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Apart from of course, the simple cases when a polynomial $f \in K[x]$ is of degree less than or equal to three. One direction is clear: If a polynomial is irreducible in $K$, it can have no roots in it. But the converse is much more bizarre. So I pose:

  • What conditions must be put on $f$ so that this happens?
  • How does information about $K$ alter this?

I do not even know where to start. Links to any research done in this area is also appreciated :) Thanks, guys!

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an obvious thing: if there are no irreducible polynomials of degree $2$ (i.e. no extensions of $K$ of degree $2$) then you have your property for polynomials of degree $4$. Etc. – user8268 Sep 7 '12 at 8:27
@user8268: why don't you post that as an answer? – Marc van Leeuwen Sep 7 '12 at 8:44
In fact, if there are no degree 2 irreducibles, property holds in degree 5 as well. – Gerry Myerson Sep 7 '12 at 9:45
What are examples of non algebraically closed fields with no degree $2$ extensions? – Georges Elencwajg Sep 7 '12 at 12:11
@Georges, I don't know, but, reasoning naively, if you start with the rationals, and throw in square roots of all rationals, and then square roots of everything in that field, and iterate to infinity, you ought to reach a field with no quadratic extensions but not algebraically closed because not containing, for example, the cube root of 2. – Gerry Myerson Sep 7 '12 at 13:04

2 Answers 2

Here are two results (to be found in Lang's Algebra, for example):

  1. Let $k$ be a field of any characteristic $\geq 0$ and $p$ an odd prime number (not supposed to equal $\operatorname{char} k$) If $x^p-a\in k(x)$ has no root in $k$, then it is irreducible and, more strongly, also $x^{p^n}-a$ is irreducible over $k$ for all integers $n\geq 0$ (Capelli).
  2. Over a field $k$ of characteristic $p\gt0$ the polynomial $f(x)=x^p-x-a\in k[x]$ is irreducible if and only if it has no root in $k$ (Artin-Schreyer).
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This is not an answer in the strict sense but a result strongly related to your question.

Let $k$ be a field and let $f(x)\in k[x]\setminus k$ be a polynomial of degree $n\gt0$. We have the equivalence:

$f(x)$ irreducible over $k\\$ $\iff$ For any extensions $k\subset K$ of degree $\leq n/2$, $f(x)$ has no root in $K$.

Proof of $\Longrightarrow:$
Any field $K$ in which $f(x)$ has a root contains a copy of $k[x]/(f(x))$ and has thus degree $\geq n$
Proof of $\Longleftarrow:$
If $f(x)$ were reducible one of its irreducible factors $g(x)$ would have degree $deg(g(x))\leq n/2$ and thus the extension $K:=k[x]/(g(x))$ would contain a root of $g(x)$ and a fortiori a root of $f(x)$, contradicting the hypothesis.

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