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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.

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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 Edoardo Oct 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 Martin Oct 18 '12 at 7:59
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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-Bailly Oct 18 '12 at 8:21
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@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 Martin Oct 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. –  zeb Oct 24 '12 at 5:36

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