Compute the Jacobson radical and the maximal semisimple quotient of the group ring $\mathbb{F}_2S_3$ of the symmetric group on three letters over the field with two elements, and compute the same for $\mathbb{F}_3S_3$.

Since the ring is left Artinian (being finite), the Jacobson radical is a nilpotent two-sided ideal. So to find it, we just need to look for a maximal nilpotent ideal.

I started by narrowing down the candidates: the Jacobson radical of $R = \mathbb{F}_2S_3$ is the intersection of the annihilators of the simple modules over $R$. The trivial representation is a simple $R$ module, and it has annihilator the augmentation ideal $\mathfrak{a}$. Therefore $J(R) \subset \mathfrak{a}$.

The only nilpotent ideal I succeeded in finding in $\mathfrak{a}$ was the ideal $I$ generated by $s = \sum_{g \in S_3}g$. Note that $rs = 0$ if $r$ has an even number of terms and $rs = s$ if $r$ has an odd number of terms. Therefore $I$ has two elements.

I haven't been able to prove that $I$ is a maximal nilpotent ideal, or that $R/I$ is semisimple.

Beyond just this particular problem, how would you approach computing the Jacobson radical or nilradical of a ring (including a noncommutative ring) in general?


1 Answer 1


The Artin-Wedderburn theorem tells us that the maximal semisimple quotient is a product of matrix rings over finite division rings, one for each irreducible representation. Furthermore, every finite division ring is a field, and the unit group of any finite field is cyclic. The only nontrivial homomorphism from $S_3$ to a cyclic group is the sign homomorphism $S_3\to\mathbb{Z}/2$. It follows that any homomorphism from $\mathbb{Z}S_3$ to a finite field lands in the prime subfield (since elements of $S_3$ can only map to $\pm 1$).

So, writing $\mathbb{F}$ for either $\mathbb{F}_2$ or $\mathbb{F}_3$, the maximal semisimple quotient of $\mathbb{F}S_3$ is a product of matrix rings $M_n(K)$ for finite extensions $K$ of $\mathbb{F}$, one for each irreducible representation, and in all the cases where $n=1$ the $K$ is just $\mathbb{F}$. The only $1$-dimensional representations are the trivial representation and the sign representation, and the sign representation is the same as the trivial representation in the case $\mathbb{F}=\mathbb{F}_2$.

For $\mathbb{F}=\mathbb{F}_3$, dimension-counting now tells us there can be no more irreducible representations: the two $1$-dimensional representations take up $2$ dimensions of the semisimple quotient, and the Jacobson radical is nontrivial since it contains $\sum_{g\in S_3} g$, so there are at most $3$ dimensions left. Another irreducible representation would give a copy of $M_n(\mathbb{F}_{3^d})$ in the semisimple quotient for some $d$ and some $n>1$, which is impossible since there aren't enough dimensions left. We conclude that the two $1$-dimensional representations are the only irreducible representations for $\mathbb{F}=\mathbb{F}_3$, and so the maximal semisimple quotient is $\mathbb{F}_3\times\mathbb{F}_3$. The Jacobson radical is then the kernel of the map $\mathbb{F}_3S_3\to\mathbb{F}_3\times\mathbb{F}_3$; explicitly, it is the set of elements $\sum_{g\in S_3} a_g g$ such that $\sum a_g=0$ and $\sum a_g \sigma(g)=0$, where $\sigma(g)$ is the sign of $g$.

Over $\mathbb{F}_2$, on the other hand, there are up to $4$ dimensions left after accounting for the single $1$-dimensional representation and the fact that the Jacobson radical is nontrivial, so there might be a $2$-dimensional irreducible representation. To find one, note that there is a permutation representation of $S_3$ on $\mathbb{F}_2^3$, and this splits as a direct sum of a trivial subrepresentation (generated by $(1,1,1)$) and a $2$-dimensional subrepresentation (consisting of $(a,b,c)$ such that $a+b+c=0$). (Note that this splitting of the permutation representation doesn't happen over $\mathbb{F}_3$, since $(1,1,1)$ is contained in the latter $2$-dimensional subrepresentation.) This $2$-dimensional representation can easily be verified to be irreducible (for another way of seeing it, note that $\mathbb{F}_2^2\setminus\{0\}$ has three elements, and every permutation of them gives a linear map, so in fact $GL_2(\mathbb{F}_2)\cong S_3$).

So over $\mathbb{F}_2$, we conclude that there is the trivial representation and also this $2$-dimensional irreducible representation; counting dimensions, we now see that we have accounted for all $6$ dimensions of $\mathbb{F}_2S_3$. We conclude that the Jacobson radical is only $1$-dimensional (generated by $\sum_{g\in S_3} g$), and the quotient is $\mathbb{F}_2\times M_2(\mathbb{F}_2)$.

  • $\begingroup$ Sorry to dig this old question up, but could you explain why there can't be more $\mathbb{F}_3$'s in the maximal semisimple quotient? $\endgroup$ Dec 5, 2016 at 22:41
  • $\begingroup$ An $\mathbb{F}_3$ in the maximal semisimple quotient corresponds to a homomorphism $S_3\to \mathbb{F}_3^{\times}=\{\pm 1\}$. The only such homomorphisms are the trivial homomorphism and the sign homomorphism. (Sorry if this didn't answer you're question; I'm not sure exactly which step you don't follow.) $\endgroup$ Dec 5, 2016 at 23:09
  • $\begingroup$ Why is there a corresponding homomorphism $S_3 \rightarrow \{\pm 1\}$ (You have almost answered my question, thanks!) $\endgroup$ Dec 5, 2016 at 23:21
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    $\begingroup$ If you have a factor of $\mathbb{F}_3$ in $\mathbb{F}_3S_3$, the projection onto this factor gives a ring-homomorphism $\mathbb{F}_3S_3\to \mathbb{F}_3$. This restricts to a group-homomorphism $S_3\to \mathbb{F}_3^\times$. Note also that different factors of $\mathbb{F}_3$ give different projection homomorphisms, and thus different group-homomorphisms $S_3\to\mathbb{F}_3^\times$ (since $S_3$ generates the ring $\mathbb{F}_3S_3$). $\endgroup$ Dec 5, 2016 at 23:25
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    $\begingroup$ (This statement is really part of the Artin-Wedderburn theorem: the factors appearing in the decomposition as a product of matrix rings over division rings are in bijection with the irreducible representations, with a factor of $M_n(D)$ corresponding to a representation whose endomorphism ring is $D$ and which is isomorphic to $D^n$ as a $D$-module. So factors of $\mathbb{F}_3=M_1(\mathbb{F}_3)$ correspond to irreducible representations which are one-dimensional $\mathbb{F}_3$-vector spaces, and such a representation is just given by a homomorphism $S_3\to\mathbb{F}_3^\times$.) $\endgroup$ Dec 5, 2016 at 23:29

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