On coefficients of $(t-x_1)(t-x_2)..(t-x_i)$

Question

I was reading Emil Artin's Galois Theory (2nd edition). On pp.39-40, Artin defines a number of things as follows.

Let $k$ be a field and $E=k(x_1,\ldots,x_n)$ be the field of all rational functions of $n$ indeterminates $x_1,\ldots,x_n$. Define a polynomial $$ f(t)=(t-x_1)\cdots(t-x_n)=t^n+a_1t^{n-1}+\cdots+a_n,\tag{3} $$ where $a_1=-\sum_ix_i,\,a_2=\sum_{i<j}x_ix_j$,... are elementary symmetric functions in $x_1,\ldots,x_n$. Then he puts $F_n(t)=f(t)$ and also for $i=n-1,\,n-2,\ldots,1$, $$ F_i(t)=\frac{f(t)}{(t-x_{i+1})(t-x_{i+2})\cdots(t-x_n)}=\frac{F_{i+1}(t)}{t-x_{i+1}}.\tag{5} $$ (If I am not mistaken, this simply means that $F_i(t)=(t-x_1)(t-x_2)..(t-x_i)$ for each $i$, but what matters below is the division on the RHS of $(5)$.) He then writes that (emphasis mine):

Performing the division [in $(5)$] we see that $F_i(t)$ is a polynomial in $t$ of degree $i$ whose highest coefficient is $1$ and whose coefficients are polynomials in the variables $a_1,a_2,\ldots,a_i$ and $x_{i+1},x_{i+2},\ldots,x_n$. Only integer enter as coefficients in these expressions.

Can someone explain the bold part? I have thought about using Euclidean algorithm, but the $a_i$s are polynomials in $x_1,\ldots,x_n$ rather than constants. I am not sure how the division is done.

Answer 1

You just perform polynomial division as if the symbols were numbers. For example, you can check that $$ (t^3+a_1t^2+a_2t+a_3)\div(t-x_3)=t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2] $$ and $$ \left(t^2+(a_1+x_3)t+[(a_1+x_3)x_3+a_2]\right)\div(t-x_2)=t+(a_1+x_2+x_3). $$