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Over an algebraically closed field, every maximal ideal is an evaluation ideal

Statement

Let $k$ be an algebraically closed field, and let $m$ be a maximal ideal of $k [x_{1}, \dots, x_{n}]$ . Then there is a unique point $a = (a_{1}, \dots, a_{n}) \in k^{n}$ such that $m = (x_{1} - a_{1}, \dots, x_{n} - a_{n}) .$

Facts & Assumptions

Given: An algebraically closed field $k$ and a maximal ideal $m \subseteq k [x_{1}, \dots, x_{n}]$ .

[L1]

A maximal ideal of a finite-type $k$ -algebra has finite residue field over $k$ (A maximal ideal of an affine algebra has finite residue field over the base field).

[L2]

Evaluation at a point has kernel $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ (Evaluation at a point has kernel (x_1-a_1, ..., x_n-a_n)).

[L3]

In an algebraically closed field, every nonconstant polynomial has a root (An algebraically closed field: every nonconstant polynomial has a root in the field).

Proof

technique · direct

1.1L1given

By [L1], the residue field $K : = k [x_{1}, \dots, x_{n}] / m$ is a finite extension of $k$ . Let ${\overset{ˉ}{x}}_{i}$ be the image of $x_{i}$ in $K$ .

2.1L3step 1.1choose

Each ${\overset{ˉ}{x}}_{i}$ is algebraic over $k$ because $K / k$ is finite. Let $m_{i} (T) \in k [T]$ be its minimal polynomial. Since $k$ is algebraically closed, [L3] gives a root $a_{i} \in k$ of $m_{i}$ . Minimality forces $m_{i} (T) = T - a_{i}$ , so ${\overset{ˉ}{x}}_{i} = a_{i}$ in $K$ .

3.1L2step 2.1∎

The quotient map $k [x_{1}, \dots, x_{n}] \to K$ is therefore evaluation at the point $a = (a_{1}, \dots, a_{n})$ . Its kernel is $m$ , while [L2] says the evaluation kernel is $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ . Hence these ideals are equal. Uniqueness of $a$ follows because the quotient remembers each coordinate class ${\overset{ˉ}{x}}_{i}$ .

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Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Corollary 13.9 (standard reference, not scraped)
Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Corollary (15.5) (standard reference, not scraped)