LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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For a matrix over a field, extending the scalar field does not change its minimal polynomial

Statement

Let $K / F$ be a field extension and let $A \in M_{n} (F)$ . Whether $A$ is viewed over $F$ or over $K$ , its minimal polynomial is the same element of $F [x] \subseteq K [x]$ . This includes $n = 0$ , when both minimal polynomials are $1$ .

Facts & Assumptions

Given: A field extension $K / F$ and a matrix $A \in M_{n} (F)$ .

[L1]

A field extension identifies $F$ with a subfield of $K$ (Field extensions, generated subrings $F [S]$ , generated subfields $F (S)$ , and simple extensions).

[L2]

A finite list is linearly independent when its only vanishing finite linear combination has all coefficients zero (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L3]

The minimal polynomial is the least-degree monic annihilator, equivalently the unique monic generator of all annihilating polynomials (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

Proof

technique · direct

1.1L1L3

Every polynomial over $F$ that annihilates $A$ still annihilates it over $K$ . Hence the $K$ -minimal polynomial divides the $F$ -minimal polynomial and has no larger degree.

1.2L1L2choose

Conversely, let $q = \sum_{i = 0}^{r} c_{i} x^{i} \in K [x]$ be a nonzero annihilator of $A$ . Choose a maximal $F$ -linearly independent sublist $d_{1}, \dots, d_{s}$ from the finite list of nonzero coefficients $c_{i}$ ; maximality makes it span all the $c_{i}$ . Write $c_{i} = \sum_{j} a_{i j} d_{j}$ with $a_{i j} \in F$ .

2.1step 1.2L2algebra

The equality $0 = q (A) = \sum_{j} d_{j} (\sum_{i} a_{i j} A^{i})$ holds entrywise. Since every entry of each inner matrix lies in $F$ and the $d_{j}$ are $F$ -independent, every matrix $\sum_{i} a_{i j} A^{i}$ is zero. At least one corresponding polynomial $q_{j} = \sum_{i} a_{i j} x^{i}$ is nonzero and has degree at most $r$ .

3.1step 1.1step 2.1L3∎

Applying step 2.1 to the $K$ -minimal polynomial gives a nonzero $F$ -annihilator of no larger degree. Thus the two minimal polynomials have equal degree; step 1.1 and monicity then force equality. For $n = 0$ , [L3] gives $1$ over either field.

Depends on

Used by

Dependency tree · two levels

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Sources

Keith Conrad, Potential Diagonalizability, Theorem 4(1) (standard reference, not scraped)