Mathematics Lecture Notes

Theorem1.1aWeak Nullstellensatz

Let $k$ be an algebraically closed field. If $\mathfrak{m}$ is a maximal ideal of $k[x_1, \ldots, x_n]$ , then

$\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$

for some $(a_1, \ldots, a_n) \in k^n$ .

Equivalently: the points of $\mathbb{A}^n_k$ are in natural bijection with the maximal ideals of $k[x_1,\ldots,x_n]$ .

Theorem1.1bNullstellensatz (Ideal form)

Let $k = \bar{k}$ and let $\mathfrak{a} \subseteq k[x_1,\ldots,x_n]$ be a proper ideal. Then $V(\mathfrak{a}) \neq \varnothing$ .

In words: a system of polynomial equations over an algebraically closed field has no solution if and only if $1 \in \mathfrak{a}$ (i.e., the ideal is the whole ring).

Theorem1.1cStrong Nullstellensatz (Hilbert)

Let $k = \bar{k}$ . For any ideal $\mathfrak{a} \subseteq k[x_1,\ldots,x_n]$ :

$I(V(\mathfrak{a})) = \sqrt{\mathfrak{a}}$

where $\sqrt{\mathfrak{a}} = \{f \mid f^r \in \mathfrak{a} \text{ for some } r \geq 1\}$ is the radical of $\mathfrak{a}$ .

RemarkThe Galois correspondence

The strong Nullstellensatz establishes an inclusion-reversing bijection:

$\left\{\begin{array}{c}\text{radical ideals}\\ \mathfrak{a} \subseteq k[x_1,\ldots,x_n]\end{array}\right\} \xleftrightarrow{\;1:1\;} \left\{\begin{array}{c}\text{algebraic sets}\\ Y \subseteq \mathbb{A}^n\end{array}\right\}$

via $\mathfrak{a} \mapsto V(\mathfrak{a})$ and $Y \mapsto I(Y)$ . Under this correspondence:

| Algebra | Geometry | |---------|----------| | Radical ideal $\mathfrak{a}$ | Algebraic set $V(\mathfrak{a})$ | | Prime ideal $\mathfrak{p}$ | Irreducible variety | | Maximal ideal $\mathfrak{m}$ | Point | | $\mathfrak{a} \subseteq \mathfrak{b}$ | $V(\mathfrak{a}) \supseteq V(\mathfrak{b})$ | | $\mathfrak{a} + \mathfrak{b}$ | $V(\mathfrak{a}) \cap V(\mathfrak{b})$ | | $\mathfrak{a} \cap \mathfrak{b}$ | $V(\mathfrak{a}) \cup V(\mathfrak{b})$ | | $(0)$ | $\mathbb{A}^n$ | | $(1) = k[x_1,\ldots,x_n]$ | $\varnothing$ |

ExampleRadical matters: V(x²) = V(x)

In $k[x,y]$ : $\mathfrak{a} = (x^2)$ and $\mathfrak{b} = (x)$ .

$V(x^2) = V(x) = \{(0,b) \mid b \in k\} = \text{the } y\text{-axis}.$

But $(x^2) \neq (x)$ . The Nullstellensatz says $I(V(x^2)) = \sqrt{(x^2)} = (x)$ . The "extra" algebraic information in $(x^2)$ — the nilpotent structure — is invisible to the zero set. Schemes recover this: $\operatorname{Spec} k[x,y]/(x^2)$ is a "thickened" $y$ -axis (a "double line").

ExampleNo solutions ⟺ ideal is (1)

Consider the system $x^2 + 1 = 0$ over $\mathbb{R}$ :

Over $\mathbb{R}$ : $V(x^2+1) = \varnothing$ , but $(x^2+1) \neq (1)$ in $\mathbb{R}[x]$ . The Nullstellensatz fails because $\mathbb{R}$ is not algebraically closed.
Over $\mathbb{C}$ : $V(x^2+1) = \{i, -i\} \neq \varnothing$ , consistent with $(x^2+1) \neq (1)$ in $\mathbb{C}[x]$ .

The algebraic closure hypothesis is essential.

ExampleIntersection via ideal sum

The intersection of the parabola $V(y - x^2)$ and the line $V(y - 1)$ in $\mathbb{A}^2$ :

$V(y - x^2) \cap V(y - 1) = V(y - x^2, y - 1) = V(x^2 - 1, y - 1) = \{(\pm 1, 1)\}.$

The ideal sum $(y - x^2) + (y - 1) = (x^2 - 1, y - 1)$ detects both intersection points. Its radical is $(x-1, y-1) \cap (x+1, y-1)$ , the intersection of the two maximal ideals.

ExampleTangent intersection and non-radical ideal

The line $y = 0$ meets the parabola $y = x^2$ tangentially at the origin:

$V(y - x^2, y) = V(x^2, y) = \{(0, 0)\}.$

The ideal $(x^2, y)$ is not radical: $x \notin (x^2, y)$ but $x^2 \in (x^2, y)$ . Its radical is $(x, y)$ , the maximal ideal of the origin. The non-radical ideal remembers that this is a tangent intersection (multiplicity 2), but $V$ forgets it.

ExampleParametric system: when do solutions exist?

For which values of $a \in k$ does the system

$x^2 + y^2 = 1, \quad x + y = a$

have a solution over $k = \bar{k}$ ? Substituting $y = a - x$ : $x^2 + (a-x)^2 = 1$ , i.e., $2x^2 - 2ax + a^2 - 1 = 0$ . This has a solution iff the discriminant $4a^2 - 8(a^2 - 1) = 8 - 4a^2 \neq 0$ or equals 0 (it's a quadratic, so always has a root over $\bar{k}$ ). In fact, $V(x^2+y^2-1, x+y-a) \neq \varnothing$ for all $a \in \bar{k}$ , since the ideal $(x^2+y^2-1, x+y-a)$ is always proper.

ExampleThe containment I(Y₁ ∩ Y₂) ⊇ I(Y₁) + I(Y₂)

Let $Y_1 = V(y - x^2)$ and $Y_2 = V(y)$ in $\mathbb{A}^2$ . Then $Y_1 \cap Y_2 = \{(0,0)\}$ .

$I(Y_1) + I(Y_2) = (y - x^2, y) = (y, x^2).$

But $I(Y_1 \cap Y_2) = I(\{(0,0)\}) = (x, y)$ .

So $I(Y_1 \cap Y_2) \supsetneq I(Y_1) + I(Y_2)$ . The inclusion can be strict! We have $I(Y_1 \cap Y_2) = \sqrt{I(Y_1) + I(Y_2)}$ , and indeed $\sqrt{(y, x^2)} = (x, y)$ .

ExampleNullstellensatz in characteristic p

The Nullstellensatz works over any algebraically closed field, including $k = \overline{\mathbb{F}_p}$ .

Over $\overline{\mathbb{F}_2}$ : $V(x^2 + x) = V(x(x+1)) = V(x) \cup V(x+1)$ . But in characteristic 2, $x + 1 = x - 1$ , so $V(x+1) = V(x-1) = \{1\}$ and $V(x) = \{0\}$ . The Nullstellensatz still holds: $I(V(x^2+x)) = \sqrt{(x^2+x)} = (x^2+x) = (x(x+1))$ , which is already radical.

RemarkKey consequences of the Nullstellensatz

Anti-equivalence: The category of affine varieties over $k$ is anti-equivalent to the category of finitely generated reduced $k$ -algebras that are domains. The functor sends $X \mapsto k[X]$ (coordinate ring) and a morphism $\varphi : X \to Y$ to $\varphi^* : k[Y] \to k[X]$ .
Function determines variety: If $I(X) = I(Y)$ , then $X = Y$ . A variety is completely determined by its polynomial functions.
Closed points are dense: In any affine variety $X = V(\mathfrak{p})$ , the closed points (= maximal ideals containing $\mathfrak{p}$ ) are dense.
Algebraic closure is necessary: Over $\mathbb{R}$ , the ideal $(x^2 + 1) \subseteq \mathbb{R}[x]$ is maximal but does not correspond to a point of $\mathbb{A}^1_\mathbb{R}$ . Over non-algebraically closed fields, one must use the scheme-theoretic $\operatorname{Spec}$ instead.

RemarkLooking ahead

The Nullstellensatz is the starting point for scheme theory. Grothendieck's key insight: replace maximal ideals (points) with all prime ideals. The spectrum $\operatorname{Spec} A$ of a ring $A$ has a point for every prime ideal, not just the maximal ones. This allows non-radical ideals (= nilpotents = infinitesimal thickenings) and non-algebraically closed fields to be handled uniformly.

See the proof of the Nullstellensatz for the classical argument via the Rabinowitsch trick.

Math Notes

Hilbert's Nullstellensatz

Statements

Examples illustrating the Nullstellensatz

Consequences