Let $k$ be an algebraically closed field, $A = k[x_1, … , x_n]$. For $Y \subseteq \mathbb A^n$, define $I(Y) = \{f \in A| f(P) = 0 \ \forall P \in Y\}$
Hartshorne's Algebraic Geometry, p. 4-5, says the following:
If $Y \subseteq \mathbb A^n$ is an affine algebraic set, we define the affine coordinate ring $A(Y)$ of $Y$ to be $A/I(Y)$…
… $A(Y)$ is a finitely generated $k$-algebra. Conversely, any finitely generated $k$-algebra $B$ which is a domain is the affine coordinate ring of some affine variety. Indeed, write $B$ as the quotient of a polynomial ring $A = k[x_1, … , x_n]$ by an ideal $\mathfrak{a}$, and let $Y = Z(\mathfrak{a})$.
As might be obvious from my recent posts, I'm trying to learn Algebraic Geometry with (currently) less-than-desired knowledge of commutative algebra. I'm picking up the relevant bits as I go (as much as I feel I need to). I have a few questions about the above passage:
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I've just learned the definition of a $k$-algebra. Am I safe to think of a $k$-algebra as a ring $R$ which is also a $k$-module?
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If I knew that $A$ was a finitely generated $k$-algebra, say $A = \sum_{i=1}^d A f_i$, then I think $A(Y)$ would be generated by $f_1 + I(Y), …. , f_d + I(Y)$. But why/is $A$ finitely generated? Some googling has shown me that if $A$ is a Noetherian module then it is finitely generated (as a $k$-module). Is this equivalent to being finitely generated as a $k$-algebra? If so, how would I show that it's a Noetherian module?
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Why can any finitely-generated $k$-algebra which is a domain be expressed as the quotient of a polynomial ring and an ideal?
Thank you.
Best Answer
You need to specify some compatibility between the ring and $k$-module structures: namely, the ring multiplication $R \times R \to R$ needs to be a map of $k$-modules.
What do you mean by $A(Y)$ for $A$ a finitely generated $k$-algebra? Why do you ask why $A$, which you stipulated to be finitely generated, is finitely generated?
Pick a finite set of generators $x_1, ... x_n$. This defines a natural surjection $k[x_1, ... x_n] \to A$ whose kernel is an ideal defining $A$ by the isomorphism theorems for rings (I can never remember which number is which). You don't need to use the fact that $A$ is a domain; that just tells you that $I$ is a prime ideal.