Give an example of a \(\CC\dash\)algebra which is not semisimple.
Solution
1. Let\[ A=\mathbb C[\varepsilon]/(\varepsilon^2). \]Then \(A\) is a two-dimensional unital \(\mathbb C\)-algebra with basis \(1,\varepsilon\) and relation \(\varepsilon^2=0\).
Every class modulo \((\varepsilon^2)\) has a unique representative \(a+b\varepsilon\) with \(a,b\in\mathbb C\).
2. The one-dimensional subspace\[ J=\mathbb C\varepsilon \]is a nonzero left ideal of \(A\).
For \(a+b\varepsilon\in A\) and \(c\varepsilon\in J\), \[ (a+b\varepsilon)(c\varepsilon)=ac\varepsilon+bc\varepsilon^2=ac\varepsilon\in J. \] Also \(\varepsilon\neq0\) in \(A\), so \(J\neq0\).
3. The ideal \(J\) has no complementary left ideal in the regular left \(A\)-module \(A\).
Suppose \[ A=J\oplus K \] for a left ideal \(K\). Since \(\dim_{\mathbb C}A=2\) and \(\dim_{\mathbb C}J=1\), one has \(\dim_{\mathbb C}K=1\). Choose a nonzero element \[ k=a+b\varepsilon\in K. \] Because \(K\) is a left ideal, \[ \varepsilon k=a\varepsilon\in K. \] If \(a\neq0\), then \(\varepsilon\in K\), contradicting \(J\cap K=0\). Thus \(a=0\), so \(k=b\varepsilon\in J\). Since \(k\neq0\), this again contradicts \(J\cap K=0\). Therefore no such \(K\) exists.
4. Hence \(A\) is not semisimple.
A ring is semisimple precisely when its regular left module is semisimple; equivalently, every left ideal is a direct summand of the regular module. By 2 and 3, the left ideal \(J\) is not a direct summand of \(A\).