# Dummit and Foote page 591

Having trouble with something that Dummit and Foote is saying. On page 591 it says "Note that over $\mathbb{Q}$ or over a finite field (or, more generally, over any perfect field) the splitting field of an arbitrary polynomial $f(x)$ is the same as the splitting field for the product of the irreducible factors of $f(x)$ taken precisely once, which is a separable polynomial."

It is the last sentence that doesn't make sense because...

take $F_p$ to be our finite field. Take some element $\alpha$ that is not in $F_p$ so that $\alpha^p$ is. Then $f(x):=x^p-\alpha^p$ is irreducible over $F_p$ because $f(x)=(x-\alpha)^p$ so if this is a product of two factors, $f(x)=(x-\alpha)^i(x-\alpha)^{p-i}$ and the $x^{i-1}$ coefficient of the first factor is $i\alpha$ which is not in $F_p$. We also see that this irreducible polynomial has multiple roots in the extension field containing $\alpha$ which is evidently a splitting field for this polynomial. So $f(x)$ is not separable.

Thus $f(x)$ is the polynomial we get by taking the irreducible factor $x^p-\alpha^p$ once. But $f(x)$ is certainly not separable as dummmit and foote says it should be.

• Why do you think such an element $\alpha$ exists? $\mathbf{F}_p$ being perfect implies that the unique $p$th root of every element of $\mathbf{F}_p$ also lies in $\mathbf{F}_p$. – Brandon Carter Dec 26 '14 at 15:54

Remember Fermat's little theorem: $a^p=a$ for all $a\in\mathbb F_p$.
In particular, for every $a\in\mathbb F_p$ we have $x^p-a=x^p-a^p=(x-a)^p$, so $a$ is a $p$-fold root of $x^p-a$, and thus the only possible root of $x^p-a$ in any field extension of $\mathbb F_p$.
So there cannot be any $\alpha$ outside $\mathbb F_p$ whose $p$th power is in $\mathbb F_p$, and your construction is impossible from the outset.
If you try to create such an $\alpha$ by force, say by adjoining a new $p$th root of $a$ to $\mathbb F_p$, what you get is a ring with nontrivial zero divisors in it (namely, $(a-\alpha)^p=0$ but $a-\alpha\ne 0$), which cannot be extended to a field.
• Oh I get it. I need the ground field to be a field that is not perfect for this construction to work. But then it will not be a finite field. It will be just an ordinary field of characteristic p. I would take $\alpha$ to be an element in $F\F^p$(set of pth powers of elements in $F$). Thanks for all the replies. – Hari Rau-Murthy Dec 26 '14 at 16:45
• For example I could take $F=F_p(t)$ with $t$ transcendental. Let $t^{1/p}$ to be the unique root of the polynomial $x^p-t$. Then the construction works with the ground field being $F(t)$ and I learn that $x^p-t$ is irreducible. So its not completely useless. Just can't use it to give a counterexample for what Dummit and Foote are saying. – Hari Rau-Murthy Dec 26 '14 at 16:56