## 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 $a(x) = a^\prime(x) x^{tn} \in F[x]$.