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 $\mathbb Q$?

  • $\begingroup$ 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 Willse 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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. )

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Thanks, quid. I'll try to be more precise in future. $\endgroup$ – P Vanchinathan Jun 4 '15 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.