# Prove coefficients of polynomial are elementary symmetric polynomials

I want to show that for the $k$-th elementary symmetric polynomial $s_k:=\sum_{i_1\lt\cdots\lt i_k}X_{i_1}\cdots X_{i_k}\in R[X_1,\ldots,X_n]$ a monic polynomial that factors $\prod_{i=1}^n (X-\alpha_i)$ has the elementary symmetric polynomials as coefficients, e.g. $$\prod_{i=1}^n (X-\alpha_i)=X^n-s_1(\alpha_1,\ldots,\alpha_n)X^{n-1}+\cdots+(-1)^n s_n(\alpha_1,\ldots,\alpha_n).$$ Is this possible by induction?

• Yes, just write it out. – aes Dec 3 '14 at 21:10

Write $$\prod_{i=1}^{n-1}(X-\alpha_i)=X^{n-1}-s_1(\alpha_1,\ldots,\alpha_{n-1})X^{n-2}+\cdots+(-1)^{n-1} s_{n-1}(\alpha_1,\ldots,\alpha_{n-1})$$ and multiply this by $X-\alpha_n$. You get $$X^n-(s_1(\alpha_1,\ldots,\alpha_{n-1})+\alpha_n)X^{n-1}+\cdots+(-1)^{n} s_{n}(\alpha_1,\ldots,\alpha_{n-1})\alpha_n.$$
Maybe this is not very convincingly (because only two coefficients are explicit), but if you look at the coefficient of $X^{n-k}$ you can see that this is $$s_{k}(\alpha_1,\ldots,\alpha_{n-1})+s_{k-1}(\alpha_1,\ldots,\alpha_{n-1})\alpha_n=s_{k}(\alpha_1,\ldots,\alpha_{n})$$ (times $\pm$, of course).
The coefficient of $X^k$ is the sum of all products of $n-k$ distinct terms of the form $-\alpha_i$. This is the elementary symmetric polynomial of degree $n-k$ in $\alpha_1,\ldots,\alpha_n$ times $(-1)^{n-k}$ as we may identify these distinct terms by the sequence of their indices in increasing order.