Let $\mathbb{F}_q$ be a field of order $q = p^m$, where $p$ is the characteristic of the field; a prime.

Consider the ring $$R_n = \mathbb{F}_q[x]/\langle x^n - 1 \rangle $$ Now, I've read that this ring is semi-simple when $p$ does not divide $n$, but no proof was given.

Does anyone know how to prove it or somewhere where I can read the proof? Thank you.


2 Answers 2


One way to look at it is that if $C=\{1, c, c^2,\ldots, c^{n-1}\}$ is the cyclic group of order $n$, then $x\mapsto c$ is an isomorphism of $R_n\to F_q[C]$, the group ring of $C$ over $F_q$.

Maschke's theorem says that if $|G|=n<\infty$, then $F[G]$ is semisimple exactly when $|G|$ is a unit in $F$, and this would be the case only when $\gcd(p,n)=1$.

Another way to prove it would be to show that $x^n-1$ is square-free when factorized over $F_q$. This says that the ring is semiprime, and since it is already obviously Artinian, that would make it semisimple. Sorry to say my cyclotomic factorization knowledge is too rusty to spell this out, but the same is true.

  • 1
    $\begingroup$ if $p$ does not divide $n$, then the derivative of $x^n-1$ does not vanish at the roots of $x^n-1$, so the monic $x^n-1$ is square-free. Reference: theorem 4.5 Jacobson BA1. (+1 by the way) $\endgroup$
    – peter a g
    Dec 3, 2018 at 14:28
  • $\begingroup$ @peterag Thanks for filling that in! $\endgroup$
    – rschwieb
    Dec 3, 2018 at 14:31
  • $\begingroup$ Why does $x^n - 1$ being square free imply that the ring is semiprime? $\endgroup$
    – the man
    Dec 3, 2018 at 16:29
  • $\begingroup$ @theman Prove it! More generally if $p(x)\in F[x]$, $F[x]/(p(x))$ is semiprime iff $p(x)$ is square free. If it is square free, you can apply the Chinese Remainder theorem, and if it isn't square free you can find a nonzero nilpotent. $\endgroup$
    – rschwieb
    Dec 3, 2018 at 16:46

I'll give a proof. First we'll need some facts.

Fact 1

A commutative ring (possibly without identity) is semisimple if and only if it is a direct sum of fields.


It is clear that a direct sum of fields is semisimple, so assume $R$ is a commutative semisimple ring. Then note that since $R$ is commutative maximal left ideals are simply maximal ideals of $R$, so every simple left $R$ module is in fact a quotient ring of $R$.

Now let $\phi : R\to \bigoplus_i S_i$ be the isomorphism demonstrating semisimplicity of $R$, with all of the $S_i$ simple. Let $m_i$ be the kernel of the composite map $R\to \bigoplus_i S_i\to S_i$, so that we can identify $\phi$ with the canonical map $R\to \bigoplus_i R/m_i$. However, this canonical map is a ring homomorphism. Since $\phi$ was a bijection, and we can identify it with this map, this canonical map is a bijection. Thus $R\cong \bigoplus_i R/m_i$.

Fact 2

An Artinian ring is a finite product of (necessarily Artinian) local rings.

Proof sketch: See e.g. Atiyah and MacDonald for more details, but the idea is to prove the following lemma: An Artinian ring with no nontrivial idempotents is local. Then this implies that we can recursively break the ring up into a product of rings until we arrive at a local ring, and this process must terminate since the ring is Artinian.

Fact 3

An Artinian local ring is either a field or has nilpotents.

Proof: Let $m$ be the maximal ideal of the Artinian local ring. Since the descending chain $m\supseteq m^2 \supseteq \cdots \supseteq m^n\supseteq \cdots $ must stabilize, for some $n$, we have $m^n=m^{n+1}$, so by Nakayama's lemma, we have $m^n=0$. Thus $m^n$ is nilpotent. Thus any nonzero element of $m$ is nilpotent, so either $m=0$ and the ring is a field, or the ring has nilpotents.


Combining facts 1, 2, and 3 we see that a commutative Artinian ring is semisimple if and only if it is reduced.

Then $k[x]/(x^n-1)$ is reduced if the characteristic of $k$ doesn't divide $n$, so being artinian, it is semisimple.


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