If P(x) is a polynomial in Q[X], is there any iff theorem that states all the roots of P(x) are rational based on the coefficients?!
In another words, what could you impose on the coefficients to make sure a polynomial in Q[x] will split over Q.

Do you have in mind some special kind of requirements (e.g. size, or something like this)? Are you happy with "all the roots you need to extract in a formula for the general solution need to be rational" for degree less than $5$? I guess not, but it would be interesting to know if you have something in mind.
– Filippo Alberto EdoardoOct 18 '12 at 7:22

Almost surely no such if-and-only-if theorem exists. Even getting nontrivial sufficient conditions for a polynomial to split over $\mathbb Q$ would be very challenging.
– Greg MartinOct 18 '12 at 7:59

16

The rational roots of a given $P$ (with integer coefficients, say) can be found effectively: if a root is written $a/b$ in lowest terms, then $b$ (resp. $a$) must divide the leading coefficient (resp. the constant term). Isn't this an answer?
– Laurent Moret-BaillyOct 18 '12 at 8:21

2

@Laurent: it is an answer, but I don't find it satisfying. As a theoretical statement, it says "a polynomial has all rational roots if and only if all of its roots are in this particular finite set"; that gives us no new information on how to characterize such polynomials using their coefficients. As a computational statement, one needs to know the factorization of $a$ and $b$ to compute the possible roots, so it doesn't lead to a polynomial-time algorithm for determining whether a polynomial has all rational roots.
– Greg MartinOct 24 '12 at 5:20

Is jstor.org/stable/2691297 helpful? For cubic polynomials, at least, it shows that you can check whether all the roots are rational by checking that the discriminant is a square and that a certain element of Q[w] is a cube, for w a cube root of unity.
– zebOct 24 '12 at 5:36