Definition of a basic algebra over a field $K$

Question

Let $K$ be an algebraically closed field and let $A$ be a $K$-algebra with a complete set $\{e_1,…,e_n\}$ of primitive orthogonal idempotents. So, the algebra $A$ is called basic if $$e_iA \cong e_jA \implies e_i=e_j.$$ (In this context $\cong$ means isomorphic as $A$-modules)

But I have a very stupid question: how can $e_iA$ and $e_jA$ not be isomorphic as $A$-modules?

Can't the function $$f:e_jA \to e_iA$$ that associates (for all $a \in A$) $e_i a$ to $e_ja$ be extended to an isomorphism?

Answer 1

If there is a function $f:e_jA\to e_iA$ such that $f(e_ja)=e_ia$ for all $a$ in $A$, then whenever $a\in A$ is such that $e_ja=0$ we have that $e_ia=f(e_ja)=f(0)=0$, and therefore $$\{a\in A:e_ja=0\}\subseteq\{a\in A:e_ia=0\}.$$ In general this is not true. For example, if $A=\mathbb Q\times\mathbb Q$, $e_1=(1,0)$ and $e_2=(0,1)$, you can easily check that that inclusion does not holdf.