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This is question is in light of comments in this question and another question I asked few days back.

Quoting Hayden from the first link,

You can use Rational Root Theorem to show a polynomial is irreducible (over $\mathbb Q$) when the degree is 2 or 3, but for any higher degree you would need to do more than just that. .. .For example, $(x^2+1)^2$ doesn't have any rational roots, but that doesn't mean it's irreducible.

which is true.

Now, in a general case, if a polynomial has a root in a field $\mathbb F$, then it can easily be concluded that it is not irreducible over $\mathbb F$.

My question is,

1) Are there any established results with conditions under which a polynomial's irreducibility over $\mathbb F$ (Or in particular well known Fields like $\mathbb Q$ or $\mathbb R$ etc) is only dependent on it having no roots in $\mathbb F$? (As Hayden mentioned, for polynomials of degree 2,3 in $\mathbb Q$, irreducibility over $\mathbb Q$ can be proved by showing it has no rational roots - are there any such similar conditions for higher polynomials, in general case?)

2) For irreducibility over $\mathbb Q$, are there certain classes of polynomials identified (like $(x^n+1)^k,n\ge2,k\ge2$) which are not irreducible even if they don't have roots in $\mathbb Q$?

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  • $\begingroup$ What sort of conditions are allowed? It's not hard to see the condition "$p$ does not have a quadratic factor" is sufficient for the equivalence to hold when $\deg p$ is $4$ or $5$, but I assume this is more trivial than what you're looking for. $\endgroup$ – Travis Jun 3 '15 at 12:27
  • $\begingroup$ @Travis, yours obviously fits in! I think I did the mistake of asking something too broad, and should have narrowed it down a bit more. Thanks anyway! $\endgroup$ – Jesse P Francis Jun 3 '15 at 13:50
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Suppose that every root-free polynomial in $F$ is irreducible. Then $F$ is algebraically closed.

To show this, suppose by contradiction that exists some $g \in F[x]$ irreducible of degree $\ge 2$. Then $g^2 \in F[x]$ has no roots, so it is irreducible. And this is clearly a contradiction. So every irreducible polynomial has degree $\le 1$, i.e. $F$ is algebraically closed.

In general it is very easy to construct reducible root-free polynomials: simply take any product of root-free polynomials.

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  • $\begingroup$ First of all, thank you. But for (2), I was more looking for special general classes (or examples, like one I quoted): its easy to construct a random reducible root free polynomial! $\endgroup$ – Jesse P Francis Jun 3 '15 at 14:39
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    $\begingroup$ As for (2): I hardly believe that a satisfactory answer exists. I suggest you to consider all polynomials of the type $f(x^k)$ where $k \ge 2$ is an integer, and $f$ is a polynomial whose roots are not $k$-th powers in $\Bbb{Q}$. For example, consider $f(x) = x^2+x-6=(x-2) (x+3)$. Then $f(x^k)=x^{2k}+x^k-6=(x^k-2)(x^k+3)$ is reducible and has no roots. $\endgroup$ – Crostul Jun 3 '15 at 14:56
  • $\begingroup$ I guessed, asked just in case there are some well known examples! I think you should add that to the answer as well! Thanks again! :) $\endgroup$ – Jesse P Francis Jun 3 '15 at 16:16
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Start with two irreducible polynomials $f(x), g(x)$ with coefficients in some field $F$, of any degree $m$ and $n$. The polynomial $h(x)$ defined as the product of $f(x)$ and $g(x)$ is reducible but has no roots in $F$.

EDIT: Adding precision to the above. Choose the degrees $f(x)$ and $g(x)$ to be positive integers that are at least 2. (It is implicitly assumed the field is not algebraically closed. )

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    $\begingroup$ Surely, you need to have $g$ and $f$ to have no roots. $\endgroup$ – quid Jun 3 '15 at 12:33
  • $\begingroup$ As they are irreducible by choice they have no roots. $\endgroup$ – P Vanchinathan Jun 3 '15 at 16:34
  • $\begingroup$ $X-1$ is irreducible. $\endgroup$ – quid Jun 3 '15 at 16:38
  • $\begingroup$ @quid: You have succeeded in proving I am wrong. I concede. But the spirit of question can be guessed. I stand corrected. Choose the degrees of $f(x), g(x) > \pi/2$ $\endgroup$ – P Vanchinathan Jun 3 '15 at 16:44
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    $\begingroup$ Thanks, quid. I'll try to be more precise in future. $\endgroup$ – P Vanchinathan Jun 4 '15 at 0:34

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