# Monday April 13th

## Nullstellensatz

Let $k$ be a ring (later a field) and $R_n \definedas k[t_1, \cdots, t_n]$.
If $x \in k^n$ and $f\in R_n$, we can evaluate $f$ at $x$ since $f(x) \in k$, and we thus get an evaluation map
\begin{align*}
E: R_n &\to k^{k^n}\\
f &\mapsto (x\mapsto f(x))
\end{align*}

which is a ring homomorphism, where the RHS is regarded as a large direct product of rings.

When can we identify polynomials (abstract elements of a ring) with the corresponding polynomial *function*?

Exercise
:   Suppose $k$ is a domain.

    a. If $k$ is infinite, then $E$ is injective but not surjective.
    b. If $k$ is finite, the $E$ is surjective but not injective.

Definition (Incidence Relation)
: Put a relation on $R \cross k^n$ where $f\sim x \iff f(x) = 0_R$.

We'll define an antitone Galois connection
\begin{align*}
\mci(R) \stackrel[I]{V}{\tofrom} 2^{k^n}
\end{align*}

where the RHS is the powerset.
We define $V$ which takes an ideal and yields a subset of $k^n$,
\begin{align*}
V(J) &= \theset{x\in k^n \suchthat \forall f\in J,~ f(x) = 0} \\
I(S) &= \theset{f\in R \suchthat \forall x\in S,~ f(x) = 0}
.\end{align*}

> That this forms an ideal is immediate.

The pair $(I, V)$ are antitone (order-reversing) and fits into the (extremely general formal) framework of Galois connections.
This can be seen by writing $V(J) = \intersect_{f\in J} V(f)$ and $I(S) = \intersect_{x\in S} I(\theset{x} )$.
Therefore (by the formalism) there are associated closure operators:

1. $2^{k^n} \selfmap, S \mapsto \bar S \definedas V(I(S))$
2. $\mci(R) \selfmap, J\mapsto \bar J \definedas I(V(J))$.

These are both *isotone* maps (order-preserving) and are closure operators on posets, i.e. they satisfy

- $(X, \leq) \mapsto (X, \leq), x\leq y\implies C(x) \leq C(y)$ (isotonicity)
- $x\leq C(x)$
- $C(C(x)) = C()$ (idempotence)

By construction, $\bar S = \theset{x\in k^n \suchthat f(s) = 0 \implies f(x) = 0}$ and $\bar J = \theset{f\in k^n \suchthat g(x) = 0 \forall g\in S \implies f(x) = 0 }$.

Example:
Take $k$ a field, $S \subset k$.
Then 
\begin{align*}
\bar S = 
\begin{cases}
S & S \text{ is finite} \\
k & \text{else}
\end{cases}
.\end{align*}

Exercise
: For $k$ a field, show that for all $S_1, S_2 \subset k$, we have $\bar{S_1 \union S_2} = \bar S_1 \union \bar S_2$.

> This mirrors what closures in a topological space would do, so $S \to \bar S$ is in fact a *Kuratowski* closure operator and therefore is the closure operator for a unique topology (here, the Zariski topology, where the closed subsets are given by $V(I)$).
> For $n=1$, the Zariski topology on $k$ is the cofinite topology.
> If $k$ is finite, this is discrete, and is very coarse and non-Hausdorff when $k$ is infinite since any two open sets intersect.

These maps satisfy a *tridempotence* relation, i.e. $VIV =V$ and $IVI = I$, and are antitone *bijections* when restricted to closed sets.

Exercise
: The closure operator on ideals never satisfies the Kuratowski property.

So we don't have a topology on the ring/ideal side, since even unions of ideals may fail to be ideals.
We take the description sending $S\to \bar S$ to be fairly explicit, but what is the corresponding description for $J\to \bar J$?

For all $S\in k^n$, $I(S) \normal k[t_1, \cdots, t_n]$, we use the fact that $f^n(x) = 0 \implies f(x) = 0$ and thus $I(S) = \rad I(S) = \intersect_{\mfp \supset I(S)} \mfp$, and $V(J) = V(\rad J)$.
Moreover, for all $J \normal R$, we have $\rad J \subset \bar J$.

Passing from an ideal to its radical is a closure operator, so does $\bar J = \rad J$ for all $J\normal R$?
Not for all $k$, e.g. take $k = \RR, n= 1, J = \gens{t^2 + 1}$.
Then $V(J) = \emptyset$, but $\bar J = I(\emptyset) = \RR[t]$ vacuously.
Note that $\RR/J \cong \CC$, so $J$ is maximal and hence radical, so $\bar J \supsetneq \rad J = J$.

What went wrong? Potentially $\bar k\neq k$.

Exercise
: Let $k$ be a field, fix $n \in \ZZ^+$, and show that if all $\mfm \in \maxspec(R)$ satisfy $\bar \mfm = \rad \mfm$, then $L$ is algebraically closed.

Exercise
: Suppose $k = \bar k$ is algebraically closed and $n=1$, show that for all $J\normal k(t)$, we have $\bar J = \rad J$.

Theorem (Hilbert's Nullstellensatz)
:   If $k = \bar k$  and for all $n\in \ZZ^+$ and $J \normal R = k[t_1, \cdots, t_n]$, $\bar J = I(V(J)) = \rad J$
    Therefore $I, V$ induce a galois correspondence
    \begin{align*}
    \correspond{\text{Radical ideals} \\ \text{of } k[t_1, \cdots, t_n]} \iff \correspond{\text{Zariski closed subets} \\ \text{of } k^n}
    .\end{align*}

Exercise
:   If $k = \bar k$, show that the above theorem implies the *weak Nullstellensatz*, i.e. the map   
    
    \begin{align*}
    I: k^n &\mapsvia{\cong} \maxspec k[t_1, \cdots, t_n]  \\
    \vector x &\mapsto \mfm_x = I(\theset {\vector x}) = \gens{t_1 - x_1, \cdots, t_n - x_n}
    \end{align*}
    
    is a bijection.

> Try without using the formalism of Galois connections.

Proof sketch: we'll spend most of our time proving the weak Nullstellensatz, and use a trick to bootstrap up to the full theorem.
The closure operator is only the radical for algebraically closed -- what is the closure operator for other fields.