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Given that we shouldn't say that "$f(z)$ is a function", shouldn't we also not write "$p \in k[X_1, \ldots, X_n]$ is a polynomial"? Along those lines, I usually write $p(X_1, \ldots, X_n) \in k[X_1, \ldots, X_n]$ in order to balance the "free variables" on both sides of the relation, but that gets unwieldy when you start dealing with iterated polynomial rings. My question is: Is there a notation for polynomial rings which allow us to talk about polynomials without explicitly naming the indeterminates? Consider, for an analogy, vector spaces $\mathbb{R}^n$. These have a canonical basis, but the notation $\mathbb{R}^n$ does not commit me to naming the canonical basis, unlike, say, the notation $\operatorname{span} \{ e_1, \ldots, e_n \}$.

I suppose I should fix a definition for polynomial rings. For simplicity let's work in the category $\mathbf{CRing}$ of commutative rings with 1. Let $U: \mathbf{CRing} \to \mathbf{Set}$ be the forgetful functor taking rings to their underlying sets. A polynomial ring in a set of indeterminates $\mathcal{S}$ over a ring $A$ is a ring $R$ together with an inclusion map $\iota: A \hookrightarrow R$ and a set-map $x: \mathcal{S} \hookrightarrow UR$, and has the universal property that for every ring $B$, homomorphism $\phi: A \to B$, and set-map $b: \mathcal{S} \to B$, there is a homomorphism $\epsilon: R \to B$ such that $\epsilon \circ \iota = \phi$ and $U\epsilon \circ x = b$.

If we write $A[\mathcal{S}]$ for such a ring $R$, then we could write, for instance, $A[5]$ for the ring of polynomials in 5 variables over $A$, but that would, I imagine, be extremely confusing. Yet, on the other hand, if we have a bijection $\mathcal{S} \to \mathcal{S}'$, then this lifts to an isomorphism of $A[\mathcal{S}] \to A[\mathcal{S}']$, so it is all the more tempting to write $A[\kappa]$, $\kappa = |\mathcal{S}|$, for the canonical representative of this isomorphism class.

If $\mathcal{S} = \{ 1, \ldots, n \} \subset \mathbb{N}$ and $\phi: A \to B$ is given, I write $\phi p(b_1, \ldots, b_n)$ for the image of $p \in A[\mathcal{S}]$ under $\epsilon$ for $b(m) = b_m, m \in \mathcal{S}$. When the choice of homomorphism $\phi$ is clear I'll omit it in writing. This justifies my notation $p(X_1, \ldots, X_n) \in A[X_1, \ldots, X_n]$, since I would like to regard $A[X]$ as being analogous to $\mathbb{Z}[\pi]$, i.e. it's a ring with a transcendental element adjoined so is isomorphic to a polynomial ring, but doesn't come with evaluation maps attached. But following this line of thought, how should I denote the object that $p$ itself belongs to?

I recently started attending an algebraic geometry course and at one point the lecturer wrote $k[\mathbb{A}^n]$ for the ring of polynomials in $n$ indeterminates over $k$. This seems like a reasonable solution, but there are some problems:

  1. It feels suspiciously like a function ring, but in general the map taking formal polynomials to polynomial functions is neither injective nor surjective.
  2. The notation makes it look like a ring with $\mathbb{A}^n$ adjoined, but that doesn't seem to make sense. (Is there a way to make sense of it, e.g. by defining ring operations on $\mathbb{A}^n$?)
  3. Is it standard notation? I have seen $k[V]$ in some algebraic geometry textbooks for the coordinate ring of the (affine) variety $V$, but never for $V = \mathbb{A}^n$. (I have similar reservations about the notation $k[V]$, but not as strongly.)
  4. Would it make sense to write, say, $\mathbb{Z}[\mathbb{A}^n]$?

A related problem arises from the following: let $p(X)$ and $q(X)$ be formal polynomials in $k[X]$, with $p(X) = q(X^2)$. It's clear that $\operatorname{deg} p = 2 \operatorname{deg} q$... but this shows that, in a certain sense, the degree depends on the ambient polynomial ring: if $p(X)$ were considered as a formal polynomial in $k[X^2]$, its degree would be the same as $q$, since, after all, $p(X) = q(X^2)$. It is clear that we should have $k[X^2] \subset k[X]$, but if we obviate the indeterminates and reduce polynomials to their bare skeletons, then the "inclusion" map $k[X^2] \hookrightarrow k[X]$ is no longer a set-theoretic inclusion map. Is there a coherent way of thinking about polynomials and polynomial rings which resolves this ambiguity, and what is the notation that goes with it?

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I frequently say "$f(z)$ is a function", and I would almost always write a polynomial in the form $p(x_1,\ldots,x_n)$ (or whatever the indeterminates are). It is very occasionally inconvenient to name the indeterminates, but typically it is actually quite convenient (and eliminates any confusion of the $p(X)$ v.s. $q(X^2)$ type). Why the desire to remove the notation for indeterminates (or, in older terminology, independent variables) from polynomials, or, more generally, from functions? –  Matt E Jan 24 '11 at 4:04

2 Answers 2

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The other answer is bogged down with a lot of comments, so let me give a fresh answer to the revised question. The short answer is as follows: $k[\mathbb{A}^n]$ is the notation I support because $k[V]$ is perfectly standard. The fact that it looks like adjoining $V$ is irrelevant; what you are adjoining is coordinate functions on $V$. (That is, you should think of it as analogous to the notation $C(X)$ for the ring of real-valued functions on a topological space $X$, except that $k(V)$ is already taken; it means the field of functions on $V$ when $V$ is irreducible.) Finally, the fact that it looks like a function ring is also irrelevant because, in the right category, it is a function ring.

The long answer is as follows: your concern that

It feels suspiciously like a function ring, but in general the map taking formal polynomials to polynomial functions is neither injective nor surjective.

comes from the fact that the functor $\text{Hom}(\text{Spec } k, -) : k\text{-Alg}^{op} \to \text{Set}$ is not faithful, but you don't have to think about this functor. You can in fact work directly in $k\text{-Alg}^{op}$, and in this category $k[V]$ is precisely the ring of functions $\text{Hom}(V, k)$ where $k$ is shorthand for the affine line $\mathbb{A}^1(k) \simeq \text{Spec } k[x]$. One need not make a distinction between polynomials and the functions they define in this case.

Here's why. Suppose $V$ is given by an ideal $I$ in $k[x_1, ... x_n]$ for some $n$, so that $k[V]$ denotes the ring $k[x_1, ... x_n]/I$. (This is by definition.) Then a morphism $\text{Hom}(V, k)$ (in $k\text{-Alg}^{op}$, which is emphatically not the naive category of affine varieties over $k$) is precisely a morphism $k[x] \to k[x_1, ... x_n]/I$ of $k$-algebras. By the universal property of $k[x]$, such a morphism is freely determined by the image of $x$, so $\text{Hom}(k[x], -)$ represents the forgetful functor $k\text{-Alg} \to \text{Set}$. In particular such morphisms are in one-to-one correspondence with elements of $k[V]$, and this may therefore be taken as a definition of $k[V]$. (The details of how to get the ring operations are discussed, as I said, in this blog post.)

When $k$ is algebraically closed and we restrict to the opposite of the category of finitely-generated reduced $k$-algebras, then the functor $\text{Hom}(\text{Spec } k, -)$ is faithful by the Nullstellensatz, so we can define the ring of regular functions $k[V]$ in a naive way by regarding $V$ as a set of points in $\mathbb{A}^n(k)$. The point I am trying to make above is that, as long as we switch to a more sophisticated category, we can still do this for $k$ arbitrary (and not necessarily a field) and $V$ an arbitrary $k$-scheme rather than just a variety. In particular, if $V$ is defined over $\mathbb{Z}$, then the notation $\mathbb{Z}[V]$ makes perfect sense.

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Whenever I see $k[V]$, I don't know whether I am looking at polynomial functions on $V$, or at $\bigoplus \mathrm{Sym}^n(V)$, which is polynomial functions on $V^*$. If you use this notation, please clarify this point! –  David Speyer Jan 24 '11 at 15:28
    
@David: the former interpretation is the one I've seen in textbooks, and it's the one I mean here. I'm not sure what "polynomial functions on V*" means if V is not a vector space. –  Qiaochu Yuan Jan 24 '11 at 15:43
    
I think I need to immerse myself in this commutative algebra / abstract nonsense more... I can convince myself that your construction works, but it seems contrived. Are there other reasons to work in $(k\text{-Alg})^\text{op}$? I imagine there might be a functor from the category of affine varieties over $k$ into $(k\text{-Alg})^\text{op}$, but I can't quite put my finger on it yet. (Probably because I'm stuck in concrete mode where arrows are set-functions.) –  Zhen Lin Jan 24 '11 at 16:00
    
@Zhen: which part is contrived? It is exactly the generalization of the natural construction, but in a category which makes it work. The category of affine varieties over k, at least when k is algebraically closed, is the subcategory where the objects are finitely-generated reduced k-algebras (so yes, there is an inclusion functor). When k is not algebraically closed, k-points do not fully describe a variety; a "concretization" in this case is to send a k-algebra to its set of points over the algebraic closure of k, which is equipped with an action of the absolute Galois group of k. –  Qiaochu Yuan Jan 24 '11 at 16:16
    
@Zhen: the lecturer will cover this at some point, but if you don't already know, the Nullstellensatz says that, when k is algebraically closed, the Galois connection we described in class between ideals of k[x_1, ... x_n] and varieties in A^n restricts to an equivalence between radical ideals and affine varieties. Radical ideals are precisely those for which k[x_1, ... x_n]/J is reduced, so that's where the abstract characterization (which does not require an embedding into affine space) comes from. –  Qiaochu Yuan Jan 24 '11 at 16:18

Last question first: degree is not (depending on how you look at it) a property of elements of polynomial rings. It's a property of elements of graded polynomial rings, and the easiest way to choose a grading is to choose a set of generators. Your example is a little confused: when you consider $k[x^2]$ you are implicitly considering $x^2$ to have degree $1$, but you don't need to adopt this convention; you can declare that it has degree $2$ instead. (In any case, the notation $k[x^2]$ is misleading: when you write this you are really talking about the entire inclusion map $k[x^2] \to k[x]$, so everything depends on whether you want this to be a map of rings or a map of graded rings.)

Your point 3 seems to answer your first question. I'm not sure why you find $k[\mathbb{A}^n]$ objectionable but $k[V]$ not given that I assume you think $\mathbb{A}^n$ is a variety. I think this is a fine way to refer to a polynomial ring without naming its generators. Your worry about the distinction between polynomials and the functions they induce can be ignored if $k$ is algebraically closed, and otherwise you should just remind yourself that in the remaining cases the functor from $k$-varieties to $\text{Set}$ isn't faithful, so the answer is not to take the set-theoretic picture too seriously in the first place and work directly with the opposite of the category of finitely-generated reduced $k$-algebras. In this category I describe exactly how to recover the ring of functions geometrically in this blog post. (I should be more explicit about what I mean here: if you work in the right category, the polynomial ring in $n$ variables over $k$ is the space of functions on $\mathbb{A}^n$.)

I also don't understand your first two sentences; they seem to be inconsistent with each other. (Off-topic: I don't know what you look like, but because of my Gravatar you know what I look like. I'm the guy sitting in the back of class on his laptop, so if you'd like to introduce yourself that would be cool.)

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Why do you sit on your laptop?! –  Mariano Suárez-Alvarez Jan 23 '11 at 0:57
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@Qiaochu: I think Mariano is making the point that you are probably sitting behind your laptop, or in front of your laptop, rather than on your laptop. The latter would likely require the laptop to be on the chair with you on top of it; that would make taking notes in LaTeX extremely difficult, I would think. (-: –  Arturo Magidin Jan 23 '11 at 1:15
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@Arturo: ah, I see. @Mariano: in case this was a language issue, I'm using "on" here in the same sense as I would say "I'm on the internet." –  Qiaochu Yuan Jan 23 '11 at 1:16
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That's always nice when people accidentally meet online people from their class :) –  Pandora Jan 23 '11 at 17:05
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@Qiaochu, it was a silly joke---I of course understood what you meant (But surely nowadays that more and more people have their notebooks in class it would make it considerably easier to identify you if you were actually sitting on yours! :P ) –  Mariano Suárez-Alvarez Jan 24 '11 at 2:12

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