= Solution
An <idempotent> $e$ gives a <projective module> $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal <idempotents>. This is equivalent to indecomposability of $Ae$: a projection onto a <direct summand> is an <idempotent> of $\operatorname{End}_A(Ae)\cong(eAe)^{\mathrm{op}}$, and conversely such an <idempotent> splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thus
$$
\boxed{1=e_1+\cdots+e_m,\qquad {}_AA=\bigoplus_{\nu=1}^mAe_\nu,}
$$
with orthogonal <primitive idempotents> and indecomposable <projective modules>. Each summand is generated by its <idempotent>.
Let $D=\operatorname{Hom}_k(-,k)$. A <Frobenius algebra> is a finite-dimensional algebra with $A\cong D(A)$ as left <modules> over $A$, where $(a\phi)(x)=\phi(xa)$. Equivalently there is a <linear functional> $\lambda:A\to k$ for which $(x,y)\mapsto\lambda(xy)$ is nondegenerate. A <symmetric algebra (Frobenius algebra)> satisfies the stronger <bimodule> isomorphism $A\cong D(A)$, with $(a\phi b)(x)=\phi(bxa)$; equivalently the functional can be chosen with $\lambda(xy)=\lambda(yx)$. This is the associative-algebra meaning of symmetric, rather than the <symmetric algebra> of a <vector space>.
For a finite-dimensional <module>,
$$
\boxed{\operatorname{Soc}M=\sum_{S\subseteq M\ \mathrm{simple}}S,\qquad
\operatorname{Rad}M=\bigcap_{N\subsetneq M\ \mathrm{maximal}}N.}
$$
The <socle> is its largest semisimple <submodule>, and the <radical of a module> gives its largest semisimple quotient. With $J=J(A)$ one has $\operatorname{Rad}M=JM$ and $\operatorname{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every <simple module>.
Now let $P=Ae$ for a <primitive idempotent>, allowing the <field> $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairing
$$
eA\times Ae\longrightarrow k,\qquad (u,v)\longmapsto\lambda(uv).
$$
It is nondegenerate on these corners. If $v=ve\ne0$, choose $a$ with $\lambda(av)\ne0$; then $\lambda(eav)=\lambda(ave)=\lambda(av)\ne0$. The other variable is treated similarly. The pairing respects the left $A$-action, so
$$
\operatorname{Soc}(Ae)\cong\operatorname{Soc}D(eA)=D(eA/eAJ).
$$
The right-module quotient $eA/eAJ$ is simple. A <primitive idempotent> remains primitive modulo the nilpotent radical: <idempotents> lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the <semisimple algebra> $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same <matrix> block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed <field>. Hence the <head-socle identity for a symmetric algebra> is
$$
\boxed{\operatorname{Soc}P\cong P/\operatorname{Rad}P.}
$$
This is an isomorphism of left <modules>, and need not be canonical.
For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the <group algebra> is symmetric and nondegenerate, so the identity just proved applies.
In characteristic two, the <central idempotents> are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_0s$ and is $kC_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and use
$$
r\longmapsto\begin{pmatrix}\omega&0\\0&\omega^2\end{pmatrix},
\qquad s\longmapsto\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
The two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal <matrix units>, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:
$$
\boxed{kS_3\cong k[u]/(u^2)\oplus M_2(k)\quad(\operatorname{char}k=2).}
$$
There are two <simple modules>, the trivial one-dimensional <module> $T$ and the two-dimensional <module> $V$; the sign <module> equals $T$. Their <projective covers> are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular <module> is $P_T\oplus V^{\oplus2}$, and the <Cartan matrix of a group algebra> is $\operatorname{diag}(2,1)$. Every <module> is a sum of $T,P_T,V$: the first block is classified by <Jordan chains> for a <nilpotent linear map> of lengths one and two, while the <matrix> block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the <2-modular blocks of S3>.
In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong kC_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign <modules> $S_+,S_-$. Put $e_\pm=(1\pm s)/2$. The projectives $P_\pm=Ae_\pm$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:
$$
v^3=0,\qquad svs=-v,\qquad v=-u-u^2,\qquad v^2=u^2.
$$
The basis $e_\pm,ve_\pm,v^2e_\pm$ gives the radical layers
$$
\boxed{P_+:\ S_+,S_-,S_+;\qquad P_-:\ S_-,S_+,S_-.}
$$
Thus $A=P_+\oplus P_-$ and its <Cartan matrix of a group algebra> is
$$
\boxed{\begin{pmatrix}2&1\\1&2\end{pmatrix}.}
$$
The elements $e_-ve_+$ and $e_+ve_-$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are <linearly independent>. Hence $A$ is the <path algebra> of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected <quiver> also proves there is a single <block of S3 in characteristic three>.
Any <module> is a $\mathbb Z/2$-graded <vector space> given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing <Jordan chains> with homogeneous bases in the <kernel> and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a <uniserial module> and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_\pm/J^\ell P_\pm$, $\ell=1,2,3$.
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