M commutative-algebra-notes.pdf → commutative-algebra-notes.pdf

M commutative-algebra-notes.tex → commutative-algebra-notes.tex

@@ -1361,7 +1361,24 @@ It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$.
 
 Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie.
 $$\frac{k[X_1, \ldots, X_n]}{m}=k$$
-Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown at \ref{5.2}).
+
+
+Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, we have the natural map
+$$\psi: k \longrightarrow \frac{k[X_1, \ldots, X_n]}{m}$$
+so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $k[X_1, \ldots, X_n]/m$ must be an element of $k$).\\
+Then, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
+Thus,
+$$a_i = X_i +m$$
+$$\Longrightarrow X_i - a_i \in m$$
+
+  % So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
+  % $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
+  % $$\Longrightarrow x_i - a_i \in m$$
+
+The ideal generated by these terms is a subset of $m$:
+$$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
+
+Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown also at \ref{5.2}).
 
 \vspace{0.4cm}
 Then:
@@ -1431,16 +1448,19 @@ let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
 
   By fundamental property of maximal ideals, $K=A/m$ is a field.
 
-  Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $x_i' = x_i +m$.
+  Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $\psi(x_i) = x_i' = x_i +m$ (where $\psi: k \longrightarrow A/m$).
 
   By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
 
   Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
   \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
   
-  So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
-  $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
-  $$\Longrightarrow x_i - a_i \in m$$
+  \vspace{0.3cm}
+  Recall that since $A$ is a $k$-algebra, we have the natural map $\psi: k \longrightarrow A/m$, so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $A/m$ must be an element of $k$).\\
+  Also, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
+  Thus,
+  $$a_i = X_i +m$$
+  $$\Longrightarrow X_i - a_i \in m$$
 
   The ideal generated by these terms is a subset of $m$:
   $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
@@ -1452,12 +1472,15 @@
   \vspace{0.3cm}
   Let
   \begin{align*}
-    \psi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
-    \psi: x_i &\longmapsto a_i
+    \phi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
+    \phi: x_i &\longmapsto a_i
   \end{align*}
 
-  Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
-  $$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$
+  Which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
+  \vspace{0.2cm}
+
+  Since $\phi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
+  $$\phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n))= f(a_1, \ldots, a_n)$$
 
   Thus there is a one-to-one correspondence: