A proof using composition series

While reading Elements of the Representation Theory of Associative Algebras by Andrzej Skowronski, Daniel Simson, and Ibrahim Assem, I came across a step in a proof that I didn’t find immediately obvious. I proved this step using composition series and the Jordan-Hölder theorem.

The step appears in the proof of Corollary 3.4 in Chapter 2:

If \(A\) is a basic and connected finite dimensional algebra, then the quiver \(Q_A\) of \(A\) is connected.

Proof as given in the textbook

The proof in the book is as follows: if this is not the case, then the set \((Q_A)_0\) of points of \(Q_A\) can be written as the disjoint union of two nonempty sets \(Q_0'\) and \(Q_0''\) such that the points of \(Q_0'\) are not connected to those of \(Q_0''\). We show that, if \(i \in Q_0'\) and \(j \in Q_0''\), we have \(e_iAe_j = 0\) and \(e_jAe_i = 0\). Then (1.6) will imply that \(A\) is not connected, a contradiction. Because \(i \neq j\), (I.4.2) yields:

The conclusion follows at once from (3.3).

The step \((*)\) is non-trivial to me. I came up with a proof that relies on composition series and the Jordan-Hölder theorem. I will recall some facts about composition series, which can be found on page 17:

Background on Composition Series and the Jordan-Hölder Theorem

Suppose that \(A\) is a finite-dimensional \(K\)-algebra. If \(M\) is a module in \(\text{mod } A\), then there exists a chain

of submodules of \(M\) such that the module \(M_{j+1}/M_j\) is simple for \(j = 0, 1, \dots, m - 1\). This chain is called a composition series of \(M\), and the simple modules \(M_{j+1}/M_j\) for \(j = 0, 1, \dots, m-1\) are called the composition factors of \(M\).

3.10. Jordan–Hölder Theorem

If \(A\) is a finite-dimensional \(K\)-algebra and

are two composition series of a module \(M\) in \(\text{mod } A\), then \(m = n\), and there exists a permutation \(\sigma\) of \(\{1, \dots, m\}\) such that, for any \(j \in \{0, 1, \dots, m - 1\}\), there is an \(A\)-isomorphism \(M_{j+1}/M_j \cong N_{\sigma(j+1)}/N_{\sigma(j)}\).

It follows that the number \(m\) of modules in a composition series \(0 = M_0 \subset M_1 \subset M_2 \subset \dots \subset M_m = M\) of \(M\) depends only on \(M\); it is called the length of \(M\) and is denoted by \(\ell(M)\).

As an immediate consequence of Theorem 3.10, we get the following.

3.11. Corollary

(a) If \(N\) is an \(A\)-submodule of \(M\) in \(\text{mod } A\), then \(\ell(M) = \ell(N) + \ell(M/N)\).

(b) If \(L\) and \(N\) are \(A\)-submodules of \(M\) in \(\text{mod } A\), then \(\ell(L + N) + \ell(L \cap N) = \ell(L) + \ell(N)\).

Proof of \((*)\)

Now, we prove the isomorphism \(\text{Hom}_A(e_jA,e_iA) \cong \text{Hom}_A(e_jA, \text{rad }e_iA)\) for \(i \neq j\).

Let \(i \neq j\), and let \(\varphi \in \text{Hom}_A(e_jA,e_iA)\). Since the algebra \(A\) is basic, we have \(e_jA \not\cong e_iA\), meaning that \(\varphi\) cannot be an isomorphism.

Assume, for contradiction, that \(\varphi\) is surjective. Consider a composition series for \(e_jA\):

Applying \(\varphi\) to this chain gives a series (not necessarily a composition series) for \(e_iA\), due to the surjectivity of \(\varphi\):

This series can be refined to a composition series for \(e_iA\). Thus, we obtain

On the other hand, since \(\ell(e_jA) = \ell(\text{Ker } \varphi) + \ell(e_jA/\text{Ker } \varphi)\) by Corollary 3.11(a), and since \(e_jA/\text{Ker } \varphi \cong \text{Im } \varphi = e_iA\) by surjectivity of \(\varphi\), we have:

From \((1)\) and \((2)\), we deduce that \(\ell(\text{Ker } \varphi) = 0\), so \(\text{Ker } \varphi = 0\). Hence, \(\varphi\) is injective. Since we assumed that \(\varphi\) is also surjective, it follows that \(\varphi\) is an isomorphim, which contradicts the fact that \(e_jA \not\cong e_iA\). Therefore \(\varphi\) cannot be surjective.

This implies that \(\text{Im } \varphi\) is a proper submodule of \(e_iA\). By (I.4.5)(c), \(\text{rad } e_iA\) is the unique proper maximal submodule of \(e_iA\), so \(\text{Im } \varphi \subseteq \text{rad } e_iA\). This yields a natural isomorphism:

where \(\varphi \vert ^{\text{rad } e_iA}\)​ is \(\varphi\), regarded as a homomorphism with codomain \(\text{rad }e_iA\).