small polishing

.github/workflows/typos.toml

@@ -8,6 +8,8 @@ iddeal = "ideal"
 iddeals = "ideals"
 allpha = "alpha"
 fieldd = "field"
+kernetl = "kernel"
+extenaion = "extension"
 
 # strings that are not a typo:
 thm = "thm"

commutative-algebra-notes.tex

@@ -1009,12 +1009,12 @@ As in with rings, it is equivalent to say that
 
 \subsection{A-algebras and integral domains}
 
-\begin{defn}{}[A-algebra]
+\begin{defn}{}[A-algebra / k-algebra]
   An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
 
   $B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
 
-  When $A \subset B$, $B$ is an extenaion ring of $A$; denoted $\psi(A) = A' \subset B$.
+  When $A \subset B$, $B$ is an extension ring of $A$; denoted $\psi(A) = A' \subset B$.
 \end{defn}
 
 \begin{defn}{R.4.1}\label{R.4.1}
@@ -1420,7 +1420,7 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   The ideal generated by these terms is a subset of $m$:
   $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
 
-  Since $J$ is the kernetl of the evaluation map at point $(a_1, \ldots, a_n)$,
+  Since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$,
   then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
   $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
 
@@ -1544,23 +1544,29 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   \begin{enumerate}[a.]
     \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
   Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
-  Then $L=k[X_1, \ldots, X_n]/m$ is a field (by TODO ref).
 
-  By Zariski's lemma (\ref{zariski}), since $L$ is generated as a $k$-algebra by the images of the variables $x_i$, and $k$ is algebraically closed.
+  \begin{itemize}
+    \item Since $m$ maximal, $L= k[X_1, \ldots, X_n]/m$ is a field.
+    \item Since $k[X_1, \ldots, X_n]$ is a fingen $k$-algebra, $L$ is a fingen $k$-algebra.
+  \end{itemize}
+  $\Longrightarrow~$ thus $L$ is a finite field extension of $k$.
 
-  Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$.
+  (Recall: if $k$ algebraically closed and $L$ a fingen field extension of $k$, then $L=k$.)
 
-  \vspace{0.3cm}
-  Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
+  Therefore,
+  $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$
 
-  Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$.
+  Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
 
-  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$.
+  Let $a_i = \psi(x_i)$. Then $x_i - a_i \in ker(\psi) = m ~\forall~ i \in [n]$.
 
-  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$.
+  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. (as in \ref{5.2})
 
-  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\
-  $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$.
+  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$,
+  $$f \in m ~\Longleftrightarrow~ f(P)=0$$
+
+  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$, ie. every element of $J$ vanishes at $P$.\\
+  $\Longrightarrow~$ therefore $P \in V(J)$, and thus $V(J) \neq \emptyset$.
 
 \item $I(V(J)) = rad J$:\\
   \begin{align*}