Let $f(x) = s_{0} + s_{1}x +···+ s_{n−1}x^{n−1} + x^n \in k[x]$, where $k$ is a field, and suppose that $f(x) = (x − \alpha_{1})(x − \alpha_{2})···(x − \alpha_{n})$. Prove that $s_{n-1} = −(\alpha{1} + \alpha_{2} +···+ \alpha_{n})$ and that $s_{0} = (−1)^{n}\alpha_{1}\alpha_{2} ··· \alpha_{n}$. Conclude that the sum and product of all the roots of $f(x)$ lie in $k$.
Im thinking about proving the first part by induction over $deg(f(x))\geq 1$. If $deg(f(x))= 1$, then $f(x)=s_{0}+x$, where $f(x)$ has at most one root which is clearly $\alpha_{1}=-s_{0}$ and $s_{0}=(-1)^{1}\alpha_{1}$, so $f(x)=(x-\alpha_{1})=(x+s_{0})$. So induction basis is satisfied. Then I formulate my induction hypothesis for $deg(f(x))=n-1$ and $f(x) \in k[x]$ is monic. The big question is how to conclude that
$s_{n-1} = −(\alpha{1} + \alpha_{2} +···+ \alpha_{n})$ and that $s_{0} = (−1)^{n}\alpha_{1}\alpha_{2} ··· \alpha_{n}$ just by the fact this happens for $n-1$ and we have a linear factorization of $f(x) \in k[x]$??
No problems concluding this two facts implies sum and product of the respective roots of $f(x)$ lie in $k$. Thanks.
You are almost there. You can simply say $(x - \alpha_1) \ldots (x - \alpha_n) = x^{n - 1} + t_{n - 2}x^{n - 2} +\ldots + t_0$ any by the hypothesis $t_{n - 2}$ is $-(\alpha_1 + \ldots + \alpha_{n - 1}).$ Now you can simply multiply out $x^{n - 1} +t_{n - 2}x^{n - 2} + \ldots + t_0$ with $(x - \alpha_n).$ You can deduce the coefficient of $x^{n - 1}$ of this product by using degrees. Clearly the $x^{n - 1}$ term would have to arise from $t_{n - 2}x^{n - 1} - \alpha_nx^{n - 1}.$ Similarly, you can conclude the expression for $s_0.$