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\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}
