# Wednesday April 15th

Recall that we had mutually inverse maps $V, I$ and for $J\normal R$, we had a closure $\bar J = I(V(J))$ and $\bar S = V(I(S))$.

Hilbert's Nullstellensatz says that if $k = \bar k$, then for all $J$, we have $\bar J = \rad J$ and there is a Galois correspondence between radical ideals of $k[x_1, \cdots, x_n]$ and Zariski closed subset of $\AA_k^n$.
The weak Nullstellensatz gives a correspondence between maximal ideals and points of $k^n$.

For $S = \bar S = k^n$, we have $S = \union_{x\in S} \theset{x}$ and by the Galois correspondence, we find that if $J$ is radical then $J = \intersect{\mfm \supset J} \mfm$.

Definition (Jacobson)
: A ring $R$ is *Jacobson* if every radical ideal is the intersection of the maximal ideals containing it, i.e. for all $I\normal R$ we have $\rad I = \intersect_{\mfm \supset I} \mfm$

Exercise
:   For $x\in k^n$, define $$\mfm_x \definedas I(\theset{x}) = \theset{f\in R \suchthat f(x) = 0}.$$

    1. Show that $\mfm = \gens{t_1 - x_1, \cdots, t_n - x_n}$ and $R/\mfm_x = k$.
    2. Show that the following map is injective:
    \begin{align*}
    I: k^n &\to \maxspec R \\
    x &\mapsto \mfm_x
    .\end{align*}
    3. Show that $I(k^n) = \theset{\mfm \in \maxspec R \suchthat R/\mfm = k}$.
    
    > Hint: consider the images of $t_1, \cdots, t_n$ in $R/\mfm$.


The following result will imply the weak nullstellensatz:

Lemma (Zariski, 1947)
:   Let $k$ be a field, $R$ a finitely generated $k\dash$algebra, $\mfm \in \maxspec R$.
    Then $[R/\mfm : k] < \infty$.
    Equivalently, if $K/k$ is a field extension that is finitely generated as a $k\dash$algebra, then $K$ is finitely generated as a $k\dash$vector space.

Proof
:   \hfill
    Case 1: $K = k(\alpha_1, \cdots, \alpha_n)$ is algebraic.
    
    \
    In this case you get the obvious tower of extensions where each extension is finite, so $[K: k]$ is finite.
    
    \
    
    Case 2: $K/k$ is not algebraic.
    
    \ 
    
    We can choose a transcendence basis $t_1, \cdots, t_n$, and since $K$ is finitely generated as a field extension over $k$, the transcendence degree is finite.
    Then $k \subset k(t_1, \cdots, t_n) \subset K$ and $K$ is algebraic over $k(t_1, \cdots, t_n)$, so $[K: k(t_1, \cdots, t_n)] < \infty$.
    By the Artin-Tate lemma, $k(t_1, \cdots, t_n)/k$ is a finitely generated $k\dash$algebra.
    But then $k(t_1, \cdots, t_n)$ is finitely generated over $k(t_1, \cdots, t_{n-1})$, so it suffices to consider the case of one variable.
    In other words, we need to show that for all $k$, $k(t)$ is not a finitely generated $k\dash$algebra.
    \

    Supposing otherwise, let $\theset{r_i(t)}$ be a finite set of rational functions that are generators.
    Factor each $r_i$ as $f_i/g_i$
    Since $k[t]$ is a PID with infinitely many primes, i.e. infinitely many nonassociate irreducible polynomials.
    \
    
    So choose $q$ monic, irreducible, not equal to any of the $f_i$; then $1 \over q \not\in k[f_1, \cdots, f_n]$, a contradiction.

So rational function fields are not finitely generated over their base fields, or even finitely generated over their polynomial rings.

Exercise
: Show that a PID is Jacobson iff $\spec R$ is infinite.

Exercise
: Show that $R$ is Jacobson iff for every $\mfp \in \spec R$, we have $\mfp = \intersect_{\mfp \subset \mfm \in \maxspec R} \mfm$.

Proposition (Rabinovitch? Trick)
: For $k$ a field and $n\in \ZZ^{+}$, $k[t_1, \cdots, t_n]$ is a Jacobson ring.

To see why the weak Nullstellensatz and this trick imply the Nullstellensatz, we show that for $J\normal R = k[t_1, \cdots, t_n]$, 
$$
\rad J = \bar J = I(V(J)) = I(V(\rad J)) = \bar{\rad J}
,$$ 
so wlog we can assume $J$ is radical and show that $\bar J = J$.

Taking $J$ a radical ideal, 
$$
J = \intersect_{\mfp \subset J} \mfp \equalsbecause{RT} \intersect_{\mfm \supset J} \mfm \equalsbecause{WN} \intersect_{a\in k^n, \mfm_a \supset J} \mfm_a
.$$

Then $J \subset \mfm_a$ iff for all $f\in J$, $f(a) = 0$, i.e. 
$$
a \in V(J) = \intersect_{a\in V(J)} I(\theset{a}) = I(V(J)) = \bar J
.$$

Next time: proof of RT.