CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

An annihilating polynomial need not be minimal: $\sqrt{2}$ is a root of both $x^{2} - 2$ and $x^{4} - 4$

Statement refuted

Every nonzero polynomial that vanishes at an algebraic element is that element's minimal polynomial.

Counterexample

Let $a = \sqrt{2}$ , whose nonnegative real value satisfies $a^{2} = 2$ because $R$ is complete (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Then both $x^{2} - 2$ and $x^{4} - 4$ vanish at $a$ , but only the first is minimal.

Facts & Assumptions

Given: The algebraic number $a = \sqrt{2}$ over $Q$ .

[F1]

Eisenstein's criterion proves irreducibility over $Q$ under its prime divisibility hypotheses (Eisenstein criterion over the integers).

[F2]

The minimal polynomial is the unique monic irreducible generator of the evaluation kernel, and it divides every annihilating polynomial (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

The real numbers are a complete ordered field, so $\sqrt{2}$ exists and satisfies $(\sqrt{2})^{2} = 2$ (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · counterexample

1.1

Evaluation gives $a^{2} - 2 = 0$ and $a^{4} - 4 = 0$ .

F3algebra

2.1

The polynomial $x^{2} - 2$ satisfies [F1] at $2$ , so [F2] identifies it as the minimal polynomial of $a$ over $Q$ .

F1F2step 1.1

3.1

The other annihilator factors as $x^{4} - 4 = (x^{2} - 2) (x^{2} + 2)$ and has larger degree, so it is a proper multiple of the minimal polynomial.

step 2.1algebra

4.1

Thus a polynomial may annihilate an algebraic element without being its minimal polynomial.

step 1.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

T. Judson, Abstract Algebra: Theory and Applications, Extension Fields (standard reference, not scraped)

An annihilating polynomial need not be minimal: 22​ is a root of both x2−2x2−2 and x4−4x4−4