Cases where irreducibility of polynomials over $\mathbb F \iff$ it has no roots in $\mathbb F $

Question

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 $\mathbb Q$?

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. — Travis Willse, Jun 03 '15 at 12:27
@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! — Jesse P Francis, Jun 03 '15 at 13:50
Beware these results may fail when the coef ring is not a domain, e.g. a reducible cubic with no roots. — Bill Dubuque, Nov 17 '23 at 18:48

score 5 · Accepted Answer · answered Jun 03 '15 at 12:27

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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.

answered Jun 03 '15 at 12:27

Crostul

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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! – Jesse P Francis Jun 03 '15 at 14:39
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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. – Crostul Jun 03 '15 at 14:56
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! :) – Jesse P Francis Jun 03 '15 at 16:16

P Vanchinathan · Answer 2 · 2015-06-04T00:45:21.857

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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. )

edited Jun 04 '15 at 00:45

answered Jun 03 '15 at 12:30

P Vanchinathan

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Surely, you need to have $g$ and $f$ to have no roots. – quid Jun 03 '15 at 12:33
As they are irreducible by choice they have no roots. – P Vanchinathan Jun 03 '15 at 16:34
$X-1$ is irreducible. – quid Jun 03 '15 at 16:38
@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 $f(x), g(x) > \pi/2$ – P Vanchinathan Jun 03 '15 at 16:44
But such polynomials might not exit; and, they do not exist iff the field is algebraically closed. Thus, I think it important to be clear on that point. – quid Jun 03 '15 at 17:06
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Thanks, quid. I'll try to be more precise in future. – P Vanchinathan Jun 04 '15 at 00:34

score 0 · Answer 3 · answered Apr 23 '23 at 20:57

Here is a nice class of examples: for a prime $p$ and $a \in F^\times$, the polynomial $x^p - a$ is irreducible in $F[x]$ if and only if it has no root in $F$.

For $p = 2$ and $3$ this is obvious, since those are special cases of the fact that for every polynomial of degree $2$ or $3$ in $F[x]$, irreducibility is equivalent to the lack of roots in $F$. That it is true for $x^p-a$ with prime degree $p > 3$ is surprising, but true. And it works over all fields $F$, even those of characteristic $p$. This is an important step towards classifying when a general binomial polynomial $x^n - a$ is irreducible over a field.

This answer has a proof of the more general result: $\ x^n - a\ $ is irreducible over $F$ $ \iff a \not\in F^{\large p},$ for all primes $,p\mid n,,$ and $\ a\not\in -4F^4$ when $: 4\mid n.\ \ $ — Bill Dubuque, Apr 24 '23 at 18:01

Cases where irreducibility of polynomials over $\mathbb F \iff$ it has no roots in $\mathbb F $

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