Weak Nullstellensatz

Remarks

Each $\mathfrak{m}_a$ is maximal, since it is the kernel of the surjective evaluation map $f \mapsto f(a)$ onto the field $k$. Conversely, let $\mathfrak{m}$ be maximal; then $k[x_1, \ldots, x_n]/\mathfrak{m}$ is a field finitely generated as a $k$-algebra. So it is finite over $k$ by Zariski's lemma, hence equal to $k$ as $k$ is algebraically closed. If $a_i \in k$ is the image of $x_i$, then $\mathfrak{m} \supseteq \mathfrak{m}_a$, so $\mathfrak{m} = \mathfrak{m}_a$. Part 2 follows from part 1: a proper ideal lies in a maximal ideal $\mathfrak{m}_a$, and then every $f \in I$ vanishes at $a$.

Example: over $\C$, the ideal $(x^2 + y^2 - 1, x - 1)$ of $\C[x, y]$ is proper and the point $(1, 0)$ is a common zero.

The field must be algebraically closed: over $\R$, the ideal $(x^2 + 1)$ of $\R[x]$ is maximal, but it has no zero in $\R$ and is not of the form $(x - a)$.

Going deeper

For a student

The weak form is the step from which the full Nullstellensatz follows by the Rabinowitsch trick. It identifies the points of affine space with the maximal ideals of the coordinate ring. This is the starting point for the maximal spectrum and, later, for schemes.