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

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

@@ -1338,7 +1338,7 @@
   thus there exists inverse in $A$, so $A$ is a field too.
 \end{proof}
 
-\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma]
+\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski}
   let $k$ a field, $K$ a $k$-algebra which
   \begin{enumerate}
     \item is finitely generated as a $k$-algebra
@@ -1372,6 +1372,42 @@ So $A=k$; which by Noether normalization we saw that $K$ is integral over $A=k$, and by \ref{integral-implies-finite} that it is finite, thus $K$ is finite over $k$, ie. $[K:k],\infty$, and $K$ is algebraic over $k$.
 \end{proof}
 
 
+\vspace{1cm}
+
+\section{Nullstellensatz}
+
+Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue field $K=k[X_1, \ldots, X_n]/m$ satisfies the Zariski's lemma (\ref{zariski}), thus $K$ is a finite algebraic extension of $k$.
+
+\vspace{0.3cm}
+\begin{cor}{5.2} \label{5.2}
+  $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots,
+  X_n]$ is of the form
+  $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$
+
+  The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
+
+  Thus
+  \begin{align*}
+    k^n &\longleftrightarrow m-Spec A\\
+    (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n)
+    \end{align*}
+\end{cor}
+\begin{proof}
+  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 $x_i' = x_i +m$.
+
+  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 $k$.
+  $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
+  $$\Longrightarrow x_i - a_i \in m$$
+\end{proof}
 
 
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