Let $B$ be a ring extension of $A$, a commutative ring with $1$. Suppose $y$ and $z$ in $B$ are integral over $A$.
Prove that $z + y$ and $zy$ are integral over $A$ using elementary symmetric functions.
Since $y$ and $z$ are integral over $A$ there exist monic polynomials $f, g \in A[x]$ such that $f(y) = 0$ and $g(z) = 0$. Assume that $\deg(f) = n$ and $\deg(g) = m$. There is an extension, say $L$, of $A$ such that both $f$ and $g$ split into linear factors in $L$: $$ f(x) = (x-y_1)(x-y_2)\dots (x-y_{n}) \\ g(x) = (x-z_1)(x-z_2)\dots (x-z_{m}) $$ where $y_1 = y$ and $z_1 = z$.
The coefficients $s_i$ of $f$ and $s'_j$ of $g$ in $L$ are symmetric polynomials in the roots $y_i$ and $z_j$ respectively.
Define two polynomials $p$ and $q$ over $L$: \begin{align*} p(x) &= \prod_{i=1}^{n} \prod_{j=1}^{m} (x - (y_i + z_j)) \\ q(x) &= \prod_{i=1}^{n} \prod_{j=1}^{m} (x - y_i z_j) \end{align*}
I have a hint that the coefficients of these two polynomials can be expressed in terms of $s_i$ and $s'_j$ (Why?). Even if it is true, why does it imply that $p$ and $q$ are in $A[x]$?