\documentclass[11pt]{article} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \usepackage{amsmath,amscd,amsthm,amsfonts} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{exer}{Exercise} \newtheorem{exam}{Example} \theoremstyle{plain} \newtheorem{thm}{Theorem} \newtheorem{lem}{Lemma} \newtheorem{cor}{Corollary} \newtheorem{prop}{Proposition} \renewcommand{\P}{\mathbb{P}^1} \renewcommand{\k}{\kappa} \renewcommand{\H}{\mathbb{H}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\A}{\mathbb{A}} \newcommand{\G}[1]{\Gamma(#1)} \newcommand{\GO}[1]{\Gamma_0(#1)} \newcommand{\Gb}[1]{\bar{\Gamma}(#1)} \newcommand{\SL}[1]{\mathbf{SL}_2(#1)} \newcommand{\PSL}[1]{\mathbf{PSL}_2(#1)} \newcommand{\GL}[1]{\mathbf{GL}_2(#1)} \newcommand{\V}[2]{\bigl[\begin{smallmatrix}#1 \\#2\end{smallmatrix}\bigr]} \newcommand{\z}{\zeta} \renewcommand{\hom}[2]{\mathrm{Hom}_k\left(#1,#2\right)} \DeclareMathOperator{\deck}{Deck} \DeclareMathOperator{\aut}{Aut} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\gal}{Gal} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\sol}{Sol} \DeclareMathOperator{\rad}{rad} \title{Math 631, Lecture 2} \author{Notes by Bryden Cais} \date{Sept. 9, 2002} \begin{document} \maketitle Let $S,S^{'}\subset k[T_1,\ldots,T_n]$ be systems of algebraic equations. Recall that we say that $S\simeq S^{'}$ if $\sol(S;K)=\sol(S^{'};K)$ for all $k$-algebras $K$. \begin{defn} An \textbf{affine algebraic variety} $X$ over $k$, or equivalently a $k$-variety is the set of solutions of any representative of an equivalence class of systems of algebraic equations. Note that this is a good definition since it does not depend on the representative chosen. We denote by $X(K)=X\cap K^n=\sol(S;K)$ for any representative $S$ of the equivalence class. \end{defn} \begin{exam} Let $S=\{0\}.$ then $X=\A^n_k$ is affine $n$-space (over $k$). Clearly, $\A^n_k(K)=K^n$. \end{exam} \begin{exam} Let $S=\{1\}.$ then $X=\emptyset_k$ and $\emptyset_k(K)=\emptyset$. \end{exam} \begin{defn} Let $S$ be an algebraic set of equations and let $X(K)=\sol((S);K).$ Then $(S)$ is the \textbf{defining ideal} of $X$. Observe that this is also well defined since we proved last time that $s\simeq S^{'}$ if and only if $(S)=(S')$. \end{defn} Notice that we are heavily using the fact that we can take $K$ to be \textit{any} $k$-algebra. For example, $\sol(\{X\};K)=\sol(\{X^e\};K)$ for any \textit{field} $K$, but if we take $K$ to be the $k$-algebra $k[X]/x^e$ then we see that $X^e=0$ but $X\not=0,$ so the varieties above are different. \begin{thm}[Hilbert's Basis Theorem] Any Ideal $I\subset k[T_1,\ldots,T_n]$ is finitely generated. \end{thm} \begin{proof} We will induct on $n$. For $n=1$, the Theorem is clear since for any field $k$, the ring $k[T_1]$ is a principal ideal domain. Now suppose that the theorem is true for $n-1$ generators and let $I\subset k[T_1,\ldots,T_n]$. For any $F\in I,$ we write $F=b_0T_n^r+\cdots+b_n,$ where $b_i\in k[T_1,\ldots,T_{n-1}]$. We call $b_0$ the \textit{leading coefficient} of $F$ and say that $F$ is of degree $r$ in $T_n$. Now let $J_r\subset k[T_1,\ldots, T_{n-1}]$ be the set consisting of all leading coefficients of polynomials in $I$ of degree $r$ in $T_n$. It is easy to see that $J_r$ is an ideal: since $I$ is an ideal, we can multiply any degree $r$ polynomial (in $T_n$) by \textit{any} polynomial in $k[T_1,\ldots,T_n]$ and obtain another polynomial of degree $r$ in $T_n$ contained in $I$; the leading coefficient of this polynomial is therefore in $J_r$. Now observe that we can multiply any degree $r$ in $T_n$ polynomial in $I$ by $T_n$ to obtain a degree $r+1$ in $T_n$ polynomial still contained in $I$ (again since $I$ is an ideal). Thus, we see that $J_r\subset J_{r+1}$. We therefore define the ideal $$J=\bigcup_{0\leq r} J_r\subset k[T_1,\ldots,T_n].$$ Now by our induction hypothesis, the ideals $J_r$ and $J$ are finitely generated. Let $(a_{1r},\ldots,a_{m(r)r})$ be a set of generators for $J_r$ and $(a_1,\ldots,a_s)$ a set of generators of $J$. Moreover, since we certainly have $a_{ir}\in J_r$ for each $1\leq i\leq m(r)$, there exist polynomials $F_{ir}\in I$ with leading coefficient $a_{ir}$ (this follows from the definition of $J_r$). Similarly, for each $1\leq i\leq s$ there exists a polynomial $F_i\in I$ with leading coefficient $a_i$. Let the degree of $F_i$ be $d(i)$ and set $$N=\max\{d(1),\ldots,d(s)\}.$$ We claim that $I$ is generated by $$S=\left\{F_{ir},F_j:1\leq i\leq m(r),0\leq r\leq N,1\leq j\leq s \right\}.$$ We will show that any polynomial of degree $d>N$ in $T_n$ may be reduced to a polynomial of degree at most $N$ in $T_N$, and then that any such polynomial can be obtained from the $J_{ir}$. So let $G\in I$ be of degree $d>N$. Certainly, the leading coefficient $l_d$ of $G$ is in $J$ (in fact in $J_d$), so that we may write $$l_d=c_1a_1+c_2a_2+\ldots+c_sa_s$$ for $c_j\in k[T_1,\ldots,T_{n-1}]$. Now consider the polynomial $$H=G-((c_1T^{d-d(1)})F_1+(c_2T^{d-d(2)})F_2+\ldots+(c_sT^{d-d(s)})F_s).$$ Clearly $H_1\in I$ since $F_i\in I$ and $I$ is an ideal of $k[T_1,\ldots,T_n].$ Moreover, it is clear that $H$ has degree at most $n-1$. We can therefore proceed in this manner until we obtain a polynomial $H_t$ of degree at most $N$. It now suffices to show that this $H_t$ is in the ideal generated by $S$. Observe that we can use the $F_{ir}$ to kill the leading coefficient of $H_t,$ just as above, since $a_{ir}$ for a fixed $r$ generate \textit{all} leading coefficients of polynomials of $I$ of degree $r$ in $T_n$. We proceed in this way until we obtain a polynomial of degree $0$ in $T_n$, which we can obtain from the $F_{i0}'s$. This, $G$ is in the ideal of $k[T_1,\ldots,T_n]$ generated by $S$. Since $G$ was arbitrary, we are done, and $I$ is generated by $S$. \end{proof} Now suppose that $K$ is an algebraically closed field. Observe that in this case we do not have in general that $\sol(S;K)=\sol(S^{'};K)$ if and only if $(S)=(S^{'})$. For example, the ideals generated by $S=X$ and $S^{'}=X^2$ are not equal, but over any field, $\sol(X;K)=\sol(X^2;K)=\{0\}.$ However, as the following theorem tells us, this is essentially the only way in which two different ideals can correspond to the same variety. \begin{thm}[Hilbert's Nullstellensatz] For any commutative ring $A$ and any ideal $I\subset A$ define the \textbf{radical} of $I$ to be the set $$\rad(I)=\left\{x\in A:x^n\in I\quad\text{for some } n\geq 1 \right\}.$$ It is not difficult to see that $\rad(I)$ is an ideal: If $x\in \rad(I)$ and $a\in A$ then $(ax)^n=a^nx^n\in AI\subset I$ since $A$ is commutative and $I$ is an ideal. Let $K$ be algebraically closed. Then $$\sol((S);K)=\sol((S^{'});K)$$ if and only if $$\rad((S))=\rad((S^{'})).$$ \end{thm} We first prove two crucial lemmata: \begin{lem}\label{algext} Let $A$ be a finitely generated algebra over a field $k$. Then if $A$ is a field, $A$ is an algebraic extension of $k$. \end{lem} \begin{proof} Let $A$ be generated by $\{t_1,\ldots,t_r\}$. We induct on $r$. For $r=1$ we have a surjective map $$k[T]\rightarrow A$$ given by $T\mapsto t_1$ (this is just what it means for $A$ to be a $k$ algebra). CLearly the map is not injective since $k[T]$ is not a field, so it has some kernel $I$. Since $k[T]$ is a PID, $I=(F)$ for some $f\in k[T]$. We then have $A\simeq k[T]/(F)$, which realizes $A$ as an algebraic extension of $k$ (by adjoining a root of $F$). Now suppose that $r>1$ and that the lemma is false for $A$: that is, $A$ is not algebraic over $k$ so that some $t_i$, say WLOG $t_1$ is transcendental over $k$. Now observe that since $A$ is supposed a field, it contains $k(t_1)$. Moreover, $A$ is generated over $k(t_1)$ by $t_2,\ldots,t_r,$ so by induction we have that $A$ is an algebraic field extension of $k(t_1)$. In particular, each $t_i$ for $i\neq 1$ is algebraic over $k(t_1)$ and so satisfies a polynomial of the form $$a_i(d(i))t_i^{d(i)}+a_i(d(i)-1)t_i^{d(i)-1}+\ldots,$$ where $a_i\in k(t_1)$. For each $i$ we can clear denominators of the $a_i(m)$ and assume that $t_i$ satisfies a polynomial with coefficients in $k[t_1]$, say $$b_i(d(i))t_i^{d(i)}+b_i(d(i)-1)t_i^{d(i)-1}+\ldots.$$ By multiplying this polynomial through by $b_i(d(i))^{d(i)-1}$, we see that $b_i(d(i))t_i$ satisfies a monic polynomial over $k[t_1]$. Notice that since the $t_i$ generate $A$ over $k(t_1)$ so do the $b_i(d(i))t_i$ since each $b_i(d(i))\in k[t_1]$. Therefore, $A$ is generated by \textit{integral} elements over $k[t_1]$, which implies that every element of $A$ is integral over $k[t_1]$. However, $k(t_1)\subset A$ so in particular, every element of $k(t_1)$ is integral over $k[t_1]$. But now $k[t_1]$ is isomorphic to the polynomial ring $k[T]$ in one variable since $t_1$ is transcendental, whence the previous statement is absurd since, for example, $1/T$ is not integral over $k[T]$. We have therefore reached a contradiction and $A$ is in fact an algebraic extension of $k$. \end{proof} \begin{lem} Suppose that $\sol((S);K)=\emptyset$. Then $1\in (S)$. \end{lem} \begin{proof} First, observe that given any maximal ideal of $k[T_1,\ldots, T_n]/(S)$ we can construct a point of $V=\sol((S);K)$. To see this, observe that a maximal ideal $M$ of $k[T]/(S)$ gives us a homomorphism of $k[T]/(S)$ into some field $F$ via the natural projection $k[T]/(S)\longrightarrow (k[T]/(S))/M$. By Lemma \ref{algext}, $F$ is an algebraic extension of $k$ since it is finitely generated, and hence contained in $K$ since $K$ is algebraically closed. But from last lecture, we know that $\hom{K[T/(S)]}{F}$ is in bijective correspondence with $\sol((S);F)$. By hypothesis, $\sol((S);K)$ is empty, so $\sol((S);F)$ must be empty also so that there are no maximal ideals of $k[T]/(S)$. On the other hand, maximal ideals of $k[T]/(S)$ are in bijective correspondence with maximal ideals of $k[T]$ containing $(S)$; in short, we have shown that $(S)$ is not contained in any maximal ideal. However, the Hilbert Basis Theorem can be used to show fairly easily that \textit{every} ideal except for the unit ideal is contained in a maximal ideal. Therefore, $(S)$ must be the unit ideal. \end{proof} We now prove the main Theorem. \begin{proof} Given any algebraic $k$-set $V\subset K^n$ we define $$I(V)=\left\{P\in k[T_1,\ldots, T_n]:V\subset \sol(P;K)\right\},$$ that is, $I(V)$ is the set of all polynomials in $k[T_1,\ldots,T_n]$ that vanish on $V$. It is not hard to see that $I(V)$ is an ideal. In one direction, suppose that $\rad((S))=\rad((S^{'}))$ and let $a\in\sol(S;K)$. Then for any $F\in S,$ we have $F(a)=0$. Let $G\in (S^{'}).$ Then we must show that $G(a)=0$. Observe that since $G\in (S^{'})$ we clearly have $G\in\rad((S^{'}))$ and hence $G\in\rad((S))$; in particular $G^r\in (S)$ for some $r$. We therefore see that $G^r(a)=0$. But since $K$ is a field, this implies $G(a)=0$ and hence $a\in\sol((S^{'});K),$ as required. In the other direction, it suffices to show that if $V=\sol((S);K)$ then $\rad((S))=I(V).$ One direction is clear since every polynomial in $\rad((S))$ vanishes on $V$ and is therefore in $I(V)$. By Hilbert's Basis Theorem, we have $(S)=(F_1,\ldots,F_s)$ for some $F_i\in k[T_1,\ldots,T_n]$. We must therefore show that if $F$ vanishes on $V$ then we can write $$F^N=\sum A_iF_i$$ for some $A_i\in k[T_1,\ldots,T_n]$. Since $F$ vanishes on $V$ and $F_i$ for $1\leq i\leq s$ generate $(S)$ we see that the polynomials $F_1,\ldots,F_s,FT_{n+1}-1$ in $n+1$ variables have no common zeroes (since any common zero is a common zero of $F_1,\ldots,F_n$ and hence in $V$ so that $FT_{n+1}-1=-1$. Therefore, by our lemma we can write $$1=\sum_i p_i(T_1,\ldots,T_{n+1})F_i(T_1,\ldots,T_{n})+q(T_1,\ldots,T_{n+1})(F(T_1,\ldots,T_n)T_{n+1}-1).$$ Now let $T_{n+1}=1/F$ so that $$1=\sum_i p_i(T_1,\ldots,T_n,1/F)F_i(T_1,\ldots,T_{n}).$$ Since each $p_i$ is a polynomial, there exists $N(i)>0$ such that $F^{N(i)}p_i(T_1,\ldots,T_n,1/F)$ is a polynomial in $k[T_1,\ldots,T_n]$. Set $N=\max_i\{N(i)\}$. then we have $$F^N=\sum_i A_iF_i,$$ where $A_i=F^N p_i(T_1,\ldots,T_n,1/F)\in k[T].$ Hence $F\in\rad((S)),$ as required. \end{proof} \end{document}