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Suppose that $F$ is a finite field of characteristic $p$ and $f\in F[X]$.

QUESTION: Show that the quotient ring $R=F[X]/(f)$ is completely reducible Artinian ring if and only if the Frobenius map (taking elements to their $p$-th power) is an automorphism of $R$.

HINT: $f$ and its formal derivative are relatively prime if and only if $f$ has no repeated roots (in the algebraic closure).

Further...

Show $F[X]/(X^n-1)$ is a completely reducible Artinian ring if and only if $p$ does not divide $n$.

Ok... don't feel obligated to prove, I thought it was an interesting question, and would like to hear feed back about this problem. I would like to understand what is going on. Also, I know that some use semisimple for rings instead of completely reducible.

Completely reducible definition follows from Artin-Wedderburn Theorem.

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    $\begingroup$ Your last question can also be answered by applying Maschke's theorem from representation theory. When $\gcd(n,p)=1$ the group algebra $F[C_n]$ of the cyclic group is known to be semisimple. It is also isomorphic to $F[X]/(X^n-1)$ ! $\endgroup$ – Jyrki Lahtonen Mar 29 '13 at 18:26
  • $\begingroup$ @JyrkiLahtonen Nice group ring connection! $\endgroup$ – rschwieb Mar 29 '13 at 18:49
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For any $f$ with degree 1 or more, $R$ is actually finite (since its elements can only have finitely many coefficients and powers of $x$ below $\deg(f)$. So, it suffices to consider when the Frobenius map is just 1-1.

If $q(x)$ were in the kernel of the Frobenius map, then $q(x)^p=0$ implies that $q(x)$ is nilpotent. If $R$ (which is already commutative) is semisimple Artinian, then it has no nonzero nilpotent elements, so the Frobenius map is injective, hence surjective.

Assume the Frobenius map is an automorphism of $R$. Then there exists a polynomial $b(x)$ such that $b(x)^p=x$ in $R$, or in other words, $x-b(x)^p=f(x)r(x)$ in $F[x]$. Taking the formal derivative, $1=f'(x)r(x)+r'(x)f'(x)$. If $f$ and $f'$ have a common root though, evaluation gives $0=1$, which is nonsense. Thus if the Frobenius map is an automorphism, and $f$ has no repeated roots.

Finally, note that if $Y$ is a nonzero nilpotent in $R$, that means that $Y^n=f(x)r(x)$ for some $n$ and $r(x)$ in $F[x]$. We suppose that $n>1$ is minimal with that property. The left hand side says that each root of $Y$ appears $n$ times in its factorization over a splitting field. Even after one complete set of roots of $f(x)$ are cancelled away in the polynomial ring over a splitting field, there must still be a complete set of roots of $f(x)$ left to form another copy of $f(x)$. Thus $f(x)$ is a factor of $Y^{n-1}$, so $Y^{n-1}=0$ in $R$, a contradiction to $n$'s minimality. Thus there are no nonzero nilpotent elements of $R$, and $R$ is semisimple Artinian.

The second question is easy to answer assuming the connection of repeated roots proven the first part is understood. If $n=pm$, then $X^n-1=(X^m-1)^p$, and so obviously $X^m-1$ is a nonzero nilpotent in $R$. On the other hand, if $n$ and $p$ are coprime, it's clear that $X^m-1$ and its formal derivative share no roots, and so $X^m-1$ has no repeated roots. So the quotient would be semisimple Artinian.

Hope this gives you some ideas!

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  • $\begingroup$ Sorry that it's a bit clumsy: I'm just winging it. $\endgroup$ – rschwieb Mar 29 '13 at 15:13
  • $\begingroup$ Thank you, for the help. I got a lot from it. I wasn't even considering nilpotent argument. I will post the solution once the teacher gives it. If he gives it... $\endgroup$ – mccrack1985 Mar 29 '13 at 15:28
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For a different approach you can also call upon the ever useful Chinese Remainder Theorem. If $$ f(X)=\prod_{i=1}^kf_i(X)^{a_i} $$ is the factorization of $f(X)$ to a product of powers of distinct irreducible polynomials $f_i(X)\in F[X]$, then CRT tells us that $$ F[X]/(f(X))\simeq\bigoplus_{i=1}^kF[X]/(f_i(X)^{a_i}). $$

From this we see that Frobenius is an automorphism, iff $a_i=1$ for all $i$. If all the exponents $a_i$ are equal to one, then the summands here are all finite fields, and Frobenius is obviously an automorphism. OTOH, if some $a_i>1$, then the coset of $f_i(X)^{\lceil a_i/p\rceil}$ is in the kernel of the Frobenius mapping.

A sum of extension fields of $F$ is clearly a semisimple $F$-algebra (it is Artinian by virtue of being finite dimensional), so the condition, $a_i=1$ for all $i$, also implies complete reducibility. OTOH, if $a_i>1$, then the submodule $(f_i(X))/(f_i(X)^{a_i})$ has no complement, so complete reducibility is also equivalent to the assumption that $f(X)$ is square-free.

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