The sum and product of algebraic elements are algebraic

Let F be a field. Let p(x),q(x) \in F[x] be monic, and let E be an extension of F containing the roots of p and q. Say p(x) = \prod_{i=1}^n (x - \alpha_i) and q(x) = \prod_{j=1}^m (x - \beta_j). Define a(x) = \prod_{i=1}^n \prod_{j=1}^m (x - \alpha_i - \beta_j) and b(x) = \prod_{i=1}^n \prod_{j=1}^m (x - \alpha_i\beta_j) in E[x]. Prove that a(x) and b(x) are in F[x].

Note that, for each j, p(x - \beta_j) = \prod_{i=1}^n (x - \alpha_i - \beta_j). Thus a(x) = \prod_{j=1}^m p(x - \beta_j). For a moment thinking of the \beta_j as indeterminates, a(x) is fixed by any permutation of the \beta_j. In particular, each coefficient of a(x) is a symmetric polynomial in the \beta_j. Because the \beta_j are the roots of the polynomial q(x) \in F[x], every symmetric polynomial in the \beta_j is in F. So a(x) \in F[x].

Suppose none of the \beta_j is zero. Now p(x/\beta_j) = \prod_{i=1}^n (x/\beta_j \alpha_i), and so \beta_j^n p(x/\beta_j) = \prod_{i=1}^n (x - \alpha_i\beta_j). So b(x) = \prod_{j=1}^m \beta_j^n p(x/\beta_j). Again, this product is fixed by any permutation of the \beta_j, so that each coefficient of b(x) is symmetric in the \beta_j. Thus b(x) \in F[x].

If we have q(x) = x^tq^\prime(x), where none of the roots \{ \beta_j \}_{j=1}^{m-t} of q^\prime is zero, then by the previous argument the polynomial a^\prime(x) = \prod_{i=1}^n \prod_{j=1}^{m-t} (x - \alpha_j \beta_j) is in F[x]. Certainly then a(x) = a^\prime(x) x^{tn} \in F[x].

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