Show that $\left\langle \sigma_1-a_1,\dots,\sigma_n-a_n \right\rangle $ is a radical ideal

Question

Let $\sigma_i$ is the $i$-th elementary symmetric polynomial, $\mathrm{char}(k)=0$. I want to show the zero-dimensional ideal $$\left\langle \sigma_1-a_1,\dots,\sigma_n-a_n \right\rangle $$ is radical, provided that $\mathrm{char}(k)=0$ and $$ x^n-a_1x^{n-1}+\dots +(-1)^n a_n$$ is separable.

By Vieta's theorem, we know $|\mathbb{V}(I)|=n!$. It suffices to show the dimension of $k$-vector space $$ \dim_k(k[x_1,\dots,x_n]/\left\langle \sigma_1-a_1,\dots,\sigma_n-a_n \right\rangle)=n! $$

But I have no idea how to show this. Or maybe are there any other ways to show it is radical?

Thanks in advance!

Answer 1

The $k$-algebra $k[x_1,\dotsc,x_n]/\langle \sigma_1(x_1,\dotsc,x_n) - a_1, \dotsc, \sigma_n(x_1,\dotsc,x_n) - a_n \rangle$ is the universal $k$-algebra in which we have elements $x_1,\dotsc,x_n$ satisfying the equations $\sigma_i(x_1,\dotsc,x_n) = a_i$. These equations are equivalent to the polynomial equation (where $x$ is a separate variable) $$\sum_i (-1)^{n-i} a_i x^i = \prod_{i=1}^{n} (x - x_i).$$ So this is just the splitting algebra of the polynomial, which is well-known to be free of rank $n!$. The basis consists of the $x_1^{k_1} \dotsc x_n^{k_n}$ with $0 \leq k_i \leq n-i$.

T. Nagahara, On separable polynomials over a commutative ring. II., Math. J. Okayama Univ. 15 (1971/72) 149-162.