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Each projection in the primary decomposition is a polynomial in the endomorphism

Statement

In the primary decomposition $V = ⨁_{i < r} V_{i}$ associated with $μ_{T} = \prod_{i < r} f_{i}$ , the projection $E_{i} : V \to V_{i}$ along the other primary components is a polynomial in $T$ . More precisely, if $g_{i} = μ_{T} / f_{i}$ and $a_{i} f_{i} + b_{i} g_{i} = 1$ , then

$E_{i} = b_{i} (T) g_{i} (T) .$

Facts & Assumptions

Given: The primary decomposition and the polynomials $f_{i}, g_{i}$ in the Statement.

[L1]

The irreducible-power factors of $μ_{T}$ give the direct sum $V = ⨁_{i} \ker f_{i} (T)$ (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L2]

Coprime polynomials satisfy a Bézout identity (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L3]

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

Proof

technique · direct

1.1

The polynomials $f_{i}$ and $g_{i}$ are coprime, so choose $a_{i}, b_{i}$ with $a_{i} f_{i} + b_{i} g_{i} = 1$ by [L2], and set $E_{i} = b_{i} (T) g_{i} (T)$ using [L3].

L2L3chooseconstruct

2.1

On $V_{i} = \ker f_{i} (T)$ the evaluated Bézout identity gives $E_{i} = I$ . On $V_{j}$ for $j \neq i$ , the polynomial $g_{i}$ is divisible by $f_{j}$ , so $E_{i} = 0$ .

step 1.1L1algebra

3.1

By the unique decomposition in [L1], the operator described in step 2.1 is exactly projection onto $V_{i}$ along the sum of the other components.

step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 9 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, Theorem 5.19 (standard reference, not scraped)