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

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

@@ -1809,7 +1809,7 @@
 \vspace{0.5cm}
 This brings us to the \emph{Weak Nullstellensatz}:
 
-\begin{thm}{}[Weak Nullstellensatz]
+\begin{thm}{4.10}[Weak Nullstellensatz]\label{weak-nullstellensatz-again}
   $k$ a field, $K$ a $k$-algebra with
   \begin{itemize}
     \item $K$ fingen as a $k$-algebra
@@ -1835,7 +1835,105 @@ By definition of $K$ being a $k$-algebra,
 $$\exists~ \psi: k \longrightarrow \frac{K}{m} ~~~\Longrightarrow~ k \stackrel{iso}{\cong} \frac{K}{m}$$
 
 Thus, $\forall x_i \in k,~ \psi(x_i)=a_i \in k$.
-And by definition of a quotient ring, 
+And by definition of a quotient ring,
+
+$psi(x_i)$ is the coset
+\begin{align*}
+  \psi(x_i)&=x_i+m\\
+           &\Longrightarrow a_i = x_i +m\\
+           &x_i - a_i \in m
+\end{align*}
+
+Since $x_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
+
+since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
+
+(Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
+
+\vspace{0.4cm}
+We had $k \stackrel{iso}{\cong} \frac{K}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
+
+We have the quotient map
+\begin{align*}
+  \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
+  x_i &\mapsto a_i \in k
+\end{align*}
+which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
+
+Since $\phi$ is a $k$-algebra homomorphism, then
+$$\forall~ f \in K,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n)) = f(a_1, \ldots, a_n)$$
+
+Thus, $\exists$ a one-to-one correspondence:
+\begin{align*}
+  \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~ A, ~\text{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{align*}
+
+\vspace{0.5cm}
+Observe: $x_i - a_i \in ker(\psi) =m~ \forall~ i \in [n]$.
+
+We also have seen that $m=(X_1-a_1, \ldots, X_n-a_n)$.
+
+So for $P=(a_1, \ldots, a_n) \in k^n$, it is a zero for every polynomial in $m$, ie.
+$$\forall~ f \in m ~\Longrightarrow~ f(P)=0$$
+
+Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
+
+$\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
+\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
+  \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 already proven in the previous steps.
+    
+  \item $I(V(J)) = rad J$\\
+  ie. vanishing ideal of a variety = radical of the ideal definingh the variety
+
+  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.
+  $$\exists~ \underbrace{(a_1, \ldots, a_n, a_{n+1})}_{a} \in V(J')$$
+
+  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}
+
+  By the Weak Nullstellensatz (\ref{weak-nullstellensatz-again}),\\
+  \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 $g_i$.\\
+  Therefore $f^n \in J$, so $f \in rad~J$.
+  \end{enumerate}
+\end{proof}
+
 
 
 \section{Rings of fractions $S^{-1}A$ and localization}
@@ -2026,7 +2124,7 @@
       By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
       \begin{equation}
         u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
-        \tag{eq.6.4}
+        \label{eq:6.4}
       \end{equation}
       with $u,s,t \in S$.
 
@@ -2034,7 +2132,7 @@ Since $S=A \setminus P$ and $P$ is a prime ideal,\\
       the products of elements outside $P$ must also be outside of $P$.
       $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
 
-      At \ref{eq.6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
+      At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
 
       If $uab \not\in P$, then $a \not\in P$.
 
@@ -2095,10 +2193,11 @@ the quotient ring $A/I$ can be viewed as an $A$-module and
   $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
   \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
 
-  The \ref{6.6}.i, gives an isomorphism of modules
+  The \ref{cor.6.6}.i, gives an isomorphism of modules
   $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
   it's easy to see that this is a ring homomorphism.
 \end{proof}
+% TODO 6.6 (proposition and corollary)