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

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If $\gcd (f, g) = 1$ and $(f g) (T) = 0$ , then $V = \ker f (T) \oplus \ker g (T)$

Statement

Let $T : V \to V$ be an endomorphism and let $f, g \in F [x]$ satisfy $\gcd (f, g) = 1$ and $(f g) (T) = 0$ . Then

$V = \ker f (T) \oplus \ker g (T) .$

Facts & Assumptions

Given: An endomorphism $T$ and coprime polynomials $f, g$ with $(f g) (T) = 0$ .

[L1]

If $\gcd (f, g) = 1$ , Bézout's identity supplies $a, b \in F [x]$ with $a f + b g = 1$ (Bézout identity and the Euclidean algorithm for polynomials over a field, The monic greatest common divisor of two polynomials over a field).

[L2]

For two subspaces, $V = U \oplus W$ means $U + W = V$ and $U \cap W = {0}$ (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

[L3]

Polynomial evaluation sends $p (x) = \sum a_{k} x^{k}$ to $p (T) = \sum a_{k} T^{k}$ , with $T^{0} = I$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

Proof

technique · direct

1.1L1L3choose

Choose $a, b$ as in [L1]. Evaluating the identity gives $a (T) f (T) + b (T) g (T) = I$ .

2.1step 1.1L3givenalgebra

For $v \in V$ , write $v = a (T) f (T) v + b (T) g (T) v$ . The first summand lies in $\ker g (T)$ and the second in $\ker f (T)$ because polynomial evaluations commute and $(f g) (T) = 0$ . Thus the two kernels span $V$ .

3.1step 1.1L2∎

If $v$ lies in both kernels, step 1.1 gives $v = a (T) f (T) v + b (T) g (T) v = 0$ . Hence their intersection is zero, and [L2] proves the direct sum. Unit factors and the zero space satisfy the same calculation.

Depends on

Used by

Dependency tree · two levels

18 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

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5, proof of Theorem 5.19 (standard reference, not scraped)

If gcd⁡(f,g)=1gcd(f,g)=1 and (fg)(T)=0(fg)(T)=0, then V=ker⁡f(T)⊕ker⁡g(T)V=kerf(T)⊕kerg(T)