This is in the proof to Corollary 1.8 on page 34.
How do we prove that every $k$-algebra can be written in this form, for some ideal $I$?
$k$ is an algebraically closed field if that matters.
I understand that a finitely generated $k$-algebra is already in this form for $I=(0)$, is that good enough for the converse of the proof? The corollary is:
Corollary 1.8 If $k$ is an algebraically closed field and $A$ is a $k$-algebra, then $A = A(X) \equiv k[x_1, \dots, x_n]/I(X)$ for some algebraica set $X$ iff $A$ is reduced and finitely generated as a $k$-algebra.
The proof goes:
If $A = A(X)$ for some $X \subset k^n$, then $A = k[x_1, \dots, x_n]/I(X)$ is generated as a $k$-algebra by $x_1, \dots, x_n$. Since $I(X)$ is a radical ideal, $A$ is reduced. Conversely, if $A$ is a a finitely generated $k$-algebra, then after choosing generators we may write $A = k[x_1, \dots, x_n]/I$ for some ideal $I$. SInce $A$ is reduced, $I$ is a radical ideal. Thus $I = I(Z(I))$ by the Nullstellensatz, and we may take $X = Z(I)$.
I guess I'm not seeing how we can talk about some ideal $I$ in the converse proof. How does this ideal arise given a finitely generated $k$-algebra?
Let $a_1,...,a_n$ generators of $A$, define $f:k[x_1,...,x_n]\rightarrow A$ by $f(x_i)=a_i$, its kernel is $I$ and $A\simeq k[x_1,...,x_n]/I$.