## § 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, $\sqrt{(8)} = (2)$ in $\mathbb Z$.

#### § 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 $(x)$ or $(y)$ which don't exist in variety-land. This is really really cool.