Fix typo in algebra

Also checked all the points are proven.
Thanks to BB
https://stacks.math.columbia.edu/tag/00JP#comment-4531

algebra.tex

@@ -12989,8 +12989,9 @@ \section{Proj of a graded ring}
 
 \begin{proof}
 Since $D_{+}(f) = \text{Proj}(S) \cap D(f)$, these sets are open.
-Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(E)$ are
-closed.
+This proves (1). Also (2) follows as $D(ff') = D(f) \cap D(f')$.
+Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(I)$
+are closed. This proves (8).
 
 \medskip\noindent
 Suppose that $T \subset \text{Proj}(S)$ is closed.
@@ -13000,17 +13001,18 @@ \section{Proj of a graded ring}
 with $g_d \in S_d$ then $g_d \in \mathfrak p$ for all
 $\mathfrak p \in T$. Thus, letting $I \subset S$
 be the ideal generated by the homogeneous parts of the elements
-of $J$ we have $T = V_{+}(I)$.
+of $J$ we have $T = V_{+}(I)$. This proves (9).
 
 \medskip\noindent
 The formula for $\text{Proj}(S) \cap D(g)$, with $g \in S$ is direct
-from the definitions. Consider the formula for $\text{Proj}(S) \cap D(g_0)$.
+from the definitions. This proves (3).
+Consider the formula for $\text{Proj}(S) \cap D(g_0)$.
 The inclusion of the right hand side in the left hand side is
 obvious. For the other inclusion, suppose $g_0 \not \in \mathfrak p$
 with $\mathfrak p \in \text{Proj}(S)$. If all $g_0f \in \mathfrak p$
 for all homogeneous $f$ of positive degree, then we see that
 $S_{+} \subset \mathfrak p$ which is a contradiction. This gives
-the other inclusion.
+the other inclusion. This proves (4).
 
 \medskip\noindent
 The collection of opens $D(g) \cap \text{Proj}(S)$
@@ -13019,10 +13021,10 @@ \section{Proj of a graded ring}
 $\Spec(S)$. By the formulas above we can express
 $D(g) \cap \text{Proj}(S)$ as a union of opens $D_{+}(f)$.
 Hence the collection of opens $D_{+}(f)$ forms a basis for the topology
-also.
+also. This proves (5).
 
 \medskip\noindent
-First we note that $D_{+}(f)$ may be identified
+Proof of (6). First we note that $D_{+}(f)$ may be identified
 with a subset (with induced topology) of $D(f) = \Spec(S_f)$
 via Lemma \ref{lemma-standard-open}. Note that the ring
 $S_f$ has a $\mathbf{Z}$-grading. The homogeneous elements are
@@ -13040,7 +13042,7 @@ \section{Proj of a graded ring}
 $$
 \text{Proj}(S) = \bigcup\nolimits_{i = 1}^\infty D_{+}(X_i)
 $$
-does not have a finite refinement.
+does not have a finite refinement. This proves (7).
 
 \medskip\noindent
 Let $I \subset S$ be a graded ideal.
@@ -13052,14 +13054,14 @@ \section{Proj of a graded ring}
 Clearly this means that one of the homogeneous parts of $f$
 is not nilpotent modulo $I$, in other words we may (and do)
 assume that $f$ is homogeneous. This implies that
-$I S_f \not = 0$, in other words that $(S/I)_f$ is not
+$I S_f \not = S_f$, in other words that $(S/I)_f$ is not
 zero. Hence $(S/I)_{(f)} \not = 0$ since it is a ring
 which maps into $(S/I)_f$. Pick a prime
 $\mathfrak q \subset (S/I)_{(f)}$. This corresponds to
 a graded prime of $S/I$, not containing the irrelevant ideal
 $(S/I)_{+}$. And this in turn corresponds to a graded prime
 ideal $\mathfrak p$ of $S$, containing $I$ but not containing $S_{+}$
-as desired.
+as desired. This proves (10) and finishes the proof.
 \end{proof}
 
 \begin{example}