diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index a82e2eb..bd8f076 100644 Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex index 29771d4..9d98719 100644 --- a/commutative-algebra-notes.tex +++ b/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}