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Every finite-dimensional module has a projective cover, unique up to isomorphism over the target

Statement

Let $A$ be a finite-dimensional algebra and $M$ a finite-dimensional left $A$ -module. Then $M$ has a projective cover. If $π : P \to M$ and $ρ : Q \to M$ are projective covers, then there is an isomorphism $f : P \to Q$ with $ρ \circ f = π$ .

Facts & Assumptions

Given: A finite-dimensional algebra $A$ and a finite-dimensional left $A$ -module $M$ .

[F1]

A projective cover is a projective surjection with superfluous kernel (An essential epimorphism is a surjection with superfluous kernel, and a projective cover is a projective source with such a map).

[L1]

Projective modules are direct summands of free modules (Equivalent characterizations of projective modules).

Proof

technique · direct

1.1

Choose a finite $k$ -basis $m_{1}, \dots, m_{r}$ of $M$ . It also generates $M$ as an $A$ -module, so sending the standard generators to the $m_{i}$ gives a surjection $ε : A^{r} \to M$ . Among the direct summands $P$ of the finite-dimensional module $A^{r}$ for which $ε ∣_{P} : P \to M$ is surjective, choose one of minimal $k$ -dimension; the family is nonempty because it contains $A^{r}$ . Put $π : = ε ∣_{P}$ . The module $P$ is projective by [L1].

L1givenchooseconstruct

2.1

Put $K = \ker π$ , and suppose $N \leq P$ satisfies $N + K = P$ . Then $π ∣_{N} : N \to M$ is surjective. Projectivity of $P$ supplies a map $h : P \to N$ with $(π ∣_{N}) h = π$ ; write $f : P \to P$ for $h$ followed by the inclusion. Thus $π f = π$ , and hence $π f^{n} = π$ for every $n \geq 1$ .

F1L1step 1.1construct

3.1

Since $P$ is finite-dimensional, the kernels and images of the powers of $f$ stabilize. For a sufficiently large $n$ , $P = \ker (f^{n}) \oplus im (f^{n}) :$ the intersection is zero because $f^{n} (x) = 0$ for $x = f^{n} (y)$ implies $f^{2 n} (y) = 0$ and stabilization gives $f^{n} (y) = 0$ , while rank-nullity gives that the two dimensions sum to $\dim_{k} P$ . The restriction of $π$ to $im (f^{n})$ is surjective because $π f^{n} = π$ . Minimality of $P$ in step 1.1 therefore forces $\ker (f^{n}) = 0$ . Hence $f$ is injective and thus bijective, so $P = f (P) \subseteq N$ . Therefore $N = P$ , $K$ is superfluous, and $π$ is a projective cover.

F1step 1.1step 2.1algebra

4.1

Now let $π : P \to M$ and $ρ : Q \to M$ be projective covers. Both sources are finite-dimensional: lift a finite $k$ -basis of $M$ to $P$ , let $P_{0}$ be the submodule generated by those lifts, and observe that $P = P_{0} + \ker π$ ; superfluity gives $P = P_{0}$ , which is finite-dimensional because $A$ is. The same argument applies to $Q$ . Projectivity yields maps $f : P \to Q$ and $g : Q \to P$ with $ρ f = π$ and $π g = ρ$ . Then $π (1_{P} - g f) = 0$ , so $(1_{P} - g f) (P) \subseteq \ker π$ . Hence $P = g f (P) + \ker π$ , and the superfluity of $\ker π$ gives $g f (P) = P$ . Thus $g f$ is surjective, hence bijective on the finite-dimensional module $P$ . The same argument shows that $f g$ is bijective on $Q$ . Since $g f$ is bijective, $f$ is injective; since $f g$ is surjective, $f$ is surjective. Therefore $f$ is an isomorphism, and it still satisfies $ρ \circ f = π$ .

F1step 3.1algebra

5.1

Steps 1.1, 2.1, 3.1, and 4.1 prove existence and uniqueness up to isomorphism over the target.

step 1.1step 2.1step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

11 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

Peter Webb, A Course in Finite Group Representation Theory (23 Feb 2016 draft) (standard reference, not scraped)