Does the existence of a common solution to a set of polynomials over a field extension imply a common solution over the subfield

Question

My question is that if I have a finite set of polynomials in a polynomial ring $K[x_1,....,x_n]$, where $K$ is an algebraic closed field, has a common solution in some field $L$ containing $K^n$, then must these polynomials have a common solution in $K^n$? If so, what is the proof?

Additionally, does this change if I consider an infinite set of polynomials or if $K$ is not algebraically closed?

My guess is that the first answer is yes, although I haven't been able to work out a proof. I am unsure about the follow up questions. It feels like algebraic closure would be necessary (in the event that the first answer is, in fact, yes). For the finite case, this is easy to see. I could take the polynomial $X^2+1$ in $\mathbb{Q}[X]$ for example. But I don't know how this extends to a non-finite set of polynomials.

Answer 1

This of course true, because of Nullstellensatz.

First note is that, "a finite set of polynomials in $K[x_1, \dotsc, x_n]$" is no different from "a finitely generated ideal of $K[x_1, \dotsc, x_n]$". But since the ring $K[x_1, \dotsc, x_n]$ is Noetherian, every ideal is finitely generated.

(This in particular answers your question about "non-finite set of polynomials". They are no different than finite sets of polynomials, since the ideal they generate must be finitely generated.)

Hence the statement can be:

Let $K$ be algebraically closed and $I\subseteq K[x_1, \dotsc, x_n]$ be an ideal. If there is a field extension $L/K$, such that there is an element in $L^n$ which is a common zero for all polynomials in $I$, then there is already a common zero in $K^n$.

Proof: Suppose there is no common zero of $I$ in $K^n$. By Nullstellensatz, the ideal $I$ must be equal to the whole ring $K[x_1, \dotsc, x_n]$. In particular, the element $1$ is included in $I$. Hence there cannot be any zero in any extension $L/K$.