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Is it possible for two distinct irreducible polynomials with integer coefficients to have a root in common? In other words, is it possible that a root is shared by some two distinct irreducible polynomials?

I would say no, because then it could be divided out by GCD of the polynomials, which would contradict their irreducibility but maybe I'm wrong?

Conversely, if two polynomials are irreducible (with positive leading coefficient) and share a root, they must be the same (i.e. they must share all the roots)?

Ecir Hana
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  • Your question is unclear: the notion of "minimal polynomial" is usually formulated for polynomials over a field and the minimal polynomial is required to be monic. Do you mean a minimal polynomial in that sense that happens to have integer coefficients? Or do you mean a polynomial of minimal degree with integer coefficients having the given root? In the former sense, the minimal polynomial is unique, but not every algebraic number has such a minimal polynomial. In the latter sense, the polynomial always exists and is unique up to multiplication by a non-zero rational. – Rob Arthan Oct 31 '18 at 20:25
  • @RobArthan You are right, I mixed up "minimal" and "irreducible", sorry about that! But I'm not sure it changes anything w.r.t the shared root, i.e. two irreducible polynomial cannot have a common root, correct? And please, out of curiosity, how is it possible that an algebraic number does not have a minimal polynomial? I thought algebraic are required to have such a polynomial, c.f. Wikipedia ("Given an algebraic number, there is a unique monic polynomial...") – Ecir Hana Oct 31 '18 at 21:28
  • Every algebraic number has a minimal polynomial with rational coefficients, but not necessarily with integer coefficients, e.g., the minimal polynomial of $1/3$ is $x - 1/3$ and $1/3$ is not the root of any monic polynomial with integer coefficients. When I said "the former sense", I was referring to my first conjecture about what you might have meant by "minimal polynomial with integer coefficients". – Rob Arthan Oct 31 '18 at 21:41
  • If you mean "irreducible" rather than "minimal", please edit your question. – Rob Arthan Oct 31 '18 at 21:42
  • Your "conversely" doesn't match your original statement: $1 - x$ and $x - 1$ are two distinct irreducible polynomials with a common root, but they share the same set of roots. – Rob Arthan Oct 31 '18 at 21:57
  • @RobArthan I meant up to sign, i.e. positive leading coefficient. – Ecir Hana Oct 31 '18 at 22:02

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If there is a root in common, then using the Euclidean algorithm for polynomials to find the highest common factor either (a) gives a lower degree polynomial of which the root is a factor (so one of the polynomials at least is not minimal); or (b) tells you that the polynomials are the same.

(This makes assumptions about the admissible coefficients of the polynomials - the Euclidean algorithm has to apply, but this will almost certainly be the case in the circumstances you envisage)

Mark Bennet
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