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A small proposition in Ash's Algebra states that over a field $F$ of prime characteristic $p$, an irreducible polynomial $f$ is inseparable if and only if $f'$ is the zero polynomial, or equivalently, $f$ is a polynomial in $X^p$.

A later exercise asks the following. Let $f$ be an irreducible polynomial in $F[X]$ where $F$ has characteristic $p\gt 0$. Express $f(X)$ as $g(X^{p^m})$ where the nonnegative interger $m$ is as large as possible. Show that $g$ is irreducible and separable.

I get that $g$ is irreducible, since $f$ is. But if $g$ is a polynomial in $X^{p^m}$ I can write it as $$ g=a_0+a_1X^{p^m}+a_2(X^{p^m})^2+\cdots+a_n(X^{p^m})^n $$ so $$ g'=p^ma_1X^{p^m-1}+2p^ma_2X^{2p^m-1}+\cdots+np^ma_nX^{np^m-1}. $$ But then each coefficient is divisible by $p$, so isn't $g'$ identically $0$, meaning $g$ is inseparable by the proposition stated above? The provided solution states that since $m$ is maximal $g\notin F[X^p]$, (which I don't quite get, why can't $m=1$?), and so $g$ is separable, contrary to what I say. What have I done wrong here? Thanks.

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$g$ is a polynomial such that $g(X^{p^m}) = f(X),$ it is not the polynomial $g(X^{p^m}).$ Take for example the case where $f(X) =X^p + c.$ Then $g(X) = X + c$ and $g$ is irreducible and separable as desired. –  jspecter Jun 16 '11 at 20:00
    
@jspecter Oh, I think I see where I went wrong now. Thanks for the clarification. –  yunone Jun 16 '11 at 20:04

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I think you are confusing $f$ and $g$. It is $f$ that is a polynomial in $X^p$. For example, if you were working in a field of characteristic $3$, you could have $$f(x) = 1 + 2x^3 -x^6 + x^9.$$ To express it as a polynomial in $x^3$, you would write $$f(x) = 1 + 2(x^3)^1 - (x^3)^2 + (x^3)^3$$ so the polynomial $g(u)$ would be $$g(u) = 1 + 2u - u^2 + u^3,$$ which, when evaluated at $x^3$, gives $f(x)$. That is, $f(x)=g(x^3)$. But notice that $g(x) = 1+2x - x^2 + x^3$ is in fact separable.

So your $g(x)$ above should be $$g(x) = a_0 + a_1X + a_2x^2 + \cdots + a_nx^n.$$ (As written, $g(x)$ itself can be expressed as $h(x^{p^m})$, where $h(x)$ is the polynomial I have just written, so if $f(x) = g(x^{p^n})$, then in fact $f(x) = h(x^{p^{n+m}})$ and you would use $h(x)$, not $g(x)$).

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Thanks Arturo, I think I see how I misinterpreted what $g$ was. In your last paragraph or so, is the idea to let $n$ be the highest power of $p$ which divides all the exponents in $f$? And then if I were to try to express $g(X)$ as $h(X^{p^m})$, I would necessarily have to take $m=0$, so that $f(x)=h(X^{p^{n+0}})$? I hope I haven't written gibberish... –  yunone Jun 16 '11 at 20:20
    
@yunone: Yes, you are supposed to take $n$ to be the largest possible power of $p$ that you can; I was addressing the possibility that you had begun with an even higher power $f$ and hadn't actually gone "as deep as one can". –  Arturo Magidin Jun 16 '11 at 20:24
    
I see, thank you for your answer. –  yunone Jun 16 '11 at 20:27

$g$ is the polynomial in $X$, such that when $X^{p^m}$ is substituted in for $X$, we get $f$, i.e. $g(X^{p^m})=f(X)$.

Let $$f(X)=a_0+a_1X^{d_1}+\cdots+a_nX^{d_n}$$ for some $d_i\in\mathbb{N}$. Suppose $d_i=e_ip^{m+1}$ for all $i$. Then $$g(X)=a_0+a_1X^{e_1p}+\cdots+a_nX^{e_np}$$ is the polynomial such that $$g(X^{p^m})=a_0+a_1(X^{p^m})^{e_1p}+\cdots+a_n(X^{p^m})^{e_np}=$$ $$a_0+a_1X^{e_1p^{m+1}}+\cdots+a_nX^{e_np^{m+1}}=a_0+a_1X^{d_1}+\cdots+a_nX^{d_n}=f(X).$$ However, we can let $h(X)=a_0+a_1X^{e_1}+\cdots+a_nX^{e_n}$. In other words, $h$ is the polynomial (in $X$) such that $h(X^p)=g(X)$. We will have $$f(X)=g(X^{p^m})=h(X^{p^{m+1}}),$$ which contradicts the maximality of $m$.

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@yunone: My answer avoided explicitly talking about the exponents in $f$, only showing that if $g$'s exponents were all divisible by $p$ we would have a contradiction; I've edited to be more explicit. –  Zev Chonoles Jun 16 '11 at 20:35
    
Thanks for the addition Zev, it's now clear to me why $g\notin F[X^p]$ like the appendix in the books says. –  yunone Jun 16 '11 at 20:35
    
@yunone: Actually I see that in my initial answer, I should not have written $$f(X)=h(X^{p^{m+1}})=a_0+a_1X^{r_1}+\cdots+a_nX^{r_n},$$ that is incorrect; your complaint about $f$ being both $a_0+a_1X^{r_1}+\cdots+a_nX^{r_n}$ and $a_0+a_1X^{r_1p}+\cdots+a_nX^{r_np}$ was valid. –  Zev Chonoles Jun 16 '11 at 20:41

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