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port Nullstellensatz & Varieties notes (ch5) (#6)
* 5.1, Corollary 5.2 * variety def, Nullstellensatz proof, correspondences V - I * port notes on propositions 5.3, 5.7, 5.8 * add missing conclusion at proposition 5.3
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@@ -99,7 +99,7 @@
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\end{defn}
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\begin{defn}{}[prime ideal]
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if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$.
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if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$.
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\end{defn}
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\begin{defn}{}[principal ideal]
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@@ -1338,7 +1338,7 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
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thus there exists inverse in $A$, so $A$ is a field too.
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\end{proof}
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\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma]
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\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski}
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let $k$ a field, $K$ a $k$-algebra which
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\begin{enumerate}
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\item is finitely generated as a $k$-algebra
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@@ -1372,7 +1372,326 @@ Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some
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\end{proof}
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\vspace{1cm}
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\section{Nullstellensatz}
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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 $k$.
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\vspace{0.3cm}
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\begin{cor}{5.2} \label{5.2}
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$k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots,
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X_n]$ is of the form
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$$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$
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The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
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Thus
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\begin{align*}
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k^n &\longleftrightarrow m-Spec A\\
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(a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n)
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\end{align*}
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\end{cor}
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\begin{proof}
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let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
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By fundamental property of maximal ideals, $K=A/m$ is a field.
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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 $x_i' = x_i +m$.
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By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
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Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
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\hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
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So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
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$$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
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$$\Longrightarrow x_i - a_i \in m$$
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The ideal generated by these terms is a subset of $m$:
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$$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
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Since $J$ is the kernetl of the evaluation map at point $(a_1, \ldots, a_n)$,
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then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
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$J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
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\vspace{0.3cm}
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Let
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\begin{align*}
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\psi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
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\psi: x_i &\longmapsto a_i
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\end{align*}
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Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
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$$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$
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Thus there is a one-to-one correspondence:
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points in $k^n ~~~ \longleftrightarrow~~~ m-Spec A$ (maximal ideals in $k[X_1, \ldots, X_n]$
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$(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$
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\end{proof}
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\vspace{0.5cm}
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\subsection{Variety}
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\begin{defn}{5.3}[Variety]
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A \emph{variety} $V \subset k^n$:
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$$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
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\begin{itemize}
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\item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$
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\item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations.
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\end{itemize}
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\end{defn}
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\begin{prop}{5.3} \label{5.3}
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$k$ an algebraically closed field, and $A=k[X_1, \ldots, X_n]$ a fingen $k$-algebra of the form $A=k[X_1, \ldots, X_n]/J$, where $J$ is an ideal of $k[X_1, \ldots, X_n]$.
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(notation: $x_i = X_i \pmod J$)
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Then every maximal ideal of $A$ is of the form
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$$(x_1 -a_1, \ldots, x_n - a_n)$$
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for some point $(a_1, \ldots, a_n) \in V(J)$.
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Therefore, $\exists$ a one-to-one correspondence
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\begin{align*}
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V(X) &\longleftrightarrow m-Spec A\\
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\text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n)
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\end{align*}
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\end{prop}
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\begin{proof}
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the ideals of $A$ are given by ideals of $k[X_1, \ldots, X_n]$ containing $J$, since for $Q=R/I$, $\exists$ one-to-one correspondence between ideals of $Q$ and ideals of $R$ that contain $I$, ie.
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\begin{align*}
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m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\
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&\text{s.th.}~ J \subseteq m'
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\end{align*}
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Thus, every maximal ideal of $A$ is of the form
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$$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$
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\begin{enumerate}[i.]
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\item Since $k$ is algebraically closed, the maximal ideals of $k[X_1, \ldots, X_n]$ look like $m=(X_1 - a_1, \ldots, X_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$.
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Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$.
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\item for $m$ to exist in $A$, the corresponding $m'$ must contain $J$; since if it didn't contain $J$ it wouldn't "survive" the quotient process.
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\end{enumerate}
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\vspace{0.3cm}
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However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$
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$\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$.
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If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$.
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By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$.
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\vspace{0.4cm}
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Thus,\\
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every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie.
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$$m-Spec A \longleftrightarrow k^n$$
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The condition that the ideal belongs to the quotient ring $A=k[X_1, \ldots, X_n]/J$ forces that point to lie in $V(J)$, so
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\begin{align*}
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m-Spec A &\longleftrightarrow V(J)\\
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\text{maximal spectrum} &\longleftrightarrow \text{variety}
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\end{align*}
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\end{proof}
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\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
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A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$).
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So $V$ gives a map:\\
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$$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$
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correspondence going the other way:
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$$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$
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defined by taking a subset $X \subset k^n$ into the ideal
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$$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$
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\vspace{0.4cm}
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$V,~I$ satisfy reverse inclusions:
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$$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
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\end{prop}
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\vspace{0.4cm}
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\subsection{Nullstellensatz}
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\vspace{0.5cm}
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\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
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Let $k$ algebraically closed field.
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\begin{enumerate}[a.]
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\item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
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\item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
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$$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
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\end{enumerate}
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\end{thm}
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\begin{proof}
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\begin{enumerate}[a.]
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\item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
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Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
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Then $L=k[X_1, \ldots, X_n]/m$ is a field (by TODO ref).
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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.
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Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$.
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\vspace{0.3cm}
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Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
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Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$.
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Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$.
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Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$.
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Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\
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$\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$.
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\item $I(V(J)) = rad J$:\\
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\begin{align*}
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I(V(J)) &= rad J\\
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\text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety}
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\end{align*}
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where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$.
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Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
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Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
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Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
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$$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$
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Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
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$$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
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a contradiction.
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Therefore, $V(J') = \emptyset$.
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\vspace{0.4cm}
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Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\
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\hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
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Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
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$$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)} g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
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which, since $1 \in J'$,
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$$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
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substitute $Y=1/f$,
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$$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
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thus
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$$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
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multiply by $f^n$,
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$$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
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thus $f^n$ is a linear combination of $g_i$.\\
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Thus $f^n \in J$, so $f \in rad~J$.
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\end{enumerate}
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\end{proof}
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\vspace{0.4cm}
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\subsection{Irreducible varieties}
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\begin{defn}{5.7}[Irreducible variety]
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a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if
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$$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$
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\end{defn}
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\begin{prop}{5.7} \label{5.7}
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a variety $X$ is irreducible iff $I(X)$ is prime.
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\end{prop}
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\begin{proof}
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set $I=I(X)$.
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if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$.
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Define new ideals
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$$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$
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Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$
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\hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible.
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The converse is similar.
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\end{proof}
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\begin{cor}{5.8} \label{cor.5.8}
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let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences
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\begin{align*}
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\{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\
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\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}
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\end{align*}
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Therefore,
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$$Spec k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$
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\end{corollary}
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\end{cor}
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\begin{prop}{5.8} \label{5.8}
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let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field).
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Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$.
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Then there is a one-to-one correspondence
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$$Spec A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$
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\end{prop}
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\begin{proof}
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By definition, $Spec A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$.
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By Corollary \ref{cor.5.8}:
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$$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
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About varieties:
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\begin{itemize}
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\item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\
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\hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$.
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\item $I(X)$ in $A=k[X_1, \ldots, X_n]/J$, we're not looking at all possible polynomials but at the residue classes.
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\end{itemize}
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If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$).
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Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\
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\hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$.
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Since we're restricted to the ring $A=k[X_1, \ldots, X_n]/J$, the ideal $P$ are the elements of $I(X)$ viewed through the lens of the quotient\\
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$$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$
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Now, $J$ is the set of equations defining our "universe" $V(J)$.
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Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$.
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Also we have a correspondence between $A=R/J$ and $R$.
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Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie.
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\begin{align*}
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R &\longrightarrow A=R/J\\
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\mathfrak{P} &\longmapsto P\\
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I(X) &\longmapsto I(X) ~\text{mod}~J
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\end{align*}
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$\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$.
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Now, $J \susbset \mathfrak{P} = I(X)$;\\
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by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so
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$$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$
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$\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$.
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Therefore,
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$$\mathfrak{P} \in Spec A \longleftrightarrow X \subseteq V(J)$$
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where $X = V(\mathfrak{P})$.
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\end{proof}
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Reference in New Issue
Block a user