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.

  • 1
    $\begingroup$ 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 $\endgroup$ – Matt Jan 20 '12 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.

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    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Jan 20 '12 at 21:57
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    $\begingroup$ @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! $\endgroup$ – Dylan Moreland Jan 20 '12 at 22:33

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