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 $\mathbb Q$?

Answer 1

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.

Answer 2

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. )