Let $K\subseteq F$ be a field extension. $\forall a\in F^n$ let $\phi_a:K[x_1,\dots,x_n]\to F$ defined by $f\mapsto f(a)$. Let $\psi_F:F^n\to\text{spec}(K[x_1,\dots,x_n]) $ defined by $a\mapsto \ker(\phi_a)$. Prove $$A:=\text{spec}(K[x_1,\dots,x_n])=\cup_{L/K}\text{Im}(\psi_L)=:B$$
Attempt:
I've shown that $\psi _a$ is injective.
Obviousely $B\subseteq A$. We want to show $A\subseteq B$.
Let $p\in A$. We wish to find an $a$ s.t. $p=\{f\in K[x_1,\dots,x_n]:f(a)=0\}$. I thought of defining $$T:=\{a: \exists f\in p, 0=f(a)\}\ne\emptyset, \\ S:=\{a:\forall f\in p,0=f(a)\} $$ I want to show that $S\ne\emptyset$, I'm not sure how. Thanks.
Let $\mathfrak{p}\in \text{Spec}(K[x_1,\dots,x_n])$, then $R=K[x_1,\dots,x_n]/\mathfrak{p}$ is an integral domain. Let $L$ be the fraction field of $R$, we have: $$K[x_1,\dots,x_n]\to R\hookrightarrow L$$ Hence $L$ is a field extension of $K$. Let $a_1,\dots,a_n$ be the images of $x_1,\dots,x_n$ via the composition map $K[x_1,\dots,x_n]\to L$, the element $a=(a_1,\dots,a_n)\in L^n$ gives us what we need. Indeed, the map $K[x_1,\dots,x_n]\to L$ above is $\phi_a$, and $$\text{Ker}(\phi_a)=\text{Ker}(K[x_1,\dots,x_n]\to R)=\mathfrak{p}$$ Now $\mathfrak{p}=\psi_L(a)\in \text{Im}(\psi_L)$.