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Over R, not every maximal ideal is an evaluation ideal

Example

In $R [x]$ , the ideal $(x^{2} + 1)$ is maximal but is not an evaluation ideal $(x - a)$ for any $a \in R$ .

Facts & Assumptions

Given: The polynomial ring $R [x]$ .

[L1]

Evaluation at $a \in R$ has kernel $(x - a)$ (Evaluation at a point has kernel (x_1-a_1, ..., x_n-a_n)).

[L2]

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

Verification

technique · direct

1.1

The quotient $R [x] / (x^{2} + 1)$ is isomorphic to $C$ by sending the class of $x$ to $i$ . Since $C$ is a field, $(x^{2} + 1)$ is maximal.

L2given

2.1

For every $a \in R$ , [L1] says the evaluation ideal at $a$ is $(x - a)$ , and no such ideal equals $(x^{2} + 1)$ because $x^{2} + 1$ has no real root. So weak Nullstellensatz fails in point form over $R$ . The residue field extension here is $C / R$ , which is finite of degree $2$ , as [L2] allows.

L1L2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Allen B. Altman and Steven L. Kleiman, A Term of Commutative Algebra, 13th ed., Corollary (15.5) (standard reference, not scraped)