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Suppose we have an irreducible polynomial $f\in K[x]$. Is there some way to sometimes tell whether $f$ splits completely after adjoining just one root of $f$ to $K$?

I am mostly interested in the case where $K$ is a function field $\mathbb{F}_{q}(t_{1},\ldots,t_{m})$ over some finite field, so it might not be feasible to explicitly compute roots.

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You are asking if $f$ generates a Galois extension, and this is very hard to tell in general. If you replace $K$ with a local field (such as $\mathbb{Q}_p$, or $\mathbb{F}_p((x))$) then there is an answer when $f$ has a special form, an in particular when $p$ does not divide the degree, or the degree is exactly $p$. Just to say that even in this simplified setting a general answer is very hard to obtain. – Maurizio Monge Jul 11 '11 at 11:31

Here is the best I can come up with. Consider an algebraic closure $\bar K$ of $K$ and a root $\alpha \in \bar K$.

The number of roots of $f$ in $K(\alpha)$ doesn't depend on $\alpha$: call it $ r(f)$
Moreover call $s(f)$ the number of the different subfields $K(\alpha)\subset \bar K$ obtained by adjoining roots of $f$ to $K$. Then you have the pleasant equality $$deg(f)=r(f).s(f)$$ This shows in particular that the number of roots that you get by just adjoining one root divides the degree $deg(f)$ of your polynomial.
For example if $K=\mathbb Q$ and $f(x)=X^8-2$ you have $r(f)=2$ and $s(f)=4$, since the fields you get by adjoining roots of $f$ to $\mathbb Q$ are [with $\omega =\frac{1}{\sqrt 2}(1+i)$]:
$\mathbb Q(\sqrt[4]2)$
$\mathbb Q(\pm \omega \sqrt[4]2)$
$\mathbb Q(\pm \bar{\omega} \sqrt[4]2)$
$\mathbb Q(\pm i \sqrt[4]2)$

These results are due to Perlis , and although not difficult have found their way in exactly zero books, as far as I am aware.

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Still, i would say that this result is very well known in the field. It is an immediate consequence of the fact the fields $K(\alpha)$ obtained adding one root of $f$ are all isomorphic. – Maurizio Monge Jul 11 '11 at 11:36
Dear Maurizio, Perlis states in his article "An informal survey of books and colleagues indicates that the divisibility result “r(f) divides the degree” is not well known". So this is obviously subjective, as with all results "very well known in the field" [by the way, was your pun intended? :-) ]. I'm still curious to know whether there is a book mentioning the result. – Georges Elencwajg Jul 11 '11 at 12:01
Eh, no pun intended :D. In any case i agree that there is a great amount of thing which are in some way "well known" to specialists without being written anywhere, and i think it is a very important work pointing them out and clarifying them. – Maurizio Monge Jul 11 '11 at 12:46
I believe it was Frank Adams who liked to draw a distinction between well-known and well-knowable. (Certainly he was careful to distinguish between oriented and orientable.) – Tom Goodwillie Jul 11 '11 at 15:04
"Well-known" is akin to the notion of "it is generally believed that" == "I believe it and one other person known to me believes it" (see: – Suvrit Jul 11 '11 at 18:11

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