LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30

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A finitely localized polynomial ring in positive dimension is not a field

Statement

Let $k$ be a field, let $r > 0$ , and let $s \in k [t_{1}, \dots, t_{r}]$ be nonzero. Then the localization

k [t_{1}, \dots, t_{r}] [\frac{1}{s}]

is not a field.

Facts & Assumptions

Given: A field $k$ , an integer $r > 0$ , and a nonzero polynomial $s \in k [t_{1}, \dots, t_{r}]$ .

[L1]

A $k$ -algebra map out of a polynomial ring is determined by the images of the indeterminates (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[A1]

The one-variable denominator obstruction is the model case behind the specialization to $k [u]$ .

Proof

technique · direct

1.1

Choose an integer $N > 1$ larger than every exponent occurring in $s$ , and define a $k$ -algebra map $φ : k [t_{1}, \dots, t_{r}] \to k [u], t_{i} \mapsto u^{N^{i - 1}} .$ By uniqueness of base- $N$ expansion, distinct monomials of $s$ acquire distinct $u$ -degrees under $φ$ , so $φ (s) \neq 0$ .

L1givenchoose

2.1

Because $φ (s) \neq 0$ , the universal property of localization gives a ring homomorphism $Φ : k [t_{1}, \dots, t_{r}] [\frac{1}{s}] \to k [u] [\frac{1}{φ (s)}] .$ If the source were a field, its image would also be a field.

L1step 1.1algebra

3.1

The target is not a field. If $φ (s)$ is constant, then the target is just $k [u]$ , and $u$ is not invertible in $k [u]$ . If $φ (s)$ is nonconstant and $1 / (φ (s) + 1) = h / φ (s)^{m}$ in the localization, then $φ (s)^{m} = h (φ (s) + 1)$ in $k [u]$ . Dividing the left-hand side by $φ (s) + 1$ leaves remainder $(- 1)^{m} \neq 0$ , contradiction.

step 2.1algebra

4.1

Step 3.1 contradicts the conclusion of step 2.1. Therefore $k [t_{1}, \dots, t_{r}] [1 / s]$ is not a field.

step 2.1step 3.1contradiction∎

Depends on

Used by

A field finitely generated as a k-algebra is a finite extension of k Corollary

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 13.6 and Proposition 13.7 (standard reference, not scraped)