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

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The algebra $F [T]$ generated by an endomorphism is isomorphic to $F [x] / (μ_{T})$

Statement

For an endomorphism $T$ of a finite-dimensional $F$ -vector space, the evaluation map induces an $F$ -algebra isomorphism

$F [x] / (μ_{T}) ≅ F [T] : = {p (T) : p \in F [x]} .$

Facts & Assumptions

Given: An endomorphism $T$ and polynomial evaluation ${ev}_{T} : F [x] \to L (V, V)$ .

[L1]

Polynomial evaluation sends $p$ to $p (T) = \sum a_{k} T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

[L2]

Its annihilating polynomials form the ideal $(μ_{T})$ (The annihilator ideal is nonzero and has a unique monic generator; $p (T) = 0$ if and only if $μ_{T} ∣ p$ ).

[L3]

The first isomorphism theorem for rings gives $R / \ker f ≅ im f$ (First isomorphism theorem for rings: $R / \ker f ≅ im f$ ).

Proof

technique · direct

1.1L1L2algebra

The finite-sum definition in [L1] gives $(p + q) (T) = p (T) + q (T)$ , $(p q) (T) = p (T) q (T)$ , and $1 (T) = I$ , so evaluation is an $F$ -algebra homomorphism. Its image is $F [T]$ by definition, and [L2] identifies its kernel with $(μ_{T})$ .

2.1step 1.1L3∎

Apply [L3] to step 1.1. On the zero space $μ_{T} = 1$ , and both $F [x] / (1)$ and the zero endomorphism algebra are the one-element ring.

Depends on

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Sources

Keith Conrad, The Minimal Polynomial and Some Applications, closing observation in §4 (standard reference, not scraped)

The algebra F[T]F[T] generated by an endomorphism is isomorphic to F[x]/(μT)F[x]/(μT​)