcal{O}_U$ are either zero or equal
 to $\mathcal{O}_{X, x}$. Apply
 Lemma \ref{lemma-flat-stalks-flat}.
 \end{proof}
@@ -3899,7 +3899,7 @@ \section{Modules of differentials}
 has a natural structure of an $\mathcal{O}_X$-module.
 A section $s : \mathcal{O}_X \to \mathcal{A}$ of $\pi$
 is a $f^{-1}\mathcal{O}_S$-algebra map such that $\pi \circ s = \text{id}$.
-Given any section $s : \mathcal{O}_X \to \mathcal{I}$
+Given any section $s : \mathcal{O}_X \to \mathcal{A}$
 of $\pi$ and any $S$-derivation $D : \mathcal{O}_X \to \mathcal{I}$
 the map
 $$