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Suppose $K$ is a field of characteristic $p$, but that $K \neq K^p$, that is, there exists an element $a \in K$ such that there is no $b \in K$ so that $b^p = a$. We want to prove that this implies that there exists an irreducible inseparable polynomial over $K$.

Now, if we investigate $$p(x) = x^p -a$$

we see that clearly it has no zeroes in $K$.

Furthermore, we want to show that it is irreducible.

Am I correct in that: if it was reducible, we would have $$p(x) = h(x)g(x)$$ so that the constant terms $h_0$ and $g_0$ would be such that $$h_0g_0 = -a.$$

If neither $h_0$ nor $g_0$ were $a$, we would have $$(h_0g_0)^p = h_0g_0 = -a.$$

Now, this would imply that $$(-h_0g_0)^p = -(h_0g_0)^p = a$$ given that $p \neq 2$.

This is a contradiction to our assumption. So if we are not in characteristic $2$, either $g_0 = a$ or $h_0 = a$ and $h_0 = -1$ och $g_0 = -1$ resp.

But then we get $(-1+a)$ as the coefficient infront of $x^1$ which does not vanish unless $a = 1$, but if $a = 1$ then $a^p = a = 1$, contradiction!.

Since this removes all possibilities for $h(x),g(x)$ it is irreducible.

In the case of characteristic $2$, we would have $$x^2-a = h(x)g(x)$$ where $h_0g_0 = -a$ and $h(x) = x+h_0$ and $g(x) = x+g_0$ so that $2h_0g_0x = 0 \implies 2h_0g_0 = 0$ which is impossible unless $h_0 = 0$ or $g_0 = 0$. But either way we are left with $$h(x)g(x) = (x+h_0)x \neq x^2-a$$

or $$h(x)g(x) = x(x+g_0) \neq a.$$

First of all, is this reasoning correct?

Second, if it is correct, I am not sure if I really can conclude that $$p'(x) = 0$$ implies inseparability directly, if we are in $\operatorname{char} = p$.

What I can conclude is that $$p(x) = q(x^p)$$ where $q(x) = x-a$.

So it has separability degree $1$ and inseparability degree $p$, if I am not mistaken. Am I correct in that $p(x) = x^p-a$ is inseparable if and only if it´s inseparability degree is $p^0 = 1$? In that case, I believe I am done.

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  • $\begingroup$ Why would we have $(h_0g_0)^p = h_0g_0$? $\endgroup$ Nov 18, 2023 at 6:40
  • $\begingroup$ hm, right, that might not hold, correct? I might implicitly have assumed that we are in a finite field, where this holds to fermats little theorem. $\endgroup$
    – Ben123
    Nov 18, 2023 at 6:51
  • $\begingroup$ Any suggestion for how I can show irreducibility then, for the first part? And also, any comment on the second part? $\endgroup$
    – Ben123
    Nov 18, 2023 at 6:52
  • $\begingroup$ I know that in a field $F$ of characteristic $0$, we have that if $p'(x) = 0$ then $p(x) \in F[x]$ is separable. I am still not entirely sure on how to handle the case when $p'(x) = 0$ but where we are looking at a field of $\operatorname{char} = p \neq 0$. $\endgroup$
    – Ben123
    Nov 18, 2023 at 6:56

1 Answer 1

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I can't agree with the step $$(h_0g_0)^p = h_0g_0.$$

One argument may go like this:

Clearly $p(x)$ is inseperable and all the roots of $p(x)$ are equal in some splitting field $F$ of $p(x)$. That is, in $F[X]$, we may have $$p(x) = (x-b)^p,$$ for some $b\in F$. Now, suppose $$p(x) = h_1(x)h_2(x)\dots h_k(x),$$ is the expression of $p(x)$ as product of irreducible polynomials over $K$, then each of the polynomial $h_i(x)$, $1\leq i \leq k$ has got $b$ as the only root in $F[X]$. We also have $$\sum\limits_{i=1}^k deg(h_i(x)) = p.$$ As the minimal polynomial of $b$ divides each $h_i(x)$ and as each $h_i(x)$ is irreducible, we must have that $$ p(x) = (h_1(x))^p,$$ implying that $k\times deg(h_1(x)) = p$, a contradiction unless $k=1$ and $deg(h_1(x))=p$ or $k=p$ and $deg(h_1(x))=1$. But the second case is not possible as $p(x)$ has not root in $K$. Hence we must have $$deg(h_1(x))=p \implies h_1(x) = p(x).$$

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  • $\begingroup$ I don´t see why you would get $$p(x) = (h_1(x))^p.$$ Why should this be true? I am asking specifically about the $p$ in the exponent. $\endgroup$
    – Ben123
    Nov 19, 2023 at 8:53
  • $\begingroup$ It is correct though, I realized why now. You can just assume $n$, but you will find that then the order of the minimal polynomial, or if you want, h_1(x), must be of such order that deg(h_1(x)^n) = deg(h_1(x))+...+deg(h_1(x)) = p iff ndeg(h_1(x)) = p which since p is a prime is not possible unless n = p and deg(h_1(x)) = 1 or n = 1 and deg(h_1(x)) = p. $\endgroup$
    – Ben123
    Nov 19, 2023 at 9:00
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    $\begingroup$ You are correct.. $\endgroup$
    – Yathi
    Nov 19, 2023 at 14:18

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