A module which is an extension of a regular and a preinjective module is regular?

Question

In proposition 4.4 of the paper "Representations of Wild Quivers" by Otto Kerner the setting is the following:

• $A$ is a wild hereditary path algebra
• $E$ is an indecomposable regular module

He proves that the following are equivalent:

• $E$ is elementary (i.e. it is not a nontrivial extension of two regular modules)
• If $Y \neq 0$ is a regular submodule of $E$ then $E/Y$ is preinjective

I don't understand an argument in the proof. He constructs several modules with several properties:

• $K$ is a submodule of $E$ without nonzero preinjective direct summand
• $Z_2$ is preinjective
• There is a short exact sequencence $0 \to Y \to K \to Z_2 \to 0$, so in particular the dimension vectors satisfy that $\underline{\dim} K = \underline{\dim} Y + \underline{\dim} Z_2$

From that he concludes that $K$ is regular (a contradiction). Why is this?

Answer 1

Lemma: Whenever $0 \to U \to V \to W \to 0$ is a short exact sequence, we have the following:

• If $U$ and $W$ do not have indecomposable preprojective direct summands, then $V$ also does not have an indecomposable preprojective direct summand.
• If $U$ and $W$ do not have indecomposable preinjective direct summands, then $V$ also does not have an indecomposable preinjective direct summand.
• If $U$ and $W$ are regular, then so is $V$.

Proof: ($iii$) follows directly from ($i$) and ($ii$).

Assume $U$ and $W$ do not have an indecomposable preprojective direct summand and assume that $V$ does have such a summand, i.e. $V = V' \oplus P$ with $P$ preprojective. Then since there are no homomorphisms from non-preprojective indecomposable modules to preprojective ones, we have Hom$(U, P) = 0$, i.e. the map $\pi \circ \alpha$, where $\pi: V \to P$ is the canonical projection, disappears. We claim $W = \beta(V') \oplus \beta(P)$: $W = \beta(V') + \beta(P)$ follows since $\beta$ is surjective. Let $w \in \beta(V') \cap \beta(P)$, i.e. there are $v' \in V'$ and $p \in P$ such that $w = \beta(v') = \beta(p)$. Then $p - v' \in \ker(\beta) = $im$(\alpha)$, so there is $u \in U$ such that $\alpha(u) = p - v'$. It follows $0 = (\pi \circ \alpha)(u) = \pi(p - v') = p$ and thus $w = \beta(p) = \beta(0) = 0$, proving the claim. Now, $\beta$ is injective on $P$, since as $\pi \circ \alpha = 0$, $\alpha$ cannot hit an element of $P$. Thus $P \cong \beta(P)$ and so $W \cong \beta(V') \oplus P$ has a preprojective direct summand, contradicting the assumption. This proves that $V$ also does not have a preprojective direct summand, thus we proved ($i$).

Now assume $U$ and $W$ do not have an indecomposable preinjective direct summand and assume $V$ does, i.e. $V = V' \oplus I$ with $I$ preinjective. Then since indecomposable preinjective modules do not have homomorphisms to indecomposable modules which are not preinjective we have Hom$(I, W) = 0$ and so $\beta(I) = 0$. Thus we have $I \subseteq $im$(\alpha)$. We claim U = $\alpha^{-1}(V') \oplus \alpha^{-1}(I)$. That the intersection $\alpha^{-1}(V') \cap \alpha^{-1}(I)$ is zero follows from $\alpha$ being injective. Now let $u \in U$. Then $\alpha(u) = v' + i$ with $v' \in V'$ and $i \in I$. There is some $i' \in \alpha^{-1}(I)$ such that $\alpha(i') = i$ since $I \subseteq $im$(\alpha)$. Then $\alpha(u - i') = v' + i - i = v' \in V'$ and thus $u - i' \in \alpha^{-1}(V')$. This shows $u = (u - i') + i' \in \alpha^{-1}(V') + \alpha^{-1}(I')$ and thus the claim. Since $\alpha^{-1}(I) \cong \alpha(\alpha^{-1}I) = I$ it follows that $U \cong \alpha^{-1}(V') \oplus I$ has a preinjective direct summand, contradicting the assumption. $\square$

This directly shows that $K$ has to be regular, since we already know that it has no preinjective direct summands.