# Finitely-generated $k$-algebras and their relationship with affine coordinate rings

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:

1. 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?

2. 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?

3. 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.

1. 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.

2. 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?

3. 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.

• I feel like if I'd spent another 5 minutes collecting my thoughts before writing my question I'd have appeared less muddled. Your answers to 1. and 3. are perfect, thanks. As for 2, I was (until just now) unaware of what it meant for an algebra over a field to be finitely generated. It does seem a bit silly that I'm learning Algebraic Geometry before I've formally come across $k$-algebras etc
– Matt
Commented Jan 20, 2012 at 22:13

Question 1. I prefer to think of it as a ring $R$ together with a ring homomorphism $k \to R$. As Qiaochu notes, we can't get away with this definition in general, but in the land of Hartshorne it seems to be the most natural version.

Question 2. The condition that $R$ be finitely generated as an algebra ("finite type") over $k$ is looser than being finitely generated ("finite") as a module. Being a finitely generated $k$-algebra means that there is a surjective homomorphism of $k$-algebras from $k[x_1, \ldots, x_n]$, for some $n$, to $R$. In other words, you can find a finite set $r_1, \ldots, r_n$ of elements of $R$ such that every element in $R$ is a polynomial in the $r_i$ with coefficients in $k$. Note that a polynomial ring is certainly not finitely generated as a $k$-module.

My guess is that you stumbled upon the Hilbert basis theorem, which in particular implies that finitely generated algebras over a field are Noetherian. This is good, because it guarantees that the ideal $I(Y)$ is generated by finitely many polynomials.

Question 3. I believe that Qiaochu has said all that is necessary for this one.

• The first definition doesn't generalize well to noncommutative rings; in the noncommutative case you also need to require the homomorphism $k \to R$ to land in the center of $R$ or else multiplication won't be $k$-linear. Commented Jan 20, 2012 at 21:57
• @QiaochuYuan I agree, but if we're talking about Hartshorne then this definition seems safe. You could also take issue with "algebra" being defined as an associative thing as well! Commented Jan 20, 2012 at 22:33