Let $k$ be a field of characteristic $0$, $S= k[X_1,..., X_n]$ be the ring of polynomials, $L= Frac(S)$ the fraction field on $S$ and $f(T)= T^m +a_{m-1} T^{m-1}+ ...+ a_0 \in L[T]$ an irreducible polynomial. Since $char(k)=0$ this polynomial is separable and therefore has exactly $m$ different zeroes $z_1,..., z_m \in \overline{L}$ in algebraic closure $\overline{L}$ of $L$.
Let $N_L := L(z_1,..., z_m) \subset \overline{L}$ the splitting field of $f(T)$ which by definition is generated by it's zeroes $z_k$. So $L \subset N$ is a finite normal extension, therefore Galois. Therefore in $N[T]$ the polynomial $f$ splits in linear factors as $f(T)= \prod^m_{i=1} (T-z_i)$.
Then obviously all the coefficients $a_i$ of $f$ are symmetric polynomials in $z_1,..., z_m$.
Question: Why this implies that all $a_i$ are integral dependent over $S$? That is why for every $a_j$ there exist a normed polynomial $P_j(T)= T^{m_j}+ c_{j, (m_j-1)} T^{m_j-1}+ ...+c_{j, 0} \in S[T]$ with $P_j(a_j)=0$ with integral(!) coefficents.
About motivation: This argument David Mumford uses in his Red Book of Schemes on page 43 to prove Algebraic Version of Krull's Principal Ideal Theorem. Mumford also noted that this proof is due to John Tate. But I nowhere found a version of it, which might explain this step in more detail. Nevertheless here the proof
The claim is not true in general, just simply take $f(T)=T- \dfrac{1}{X_1}$ then $\dfrac{1}{X_1}$ is not integral dependent over $S$.
The result in Mumford's book, however, is correct. The essential assumption here is that, with notation in the book, $f$ is also integrally dependent over $S$, i.e, it satisfies another equation: $$Y^k + b_1Y^{k-1} + \ldots + b_k =0$$ where $b_i$ are in $S$. Therefore, all of its conjugates (zeros of the minimal polynomial $Y^n + a_1Y^{n-1} + \ldots + a_n$) are also roots of the above equation, thus, are also integrally dependent over $S$. It follows that since each $a_i$ is a symmetric polynomial on those conjugates, $a_i$ must be integrally dependent over $S$ as well.