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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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Artin's ideal generated by f(xf) for all monic nonconstant fF[x] is proper

Statement

Let S be the set of monic nonconstant polynomials in F[x], let A=F[xf:fS], and let

I=(f(xf):fS)A.

Then I is a proper ideal of A.

Facts & Assumptions

Given: A field F, the family polynomial ring A=F[xf:fS], and the ideal I displayed in the Statement.

[L1]

The family polynomial ring consists of finite sums involving only finitely many indeterminates (The polynomial ring R[xi:iI] as finitely supported coefficient families on monomials).

[L2]

A homomorphism from a family polynomial ring is obtained by assigning an image to every indeterminate, and is unique with those assignments (Universal property of a polynomial ring on an arbitrary family of indeterminates).

[L3]

A finite family of nonzero polynomials has a common splitting field (Every finite family of nonzero polynomials has a splitting field, obtained from their product).

Proof

technique · contradiction
1.1

Suppose I=A. Then 1=j=1mhjfj(xfj) for finitely many fjS and hjA.

assume-contraL1
1.2

By [L3], choose a field E/F in which all fj split, and choose a root ajE of each fj.

L3choose
2.1

Assign xfjaj for the variables occurring as generators in step 1.1 and assign every other indeterminate, including unused ones appearing in the hj, to 0. By [L2] this gives an F-algebra homomorphism Φ:AE.

step 1.1step 1.2L2construct
3.1

Applying Φ to step 1.1 gives 1=jΦ(hj)fj(aj)=0, impossible in the field E. Therefore I is proper.

step 1.1step 1.2step 2.1discharge-contradiction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 41 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources