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

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?

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 $$0 \leq k_i \leq n-i$$.