Algebraic set $V(\mathfrak{a})$ equals the set of homomorphisms from $A = k[x_1, \dots, x_n]/\mathfrak{a}$ to $k$? Let $A$ be a finitely generated $k$-algebra. Then $A = k[x_1, \dots, x_n]/\mathfrak{a}$. Why is $V(\mathfrak{a})\subset k^n$ equal to $\operatorname{Hom}_{k-alg}(A, k)$?
 A: "Equal" is perhaps too strong a word, but there's a canonical bijection between these two sets.
Let $B$ be a $k$-algebra, and consider the sequence of maps $k[x_1,\cdots,x_n]\to A \to B$, where the first map is the natural quotient. Every map $A\to B$ gives rise to a map $k[x_1,\cdots,x_n]\to B$ by composition. By the universal property of the polynomial algebra, a map $k[x_1,\cdots,x_n]\to B$ is equivalent to a choice of $n$ elements $b_i\in B$ such that $x_i\mapsto b_i$, and the condition that this map $k[x_1,\cdots,x_n]\to B$ comes from a map $A\to B$ is exactly that the $b_i$ satisfy every polynomial in $\mathfrak{a}$ by the first isomorphism theorem.
Specializing to the case of $B=k$, we can see that the data of a $k$-algebra map $A\to k$ is equivalent to picking $n$ elements $y_1,\cdots,y_n$ of $k$ so that the $y_i$ satisfy all the polynomials in $\mathfrak{a}$, i.e. if $\alpha(x_1,\cdots,x_n)\in\mathfrak{a}$ is a polynomial, then $\alpha(y_1,\cdots,y_n)=0$. But this exactly the same description as points of $V(\mathfrak{a})\subset k^n$: they're $n$-tuples of elements of $k$ so that when you plug them in to every polynomial in $\mathfrak{a}$ you get zero.
