f(rad A) ⊆ rad B?

While reading Elements of the Representation Theory of Associative Algebras by Andrzej Skowronski, Daniel Simson, and Ibrahim Assem, I came across a mistake in Exercise 1, Chapter I.

The exercise is posed as follows:

Let \(f : A \rightarrow B\) be a homomorphism of \(K\)-algebras. Prove that \(f(\text{rad } A) \subseteq \text{rad } B\).

However, the statement is incorrect.

One simple counterexample

Consider the algebras \(A := \mathbb{T}_2(K) = \begin{bmatrix} K & 0 \\ K & K \end{bmatrix}\) and \(B := \mathbb{M}_2(K) = \begin{bmatrix} K & K \\ K & K \end{bmatrix}\), with the canonical inclusion map \(f : A \hookrightarrow B\).

First, observe that \(\text{rad } A = \begin{bmatrix} 0 & 0 \\ K & 0 \end{bmatrix}\). This is because \(I := \begin{bmatrix} 0 & 0 \\ K & 0 \end{bmatrix}\) is a two-sided nilpotent ideal of \(A\), and \(A/I \cong \begin{bmatrix} K & 0 \\ 0 & K \end{bmatrix} \cong K \times K\), implying by Corollary 1.4(c) that \(\text{rad } A = I\). Next, \(\text{rad } B = 0\) by the definition of the Jacobson radical, or alternatively, by a direct application of the Wedderburn-Artin theorem, as stated in Theorem 3.4. Thus, \(f(\text{rad } A) \not\subseteq \text{rad } B\).

Sufficient condition for correctness

We can add the condition that \(f\) is surjective for the statement to hold true:

Let \(f : A \rightarrow B\) be a surjective homomorphism of \(K\)-algebras. Then \(f(\text{rad } A) \subseteq \text{rad } B\).

Indeed, let \(b' \in f(\text{rad } A)\), and let \(a' \in \text{rad } A\) be such that \(b' = f(a')\). We need to prove that \(b' \in \text{rad } B\). By Lemma 1.3, it suffices to show that for any \(b \in B\), the element \(1-b'b\) has a right inverse. Let \(b \in B\). Since \(f\) is surjective, there exists \(a \in A\) such that \(b = f(a)\). Thus, we have:

Since \(a' \in \text{rad } A\), by Lemma 1.3, there exists \(r \in A\) such that \((1-a'a)r = 1\). Therefore, \(f(r)\) is a right inverse of \(1-b'b\) in \(B\), as:

Thus, \(b' \in \text{rad } B\), completing the proof.