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Suppose there are more than one polynomials whose coefficients are not all algebraic. Could their product give a polynomial whose coefficients are only algebraic?

I'm asking with regard to factoring over $\overline{\mathbb{Q}}$ univariate or multivariate polynomials over $\overline{\mathbb{Q}}$ by factoring over $\mathbb{C}$.

I already know that values of algebraic functions of more than one non-algebraic numbers could be algebraic.

IV_
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    Well, $\frac 1ex \times ex= x^2$, for instance. – lulu Mar 15 '20 at 12:34
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    To make the question interesting, I think you should require that the two polynomials you are multiplying be monic (lead coefficient $1$). In that case the answer is no (the roots of the product are all algebraic which means the roots of the factors are algebraic which means the coefficients of the factors are algebraic). – lulu Mar 15 '20 at 13:29
  • if $fg$ has algebraic coefs and some coef of $f$ is algebraic $\neq 0$ then all coefs of $f$ and $g$ are algebraic by Dedekind's Prague Theorem – Bill Dubuque Mar 15 '20 at 15:50

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Say you have $p(x) \cdot q(x)$, monic with coefficients algebraic, and $p$ and $q$ monic. Then the factors have algebraic zeros (subsets of the zeros of the product); as their coefficients are polynomials in their zeros, they are algebraic.

vonbrand
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  • @IV_ divide by the leading coefficient. In the end, you can have polinomials $(e x- e)(x/ e + 1/e)$, which tells you nothing useful. – vonbrand Mar 15 '20 at 17:43