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?


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? $\endgroup$ – chilliBeanDream Dec 5 '16 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.) $\endgroup$ – Eric Wofsey Dec 5 '16 at 23:09
  • $\begingroup$ Why is there a corresponding homomorphism $S_3 \rightarrow \{\pm 1\}$ (You have almost answered my question, thanks!) $\endgroup$ – chilliBeanDream Dec 5 '16 at 23:21
  • 1
    $\begingroup$ 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 $\mathbb{F}_3S_3$). $\endgroup$ – Eric Wofsey Dec 5 '16 at 23:25
  • 1
    $\begingroup$ (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 $S_3\to\mathbb{F}_3^\times$.) $\endgroup$ – Eric Wofsey Dec 5 '16 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.