Let $F$ be a field (not necessarilly algebraically closed). Let $R=F[x_1,...,x_n]$ be the corresponding polynomial ring for variables $x_1,...,x_n$. Let $S$ be a subset of $R$.
We define $\mathcal{V}(S)=\lbrace \textbf{u} \in F^n : f(\textbf{u})= 0 \forall f \in S\rbrace$.
Also define the map $\mu(\textbf{u})= <x_1-u_1,...,x_n-u_n>$.
Can we identify the set $\mu(\mathcal{V}(S))$ (where we define this set to be the image of $\mathcal{V}(S)$ by $\mu$) with the set of maximal ideals of $R$ containing $S$? I'm okay with showing that anything in this set is such an ideal. But is the converse true? If this is not true, how would you describe $\mu(\mathcal{V}(S))$? Maybe geometrically?
If the field is algebraically closed, the statement is true due to Hilbert's Nullstellensatz.
Let $\mathfrak m \supset S$ be a maximal ideal of $R$. By Hilbert's Nullstellensatz, we have $\mathfrak m = \langle x_1-u_1, \dotsc, x_n-u_n \rangle$ for some $u_i \in F$. Let $u=(u_1, \dotsc, u_n)$.
$\mathfrak m \supset S$ implies $\{u\} = \mathcal V(\mathfrak m) \subset \mathcal V(S)$, i.e.
$$\mathfrak m = \mu(u) \in \mu(\mathcal V(S)).$$
This is precisely the converse direction, you were lacking.
If the field is not algebraically closed, take an irreducible polynomial $f \in F[x_1]$ of degree $\geq 2$ and consider the maximal ideal
$$\mathfrak m = \langle f,x_2,x_3, \dotsc, x_n \rangle \subset F[x_1, \dotsc, x_n].$$
Clearly $\mathfrak m$ is not contained in $\mu(\mathcal V(\emptyset))$, but $\mathcal m$ is a maximal ideal containing $\emptyset$. (If you dont like the emptyset, take $S$ to be any subset of $\mathfrak m$)