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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

1.1

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

L1L3choose

2.1

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 $V$ .

step 1.1L3givenalgebra

3.1

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.

step 1.1L2∎

Depends on

Used by

Primary decomposition: the irreducible-power factors of μ_T split V into their invariant kernels Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 18 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

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)