variety def, Nullstellensatz proof, correspondences V - I

commutative-algebra-notes.tex

@@ -99,7 +99,7 @@
 \end{defn}
 
 \begin{defn}{}[prime ideal]
-  if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$.
+  if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$.
 \end{defn}
 
 \begin{defn}{}[principal ideal]
@@ -1407,9 +1407,145 @@ Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue fie
   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 $m$:
+  $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
+
+  Since $J$ is the kernetl 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. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
+
+  \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
+  \end{align*}
+
+  Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
+  $$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$
+
+  Thus there is a one-to-one correspondence:
+
+  points in $k^n ~~~ \longleftrightarrow~~~ m-Spec A$ (maximal ideals in $k[X_1, \ldots, X_n]$
+
+  $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$
 \end{proof}
 
 
+\vspace{0.5cm}
+\begin{defn}{5.3}[Variety]
+  A \emph{variety} $V \subset k^n$:
+  $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
+
+  \begin{itemize}
+    \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$
+    \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations.
+  \end{itemize}
+\end{defn}
+
+\begin{prop}{5.3} \label{5.3}
+  TODO
+\end{prop}
+\begin{proof}
+\end{proof}
+
+\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
+  A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$).
+
+  So $V$ gives a map:\\
+  $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$
+
+  correspondence going the other way:
+  $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$
+
+  defined by taking a subset $X \subset k^n$ into the ideal
+  $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$
+
+  \vspace{0.4cm}
+
+  $V,~I$ satisfy reverse inclusions:
+  $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
+\end{prop}
+
+\vspace{0.5cm}
+\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
+  Let $k$ algebraically closed field.
+
+  \begin{enumerate}[a.]
+    \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
+    \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
+      $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
+  \end{enumerate}
+\end{thm}
+\begin{proof}
+  \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.
+
+  Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$.
+
+  \vspace{0.3cm}
+  Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
+
+  Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$.
+
+  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$.
+
+  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$.
+
+  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\
+  $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$.
+
+\item $I(V(J)) = rad J$:\\
+  \begin{align*}
+    I(V(J)) &= rad J\\
+    \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety}
+  \end{align*}
+  where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$.
+
+  Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
+
+  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
+
+  Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
+  $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$
+
+  Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
+  $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
+  a contradiction.
+
+  Therefore, $V(J') = \emptyset$.
+
+  \vspace{0.4cm}
+
+  Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\
+  \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
+
+  Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
+
+  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)} g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
+
+  which, since $1 \in J'$,
+  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
+
+  substitute $Y=1/f$,
+  $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
+  thus
+  $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
+  multiply by $f^n$,
+  $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
+
+  thus $f^n$ is a linear combination of $g_i$.\\
+  Thus $f^n \in J$, so $f \in rad~J$.
+\end{enumerate}
+\end{proof}
+
+
+
 \newpage
 
 \section{Exercises}