= Solution
The <Jacobson radical> is $J(A)=\bigcap_L L$, where $L$ runs over the <maximal right ideals> of $A$; equivalently, it is the intersection of the <annihilators> of all <simple modules>. It is a two-sided <ideal>. A <projective module> has the lifting property against every surjective <R-module homomorphism>. A <finitely generated module> is a <projective module> precisely when it is a <direct summand> of a finitely generated <free module>. An <indecomposable module> is nonzero and admits no <direct sum> decomposition into two nonzero <submodules>.
To calculate the <top of an indecomposable projective module>, use the <right Artinian ring> hypothesis, namely the <descending chain condition> on <right ideals>. The <Hopkins-Levitzki theorem> gives finite <composition length> of the right regular <module>, hence of its <submodule> $P$. Also $J=J(A)$ is a <nilpotent ideal>, and $A/J$ is a <semisimple ring>. Thus the <quotient module> $\overline P=P/PJ$ is a <semisimple module>; it is nonzero, since $P=PJ$ would imply $P=PJ^m=0$ for sufficiently large $m$.
Suppose $\overline P$ were not a <simple module>. A nontrivial <direct sum> decomposition of this <semisimple module> would give an <idempotent> $\alpha\in\operatorname{End}_A(\overline P)$ that is neither zero nor the identity. Writing $\pi:P\to\overline P$, the lifting property of the <projective module> $P$ gives $\beta\in\operatorname{End}_A(P)$ with $\pi\beta=\alpha\pi$. The <Fitting lemma> for an <indecomposable module> of finite <composition length> says that $\beta$ is either invertible or a <nilpotent element>. Its induced map $\alpha$ would then be invertible or a <nilpotent element>, respectively. Neither is possible for a nontrivial <idempotent>. Therefore \b[$P/PJ(A)$ is simple]. This argument does not assume that an embedded <projective module> automatically splits off from the ambient <module>.
A <block of an Artinian algebra> is a nonzero two-sided <direct summand> $Ae$ determined by a <primitive central idempotent> $e$: $e$ cannot be written as a sum of two nonzero orthogonal <central idempotents>. Its identity is $e$. For a finite-dimensional <associative algebra>, the <blocks of an Artinian algebra> give its unique decomposition as a finite product of indecomposable <algebras>, or equivalently as a <direct sum> of two-sided <ideals>.
For the <block of S3 in characteristic three>, put $A=kS_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The <group algebra> has <basis> $r^i,r^is$ for $0\leq i<3$. In <characteristic> three,
$$
a^3=0,\qquad sas=r^{-1}-1=-a+a^2.
$$
Consequently $I=aA$ is a two-sided <nilpotent ideal>, $I^3=0$, and $A/I\cong kC_2\cong k\times k$. A <nilpotent ideal> lies in the <Jacobson radical>, and a quotient that is a <semisimple ring> forces the reverse inclusion. Hence
$$
\boxed{J(A)=aA,\qquad \dim_k J(A)=4.}
$$
Define orthogonal <idempotents> $e_+=(1+s)/2$ and $e_-=(1-s)/2$. They sum to one, so the right regular <module> decomposes as
$$
\boxed{kS_3=e_+A\oplus e_-A,\qquad \dim_k e_+A=\dim_k e_-A=3.}
$$
The two summands are the <indecomposable projectives of S3 in characteristic three>. Each <direct summand> is a <projective module>, with <basis> $e_\pm,e_\pm r,e_\pm r^2$. Its <quotient module> modulo multiplication by $J(A)$ is one-dimensional: the <trivial representation> for $e_+A$, and the <sign representation> for $e_-A$. Each is an <indecomposable module>, since two nonzero <direct summands> would each have nonzero <quotient module> modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the <radical series of a module> are the <trivial representation>, <sign representation>, <trivial representation> for $e_+A$, and the <sign representation>, <trivial representation>, <sign representation> for $e_-A$. Indeed, modulo $J^2$ the relation $sa=-as+a^2s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.
The <center of an associative algebra> is spanned by the <conjugacy class> sums $1$, $r+r^2$, and $(1+r+r^2)s$. Since $r+r^2=-1+a^2$ and $1+r+r^2=a^2$, this <center of an associative algebra> is
$$
Z(A)=k1\oplus ka^2\oplus ka^2s.
$$
Its <vector subspace> $ka^2\oplus ka^2s$ is a <square-zero ideal>. If $d1+n$ is a <central idempotent>, then $d^2=d$ and $(2d-1)n=0$. Thus $d=0$ or $1$ and $n=0$. There is no nontrivial <central idempotent>, so \b[the whole group algebra is its single block]. The two three-dimensional <projective modules> above are a decomposition of the regular <module>, not two <blocks of an Artinian algebra>.
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