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Over an algebraically closed field, maximal ideals of an affine algebra are kernels of points

Statement

Let k be an algebraically closed field and let A be an affine k-algebra. Then every maximal ideal of A is the kernel of a k-algebra map Ak.

Facts & Assumptions

Given: An algebraically closed field k and an affine k-algebra A.

[L1]

Every affine k-algebra admits a presentation Ak[x1,,xn]/I for some ideal I.

[L2]

Maximal ideals of a quotient correspond to maximal ideals upstairs that contain the defining ideal (R/M is a field if and only if M is a maximal ideal).

[L3]

Over an algebraically closed field, maximal ideals of the polynomial ring are evaluation ideals (Over an algebraically closed field, every maximal ideal is an evaluation ideal).

[L4]

k-points of a quotient algebra are exactly its k-algebra maps to k (k-points of k[x_1, ..., x_n]/I are exactly k-algebra maps to k).

Proof

technique · direct
1.1

Choose a presentation Ak[x1,,xn]/I as in [L1], and let m be a maximal ideal of A. Its inverse image in the polynomial ring is a maximal ideal M containing I.

L1L2givenchoose
2.1

By [L3], there exists akn with M=(x1a1,,xnan). Since IM, the point a annihilates every element of I. Therefore [L4] gives a k-algebra map Ak corresponding to a, and its kernel is exactly m.

L3L4step 1.1
3.1

Hence every maximal ideal of A is the kernel of a k-point.

step 2.1

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