Assume we have a commutative algebra $A$ over a an algebraiclly closed field F of characteristic zero defined by generators $x_1, \dots, x_n$ and relations:
$f_1=0$
$f_2=0$
$f_3 =0$
where $f_1$, $f_2$, and $f_3$ are in $x_1, \dots, x_n$.
That means $A=F[x_1, \dots, x_n]/(f_1, f_2, f_3)$.
my question:
Is true that if we want to find all maximal ideals of $A$, we need to solve the system of equations:
$f_1=0$
$f_2=0$
$f_3 =0$
and why?
Can someone help me to understand this correspondence?
Let $I = (f_1, f_2, f_3)$. By the correspondence theorem, the maximal ideals of $A$ are exactly the (reductions of) maximal ideals $\DeclareMathOperator{\m}{\mathfrak{m}} \m \subseteq F[x_1, \ldots, x_n]$ with $\m \supseteq I$. As you say, by the Weak Nullstellensatz the maximal ideals of $F[x_1, \ldots, x_n]$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$, so $\m = (x_1 - a_1, \ldots, x_n - a_n)$ for some $a_1, \ldots, a_n \in F$. Now $I \subseteq \m \iff f_i \in \m$ for all $i = 1,2,3$ and $$ f_i \in (x_1 - a_1, \ldots, x_n - a_n) \iff f_i(a_1, \ldots, a_n) = 0 \, . $$ (For details, see this post.) Thus the maximal ideals of $A$ are exactly the reductions of the ideals $(x_1 - a_1, \ldots, x_n - a_n)$ such that $$ f_1(a_1, \ldots, a_n) = f_2(a_1, \ldots, a_n) = f_3(a_1, \ldots, a_n) = 0 \, . $$
