# Fields of characteristic $p$ where there exists an element $a$ that is not a p:th power.

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.

• Why would we have $(h_0g_0)^p = h_0g_0$? Nov 18, 2023 at 6:40
• 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. Nov 18, 2023 at 6:51
• Any suggestion for how I can show irreducibility then, for the first part? And also, any comment on the second part? Nov 18, 2023 at 6:52
• 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 $\operatorname{char} = p \neq 0$. Nov 18, 2023 at 6:56

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 $$deg(h_1(x))=p \implies h_1(x) = p(x).$$

• 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. Nov 19, 2023 at 8:53
• 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. Nov 19, 2023 at 9:00
• You are correct.. Nov 19, 2023 at 14:18