# Question on $Ext^1$

Given a finite dimensional algebra $$A$$ with two indecomposable modules $$M$$ and $$N$$. Define $$H(M,N)$$ as the largest number of indecomposable summands of a module $$X$$ such that there exists a non-split short exact sequence $$0 \rightarrow N \rightarrow X \rightarrow M \rightarrow 0$$. Let $$R_i$$ be a basis of $$Ext_A^1(M,N)$$ such that $$R_i$$ is representated by short exact sequences with middle terms $$X_i$$. Let $$|W|$$ denote the number of indecomposable summands of a module $$W$$.

Question: Is it true that $$H(M,N)$$ is equal to the largest number of the $$|X_i|$$ independent of the choice of a basis?

I posted this here https://math.stackexchange.com/posts/3381428/edit with no answers.

I don't think so: If $$A$$ is of finite representation type, the space $$\mathrm{Ext}^1(M, N)$$ is algebraically stratified so that each stratum corresponds to an isomorphism class (after base change to algebraic closure) of $$X$$. Consequently, there is a dense stratum in $$\mathrm{Ext}^1(M, N)$$ which gives a maximal extension (and it tends to have a minimal number of indecomposables, morally).
For example, take the quiver with vertices $$I = \{1, \ldots, 4\}$$ and with adjacency matrix $$Q = \begin{pmatrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0\end{pmatrix}$$ In other words, it is an oriented Dynkin quiver of type $$D_4$$ where the vertex $$1$$ is a sink and $$3, 4$$ are sources. Let $$M$$ be the only indecomposable representation with dimension vector $$(0, 1, 1, 1)$$ and $$N$$ the indecomposable with dimension vector $$(1, 1, 0, 0)$$. For $$i \in I$$, let $$P(i)$$ denote the projective cover of the irreducible representation $$L(i)$$ corresponding to $$i$$.
Then $$M$$ has a projective resolution $$0\to P(1)\oplus P(2)\to P(3)\oplus P(4)\to M\to 0$$. Hence $$\mathrm{Ext}^1(M, N)$$ is of dimension $$2$$. We have the following two extensions of $$M$$ by $$N$$ who have $$2$$ indecomposables: $$(1, 1, 1, 0)\oplus (0, 1, 0, 1)$$ and $$(1, 1, 0, 1)\oplus (0, 1, 1, 0)$$. They represent two lines (minus the origin) in $$\mathbf{C}^2$$. However, there obviously exists an indecomposable extension of $$M$$ by $$N$$, which is the injective hull $$I(1)$$ of $$L(1)$$. In fact any element of the complement of $$\mathbf{C}\times \{0\}\cup \{0\}\times\mathbf{C}$$ is isomorphic to $$I(1)$$. Since this complement is Zariski-dense in $$\mathbf{C}^2$$, it follows that the largest number of indecompsables can be either $$1$$ or $$2$$, depending on the choice of basis.