$K$ a field with char$(K)=0$ and $K[X_1,...,X_n]/(F_1,...,F_s)$ Artinian local. Then for any subfield $E$, $E[X_1,...,X_n]/(F_1,...,F_s)$ is local

Question

I came across the following remark in a paper I am reading: Let $K$ be a field with char$(K)=0$ and suppose $K[X_1,...,X_n]/(F_1,...,F_s)$ is Artinian local. Then for any subfield $E\subset K$, with $E$ containing the coefficients of the $F_i$, $E[X_1,...,X_n]/(F_1,...,F_s)$ is local with maximal ideal $(X_1,...,X_r)$. I feel the answer should be simple, but I am unable to see it. Would anyone be able to shed some light on this?

Answer 1

Four key facts:

1. A finitely generated algebra over a field is artinian iff it is finite-dimensional as a vector space over that field,
2. If $V$ is a vector space over $E$, then $\dim_E V = \dim_K K\otimes_E V$,
3. An artinian ring is local iff it does not decompose as a direct product.
4. A ring is a direct product iff it contains a nontrivial idempotent $e=e^2$, $e\neq 0,1$.

If $K[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$ is artinian, then $\dim_K K[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$ is finite by fact 1 and equal to $\dim_E E[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$ by fact 2, so $E[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$ is artinian by fact 1. Now if $E[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$ was not local, it would decompose as a direct product, or by fact 4, the ring has a non-trivial idempotent, an element $e\neq 0,1$ with $e^2=e$. But $e\otimes_E 1$ would then be such an element of $K[X_1,\cdots,X_n]/(F_1,\cdots,F_s)$, which cannot exist by assumption.