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I am trying to do a question from an algebra qualifying exam:

Decompose the group ring $\mathbb{F}_5[S_3]$ as a product of simple rings.

By Maschke's theorem since $\mathrm{char}(\mathbb{F}_5) \nmid |S_3|$ (i.e. $5\nmid 6$) we know $\mathbb{F}_5[S_3]$ is semisimple so by the Artin-Wedderburn theorem and the fact that the group ring is finite we have $\mathbb{F}_5[S_3]=\Pi_i M_{n_i}(K_i)$ for some fields $K_i$ containing $\mathbb{F}_5$.

Now $|\mathbb{F}_5[S_3]|=5^6$ and $\mathbb{F}_5[S_3]$ is noncommutative so we must have at least one $n_i\geq 2$ and we only have room to have $M_2(\mathbb{F}_5)$. So $\mathbb{F}_5[S_3]$ is isomorphic to $\mathbb{F}_5\times\mathbb{F}_5\times M_2(\mathbb{F}_5)$ or $\mathbb{F}_{25}\times M_2(\mathbb{F}_5)$.

But here I'm stuck. How can I tell which one is correct? I've thought about perhaps using units (you can calculate how many units there are in each of the products above) but I don't know to find all the units of $\mathbb{F}_5[S_3]$. Any ideas?

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    $\begingroup$ Don't forget that you have the trivial simple module and another "natural" $1$-dimensional module for $S_{3}.$ $\endgroup$ – Geoff Robinson Feb 14 '13 at 20:47
  • $\begingroup$ Thanks for the comment. When you say trivial simple module what do you mean? If it's a reference to some representation theory then unfortunately I don't really know any (but if you have a reference that would be good). Is there an explanation just in terms of semisimple rings and such? $\endgroup$ – Martin Leslie Feb 14 '13 at 23:01
  • $\begingroup$ consider $\mathbb{F}_5\cdot x$ where $x=\sum_{g\in S_3}g$. this is a two-sided ideal in the group algebra. $\endgroup$ – yoyo Feb 14 '13 at 23:59
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    $\begingroup$ If we view this as th decomposition of the regular representation into irreducible representations, we can have $|G|=\Sigma_i^m n_i^2$, where $m$ is the number of irreducible representations, and $n_i$ is the dimension of the i-th representation. So, as $6=2^2+1+1$, we know there are two linear characters; does this help? $\endgroup$ – awllower Feb 15 '13 at 4:47
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    $\begingroup$ @Martin: The trivial module is the $1$-dimensional module on which each group element acts as the identity transformation. The other $1$-dimensional simple module for $S_{n}$ which is defined over any field of odd (or zero) characteristic is the $1$-dimensional moodule on which each permutation $\sigma$ acts as ${\rm sigma}(\sigma)$ times the identity transformation. @ awllower: that formula is correct over an algebraically closed feld, but does not hold in general. For example, the quaternion group of order $8$ has a $4$-dimensional irreducible representation over $\mathbb{R}.$ $\endgroup$ – Geoff Robinson Feb 15 '13 at 18:38
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Here's my attempt at answering my own question based on the hints.

Let $x=\sum_{g\in S_3} g$. Then $yx=x$ and $xy=x$ for all $y\in S_3$ so $\mathbb{F}_5\cdot x=\{0,x,2x,3x,4x\}$ is a 2-sided ideal (elements of $\mathbb{F}_5[S_3]$ are sums of $y$'s so just act as a sum of identities).

But ideals of $\mathbb{F}_5[S_3]$ are $\mathbb{F}_5[S_3]$-submodules of $\mathbb{F}_5[S_3]$. Then $\mathbb{F}_5[S_3]$ is semisimple so each submodule is a direct summand, so there exists a term in the product isomorphic to $\mathbb{F}_5$.

So from the choices above we only have $\mathbb{F}_5\times \mathbb{F}_5 \times M_2(\mathbb{F}_5)$.

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