ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16

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Primary decomposition over $Q$ with one linear and one irreducible quadratic factor

Example

On $V = Q \oplus Q^{2}$ , let $T$ act by multiplication by $2$ on the first summand and by

$C = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

on the second. Then $μ_{T} = (x - 2) (x^{2} + 1)$ , the primary components are the displayed summands, and their polynomial projections are

$E_{1} = \frac{T^{2} + I}{5}, E_{2} = \frac{4 I - T^{2}}{5} .$

Facts & Assumptions

Given: The rational endomorphism $T$ in the Example.

[L1]

Irreducible-power factors of a minimal polynomial give the direct sum of their kernels (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L2]

The associated primary projections are polynomial expressions in the endomorphism (Each projection in the primary decomposition is a polynomial in the endomorphism).

[L3]

The polynomial $x^{2} + 1$ is irreducible over $R$ , hence also over its subfield $Q$ ( $x^{2} + 1$ is irreducible over $R$ ).

Verification

technique · direct

1.1L1L3algebra

The first block has minimal polynomial $x - 2$ , and $C^{2} = - I$ makes the second block's minimal polynomial $x^{2} + 1$ ; [L3] rules out a rational linear factor. The two factors are coprime, so the direct sum has minimal polynomial their product and [L1] identifies the two primary components.

2.1L2algebra∎

The polynomial $e_{1} = (x^{2} + 1) / 5$ is $1$ at $x = 2$ and is $0$ modulo $x^{2} + 1$ ; $e_{2} = 1 - e_{1} = (4 - x^{2}) / 5$ has the opposite residues. Therefore [L2] gives the displayed projections.

Depends on

Used by

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Sources

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

Primary decomposition over QQ with one linear and one irreducible quadratic factor