A Universe of Sorts

§ Weak and Strong Nullstllensatz

§ Weak Nulstellensatz: On the tin

For every maximal ideal

m \subset k[T_1, \dots, T_n]

there is a unique

a \in k^n

such that

m = I(\{ a \})

. This says that any maximal ideal is the ideal of some point.

§ Weak Nullstellensatz: implication 1 (Solutions)

every ideal, since it is contained in a maximal ideal, will have zeroes. Zeroes will always exist in all ideas upto the maximal ideal.

It simply says that for all ideals $J$ in $\mathbb C[X_1, \dots, X_n]$ , we have $I(V(J)) = sqrt J$
Corollary: $I$ and $V$ are mutual inverses of inclusions between algebraic sets and radical ideals.

§ Weak Nullstellensatz: Implication 2 (Non-solutions)

If an ideal does not have zeroes, then it must be the full ring. Hence, 1 must be in this ideal. So if

I = (f_1, f_2, \dots, f_n)

and the system has no solutions, then

I

cannot be included in any maximal ideal, hence

I = \mathbb C[X_1, \dots, X_n]

. Thus,

1 \in I

, and there exist

c_i \in \mathbb C[X_1, \dots, X_n]

such that

1 = sum_i f_i c_i

§ Strong Nullstellensatz: On the Tin

For every ideal

J

, we have that

I(V(J)) = |_0^\infty\sqrt J

. I am adopting the radical (heh) notation

|_0^\infty \sqrt x

for the radical, because this matches my intuition of what the radical is doing: it's taking all roots , not just square roots . For example,

\sqrt{(8)} = (2)

\mathbb Z

§ Strong Nullstellensatz: Implication 1 (solutions)

Let

J = (f_1, \dots, f_m)

. If

g

is zero on

V(J)

, then

g \in \sqrt J

. Unwrapping this,

\exist r \in \mathbb N, \exists c_i \in \mathbb C[X_1, \dots, X_n], \sum_i f_i c_i = g^r

§ Weak Nullstellensatz: Proof

Let $m$ be a maximal ideal.
Let $K$ be the quotient ring $K \equiv \mathbb C[X_1, \dots, X_n] / m$ .
See that $K$ is a field because it is a ring quotiented by a maximal ideal.
Consider the map $\alpha: \mathbb C[X_1, \dots, X_n] \rightarrow K$ , or $\alpha : \mathbb C [X_1, \dots, X_n] \rightarrow \mathbb C[X_1, \dots, X_n] / m$ by sending elements into the quotient.
We will show that $\alpha$ is an evaluation map, and $K = \mathbb C$ . So we will get a function that evaluates polynomials at a given point, which will have a single point as a solution.
Core idea: See that $\alpha(\mathbb C) = \mathbb C \subset K$ . Hence $K$ is a field that contains $\mathbb C$ . But $\mathbb C$ is algebraically closed, hence $K = C$ .
First see that $\mathbb C \subset K$ , or that $\alpha$ preserves $\mathbb C$ [ie, $\alpha(\mathbb C) = \mathbb C$ ]. note that no complex number can be in $m$ . If we had a complex number $z$ in $m$ , then we would need to have $1 = 1/z \cdot z$ in $m$ (since an ideal is closed under multiplication by the full ring), which means $1 \in m$ , due to which we get $m$ is the full ring. This can't be the case because $m$ is a proper maximal ideal.
Hence, we have $\mathbb C \subseteq K$ or $K = \mathbb C$ .
Thus the map we have is $\alpha: \mathbb C[X_1, X_2, \dots, X_n] \rightarrow \mathbb C$ .
Define $z_i = \alpha(X_i)$ . Now we get that $\alpha(\sum_{ij} a_{ij} X_i^j) = \sum_{ij} a_{ij} z_i^j$ . That is, we have an evaluation map that sends $X_i \mapsto z_i$ .
CLAIM: The kernel of an evaluation map $\alpha$ is of the form $(X_1 - z_1, \dots, X_n - z_n)$ .
PROOF OF CLAIM:TODO
The kernel is also $m$ . Hence, $m = (X_1 - z_1, \dots, X_n - z_n)$ , and point that corresponds to the maximal ideal is $(z_1, z_2, \dots, z_n)$ .

§ Strong Nullstellensatz: Proof

We use the Rabinowitsch trick .

Suppose that wherever $f_1, \dots, f_m$ simultaneously vanish, then so does $g$ . [that is, $g \in I(V(J))$ where $J = (f_1, \dots, f_m)$ ].
Then the polynomials $f_1, \dots, f_m, 1 - Yg$ have no common zeros where $Y$ is a new variable into the ring.
Core idea of why they can't have common zeros: Contradiction. assume that $1 - Yg$ , and all the $f_i$ vanish at some point. Then we need $1 - Yg = 0$ which mean $Y = 1/g$ , so $g$ cannot vanish, so $g \neq 0$ . However, since all the $f_i$ vanish, $g$ also vanishes as $g \in (V(J))$ . This is contradiction.
Now by weak Nullstellensatz, the ideal $J = (f_1, \dots, f_m, (1-Y)g)$ cannot be contained in a maximal ideal (for then they would simultaneously vanish). Thus, $J = R$ and $1 \in J$ .
This means there are coefficients $c_i(Y, \vec x) \mathbb C[X_1, \dots , X_n, Y]$ such that

1 = c_0(Y, \vec x) (1 - Yg(\vec x)) \sum_{i=1}^m c_i(Y, \vec x) f_i(\vec x)

Since this holds when

\vec x, Y

are arbitrary variables, it continues to hold on substituting

Y = 1/g

, the coefficient

c_0(1-Yg) = c_0(1 - g/g) = c_0(1 - 1) = 0

disappears. This gives:

1 = \sum_{i=1}^m c_i (Y, \vec x) f_i(\vec x)

since

Y = 1/g

, we can write

c_i(Y=1/g, \vec x) = n_i(\vec x)/g^r_i(\vec x)

. By clearing denominators, we get:

1 = \sum{i=1}^m n_i(\vec x) f_i(\vec x)/ g^R(\vec x)

This means that

g^R(\vec x) = \sum_{i=1}^m n_i(\vec x) f_i(\vec x)

§ Strong Nullstellensatz: algebraic proof

We have $g \in I(V(J))$ .
We want to show that $g \in \sqrt{J}$ in $R$ .
This is the same as showing that $g \in \sqrt{0}$ in $R/J$ . ( $J \mapsto 0$ in the quotient ring).
If $g$ is nilpotent in $R/J$ , then the $(R/J)_g$ becomes the trivial ring $\{ 0 \}$ . [Intuitively, if $g$ is nilpotent and a unit, then we will have $g^n = 0$ , that is unit raised to some power is 0, from which we can derive $1 = 0$ ].
Localising at $g$ is the same as computing $R[Y]/(1 - Yg, J)$ .
But we have that $V(1 - Yg, J) = \emptyset$ . Weak Nullstellensatz implies that $(1 - Yg, J) = (1)$ .
This means that $R[Y]/(1 - Yg,J) = R[Y]/(1) = \{ 0 \}$ . Thus, $(R/J)_g$ has $g$ as nilpotent, or $g \in \sqrt J$ in $R$ .

§ Relationship between strong and weak

Strong lets us establish what functions vanish on a variety . Weak let us establish what functions vanish at a point .

§ Strong Nullstellensatz in scheme theory

Same statement: $I(V(J)) = \sqrt J$ .
$V(J)$ is the set of points on which $J$ vanihes. Evaluation is quotienting. So it's going to be set of prime ideals $p$ such that $J \xrightarrow{R/p} 0$ . So $J \subset p$ . This means that $V(J) = \{ p \text{prime ideal in } R, J \subseteq p \}$ .
$I(V(J))$ is the set of functions that vanish over every point in $V(J)$ . The functions that vanish at $p \in V(J)$ are the elements of $p$ . So the functions that vanish over all points is $I(V(J)) = \cap V(J)$ .
Unwrapping, this means that $I(V(J))$ is the intersection of all ideals in $V(J)$ , which is the intersection of all primes that contains $J$ , which is the radical of $J$ .

Holy shit, scheme theory really does convert Nullstellensatz-the-proof into Nullstellensatz-the-definition! I'd never realised this before, but this.. is crazy. Not only do we get easier proofs, we also get more power! We can reason about generic points such as

(x)

(y)

which don't exist in variety-land. This is really really cool.

Reference video