# The Nullstellensatz (Tuesday, August 25)

## Radicals, Degrees, and Affine Varieties

Given $f\in k[x_1, \cdots, x_n]$, we'll denote by $f(a)$ the value of $f$ at the point $(a_1, \cdots, a_n)$.
Let $k = \bar k$ and $R$ a ring containing ideals $I, J$.
Recall the definition of the *radical* of an ideal:

:::{.definition title="Radical"}
The *radical* of an ideal $I \normal R$ is defined as
\[  
\sqrt{I} = \ts{r\in R \st r^k\in I \text{ for some } k\in \NN}
.\]
:::

:::{.example}
Let 
\[
I &= (x_1, x_2^2) \subset \CC[x_1, x_2] \\
  &= \ts{ f_1 x_1 + f_2 x_2 \st f_1, f_2 \in \CC[x_1, x_2]}
\]

Then $\sqrt{I} = (x_1, x_2)$, since $x_2^2 \in I \implies x_2 \in \sqrt{I}$.
:::


:::{.definition title="Degree of an element of {$k[x_1, x_2, \cdots, x_n]$}"}
Define $\deg(f)$ as the largest value of $i_1 + \cdots + i_n$ such that the coefficient of $\prod x_j ^{i_j}$ is nonzero.
:::

:::{.example}
$\deg(x_1 + x_2^2 + x_1 x_2^3) = 4$
:::

:::{.definition title="Affine Variety"}
\envlist

1. Affine $n\dash$space[^affine_variety_remark] $\AA^n = \AA_k^n$ is defined as 
\[
\AA^n \da \theset{(a_1, \cdots, a_n) \suchthat a_i \in k}
\]

2.  Let $S\subset k[x_1, \cdots, x_n]$ be a **set** of polynomials.[^not_nec_ideal]
    Then define the **affine variety** of $S$ as 
  \[
  V(S) \da \ts{x\in \AA^n \st f(x) = 0} \subset \AA^n
  \] 

[^affine_variety_remark]: 
Not $k^n$, since we won't necessarily use the vector space structure (e.g. adding points).


[^not_nec_ideal]: 
We don't necessarily require $S$ to be an ideal in this definition.
We will shortly show that taking the ideal it generates yields the same variety.

:::

:::{.example title="Examples of affine varieties"}
\envlist

- Let $f(x) = 0$, then $\AA^n = V\qty{\ts{f}}$ is an affine variety.
- Any point $(a_1, \cdots, a_n)\in \AA^n$ is an affine variety, uniquely determined by 
\[
V(x_1 - a_1, \cdots, x_n - a_n) = \theset{a_1, \cdots, a_n}
\]
- For any finite set $r_1, \cdots, r_k \in \AA^1$, there exists a polynomial $f\in k[x_1]$ whose roots are $r_i$.
:::

:::{.remark}
We may as well assume $S$ is an ideal by taking the ideal it generates, 
\[
S\subseteq \gens{S} = \ts{\sum g_i f_i \suchthat g_i \in k[x_1, \cdots, x_n],\, f_i\in S}
.\]

:::{.claim}
\[  
V(S) = V\qty{\gens{S}}
.\]
:::

It's clear that $V(\gens{S}) \subset V(S)$.
Conversely, if $f_1, f_2$ vanish at $x\in \AA^n$, then $f_1 + f_2$ and $gf_1$ also vanish at $x$ for all $g\in k[x_1, \cdots, x_n]$.
Thus $V(S) \subset V(\gens{S})$.
:::

## Ideals and Properties of $V(\wait)$

See \cref{useful-algebra-facts} for a review of properties of ideals.

:::{.proposition title="Properties of $V$"}
\[  
S_1 \subseteq S_2 &\implies V(S_1) \supseteq V(S_2) \tag{1}\\
V(S_1) \union V(S_2) &= V(S_1 S_2) = V(S_1 \intersect S_2) \tag{2} \\
\Intersect V(S_i) &= V\qty{\Union  S_i} \tag{3}
.\]
:::

We thus have a map
\[  
V: \ts{\text{Ideals in } k[x_1, \cdots, x_n]} \to \ts{\text{Affine varieties in } \AA^n_{/k}}
.\]

:::{.definition title="The Ideal of a Set"}
Let $X\subset \AA^n$ be any set, then *the ideal of $X$* is defined as
\[  
I(X) \da\ts{f\in k[x_1, \cdots, x_n] \st f(x) = 0\, \forall x\in X}
.\]
:::

:::{.example}
Let $X$ be the union of the $x_1$ and $x_2$ axes in $\AA^2$, then 
\[
I(X) = \gens{x_1 x_2} = \ts {g x_1 x_2 \st g\in k[x_1, x_2]}
.\]
:::

:::{.proposition title="$I$ is inclusion-reversing"}
\[  
X_1 \subset X_2 \implies I(X_1) \supset I(X_2)
.\]
:::

:::{.proof title="?"}
If $f\in I(X_2)$, then $f(x) = 0$ for all $x\in X_2$. 
Since $X_1 \subset X_2$, we have $f(x) = 0$ for all $x\in X_1$, so $f\in I(X_2)$.
:::


:::{.proposition title="The Image of $I$ is Radical"}
\[  
I(X) = \sqrt{I(X)}
,\]
i.e. $I(X)$ is a radical ideal.
:::

:::{.proof title="?"}

It's clear that 
\[
I(X) \subset \sqrt{I(X)} \da \ts{f\in \kx{n} \st f^k \in I(X)}
\]
since we can simply take $k=1$ in this definition.

:::{.claim}
For a fixed $f\in \kx{n}$ and any $k\in \NN$, 
\[  
f(x)^k = 0\,\, \forall x\in X \implies f(x) = 0\,\, \forall x\in X
.\]
:::

Granting this claim, if $f\in \sqrt{I(X)}$ then $f^k \in I(X)$ and thus $f\in I(X)$, completing the proof.

:::{.proof title="?"}
Without loss of generality, we can take $n = 1$ and consider $\kx{n}$
Toward a contradiction, fix a $k$ suppose $f(x) \neq 0$ but $f(x)^k = 0$.
Then writing $f(x) = \sum_{j=1}^d \alpha_j x^j$ where $d \da \deg(f)$, we have $\alpha_d \neq 0$ and
\[  
f(x)^k = \alpha_d^k x^{dk} + \cdots
.\]
Equating coefficients, we have $\alpha_d^k = 0$ in the base field.
But fields have no nonzero nilpotents, so we arrive at a contradiction. 
:::

:::

These maps thus yield correspondences 
\[  
\correspond{\text{Ideals in } k[x_1, \cdots, x_n]} &\mapsvia{V} \correspond{\text{Affine Varieties}} \\
\correspond{\text{Radical Ideals}} &\fromvia{I} \correspond{\text{Affine Varieties}}
.\]

We'll find that if we restrict to radical ideals, this will yield a bijective correspondence.


## The Nullstellensatz: Statement and Proof

:::{.theorem title="Hilbert Nullstellensatz (Zero Locus Theorem)"}
\envlist

a. For any affine variety $X$, 
\[
V(I(X)) = X
.\]

b. For any ideal $J \subset k[x_1, \cdots, x_n]$, 
\[
I(V(J)) = \sqrt{J}
.\]

Thus there is a bijection between radical ideals and affine varieties.
:::

Recall the Hilbert Basis Theorem (\cref{thm:hilbert_basis}):
any ideal in a finitely generated polynomial ring over a field is again finitely generated.
We need to show 4 inclusions, 3 of which are easy.

:::{.proof title="$X \subset V(I(X))$ (a)"}
If $x\in X$ then $f(x) = 0$ for all $f\in I(X)$.
So $x\in V(I(X))$, since every $f\in I(X)$ vanishes at $x$.
:::

:::{.proof title="$\sqrt{J} \subset I(V(J))$ (b)"}
If $f\in \sqrt{J}$ then $f^k \in J$ for some $k$.
Then $f^k(x) = 0$ for all $x\in V(J)$.
So $f(x) = 0$ for all $x\in V(J)$.
Thus $f\in I(V(J))$.
:::

:::{.proof title="$V(I(X)) \subset X$ (c)"}
Need to now use that $X$ is an affine variety.
Counterexample: $X = \ZZ^2 \subset \CC^2$, then $I(X) = 0$.
But $V(I(X)) = \CC^2 \not\subset \ZZ^2$.
By (b), $I(V(J)) \supset \sqrt{J} \supset J$.
Since $V(\wait)$ is order-reversing, taking $V$ of both sides reverses the containment.
So $V(I(V(J))) \subset V(J)$, i.e. $V(I(X)) \subset X$.
:::

Thus the hard direction that remains is (d),

\[
I(V(J)) \subset \sqrt{J}
\]

:::{.warnings title="Hard theorem from commutative algebra."}
We'll need Noether Normalization (\cref{thm:noether_normalization}), which is perhaps more important than the Nullstellensatz!
:::

:::{.theorem title="1st Version of Nullstellensatz" ref="thm:nullstellensatz_one"}
Suppose $k$ is algebraically closed and uncountable[^countable_case].
Then the maximal ideals in $k[x_1, \cdots, x_n]$ are given by 
\[  
\mspec\qty{\kx{n}} = \ts{ \gens{x_1 - a_1, \cdots, x_n - a_n} \st a_j\in k}
.\]


[^countable_case]: 
Still true in countable case by a different proof.

:::

:::{.proof}
Let $\mfm$ be a maximal ideal, then by the Hilbert Basis Theorem (\cref{thm:hilbert_basis}), 
$\mfm = \gens{f_1, \cdots, f_r}$ is finitely generated.
Let $L = \QQ[\ts {c_i}]$ where the $c_i$ are all of the coefficients of the $f_i$ if $\ch(K) = 0$, **or** $\FF_p[\ts {c_i}]$ if $\ch(k) = p$.
Then $L\subset k$.
Define 
\[
\mfm_0 = \mfm\intersect L[x_1, \cdots, x_n]
\]
Note that by construction, $f_i \in \mfm_0$ for all $i$, and we can write $\mfm = \mfm_0 \cdot k[x_1, \cdots, x_n]$.

:::{.claim}
$\mfm_0$ is a maximal ideal.
:::

If it were the case that 
\[  
\mfm_0 \subsetneq \mfm_0' \subsetneq L[x_1, \cdots, x_n]
,\]
then
\[  
\mfm_0\cdot k[x_1, \cdots, x_n] \subsetneq \mfm_0'\cdot k[x_1, \cdots, x_n]  \subsetneq k[x_1, \cdots, x_n]
.\]

So far, we've constructed a smaller polynomial ring and a maximal ideal in it.
Thus $L[x_1, \cdots, x_n]/\mfm_0$ is a field that is finitely generated over either $\QQ$ or $\FF_p$.
So $L[x_1, \cdots, x_n]/\mfm_0$ is finite over some $\QQ(t_1, \cdots, t_n)$, and since $k$ is uncountable, there exists an embedding $\QQ(t_1, \cdots, t_n) \injects k$.[^countable_polynomials]
\

This extends to an embedding of $\phi: L[x_1, \cdots, x_n]/\mfm_0 \injects k$ since $k$ is algebraically closed.
Letting $a_i$ be the image of $x_i$ under $\phi$, then $f(a_1, \cdots, a_n) = 0$ by construction, $f_i \in (x_i - a_i)$ implies that $\mfm = (x_i - a_i)$ by maximality.

[^countable_polynomials]: 
Here we use the fact that there are only countably many polynomials over a countable field.

:::