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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?2011-01-24

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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!2011-01-24
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    @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.2011-01-24
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    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.)2011-01-24
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    @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.2011-01-24
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    @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.2011-01-24
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    @Zhen: as for reasons to work in (k-Alg)^{op}, there is a general philosophy due to Grothendieck that it is better to work in a nice category with nasty objects than to work in a nasty category with nice objects. (k-Alg)^{op} is a nice category, even if some of its objects are difficult to visualize. A concrete reason to work in a larger category is to get more representable functors: for example, the functor sending a variety to its tangent bundle is represented by Spec k[x]/x^2, and this gives a very straightforward definition of the Zariski tangent space.2011-01-24
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    @Zhen: some form of (k-Alg)^{op} is also the natural setting to work if k is not algebraically closed. For example, a variety over F_p is not even close to being well-represented by its set of F_p-points; you want to consider its F_{p^n}-points for all n, and this data is concisely organized into the family of functors Hom(Spec F_{p^n}, -) in (F_p-Alg)^{op} (which, in my opinion, is more natural than the "concretization" I gave above). Thus this is a natural language for doing arithmetic geometry, for example.2011-01-24
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    @Qiaochu: Thank you for the exposition, but it seems the set of things you think I know is really the complement of the set of things I actually know! (e.g. I know the Nullstellensatz, but I don't know what a reduced algebra is; I know what $k[V]$ is, but I only have passing familiarity with Spec R.) Your point about Grothendieck's philosophy is the answer I was looking for regarding the contrived-ness (artificiality?) of $(k\text{-Alg})^\text{op}$. I'll keep it in mind for the future.2011-01-24
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    @Zhen: whoops. I am extremely loose with my use of the term Spec R; here it literally just means R, but regarded as an object in (k-Alg)^{op}. (Sometimes I use it to mean the set of prime ideals, but that's not my usage here.) The construction of an affine scheme as a locally ringed space gives a category equivalent to this category, but one does not need to be familiar with this construction to work in this category (certainly I'm not). A reduced algebra is one with trivial nilradical (hence the only element which vanishes modulo all primes is zero).2011-01-24
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    @Qiaochu: Ick. I hadn't even realised there were multiple meanings! I was happily thinking $\operatorname{Spec} k$ meant a one-point set (consisting of the (0) ideal), which is a trivial $k$-algebra... but I guess this shows I'm still trying to think of $(k\text{-Alg})^\text{op}$ concretely. Hmmm. I'll have to give this more thought.2011-01-24
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    @Zhen: unless you are willing to attach the structure sheaf to that point, I think this is an occasionally useful but ultimately misleading thing to do. The functor sending a k-algebra to its set of prime ideals is _also not faithful._ For example, Spec k[x]/x^2 (as a set) also consists of one point, but as a scheme it has nontrivial automorphisms. If k is not algebraically closed, then you can't even distinguish any of the field extensions of k by their prime ideals; at the very least you need to think of them as points with the field attached.2011-01-24
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    @Zhen: here is a fairly concrete reason to care about (k-Alg)^{op}. If you work on the level of points, you get the wrong notion of intersection. For example, y = x^2 and y = -x^2 ought to intersect at (0, 0) with multiplicity 2, but on the level of points, all you see is the point (0, 0). The correct intersection is Spec k[x, y]/(y = x^2, y = -x^2) = Spec k[x]/x^2, and the extra structure you get from going behind points precisely records the multiplicity of intersection.2011-01-25
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    @Qiaochu: I can convince myself that this does give the correct notion of intersection. (I presume the intersection of $k[X, Y]/(X^2 - Y)$ and $k[X, Y]/(X^2 + Y)$ in $(k\text{-Alg})^\text{op}$ corresponds to the pushout in $k\text{-Alg}$ of the quotient maps from $k[X, Y]$.) Unfortunately $(X^2)$ not being radical means I don't really see how to interpret it geometrically. (The only tool I have at the moment is to look at $\text{Hom}_{k\text{-Alg}}(-, k)$ which gives me an idea of how many points there are, and obviously both $k[X]/(X)$ and $k[X]/(X^2)$ have only one point...)2011-01-25
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    @Zhen: I might mess this up, but I think this is how the story goes (and you can probably check Fulton's _Algebraic curves_ for the correct statements). The structure sheaf lets you define a _local ring_ at the point (0, 0). This local ring is a discrete valuation ring, and its discrete valuation is an algebraic analogue of the order of the pole of a (germ of a) meromorphic function at (0, 0). This valuation should be how you find the multiplicity of the intersection (though again, I'm not sure).2011-01-25
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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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    Regarding the last question: the degree of an element in k[x] is well-defined up to *all* automorphisms (because there are not many of these---$k$-algebra automorphisms, that is) This breaks down for two or more variables, though.2011-01-23
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    Why do you sit on your laptop?!2011-01-23
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    I have the same reservations about $k[V]$ as for $k[\mathbb{A}^n]$, but not as strongly for some reason. As for writing $p(X) \in k[X]$, in my mind, this (implicitly) declares a polynomial "skeleton" $p$ in one variable. This allows me to treat $p$ *syntactically* like a function, even though it isn't really one; whereas if I write $p \in k[X]$, it feels inconsistent to write $p(x)$ for the result of substituting $X$ by $x$.2011-01-23
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    @Mariano: I take notes in LaTeX. @Zhen: I am not sure what you mean by "treat p syntactically like a function, even though it really isn't one": are you referring to evaluation? Evaluation has a totally algebraic (that is, not set-theoretic) definition: it is just the image of p in a quotient k[A^n]/m where m is the maximal ideal corresponding to the point you're evaluating at. I'm also not sure what you mean by a polynomial "skeleton."2011-01-23
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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. (-:2011-01-23
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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."2011-01-23
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    @Qiaochu: I'm a bit surprised that Mariano would make that (ever so gentle) joke, since from personal experience I know that Spanish speakers often have a hard time with "in" vs. "on" and other propositions. (-;2011-01-23
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    @Qiaochu: My notion of a polynomial skeleton is exactly that of a formal polynomial defined as a sequence/matrix of coefficients, without reference to indeterminates. If I were to be pedantic, I would call $p(X)$ a polynomial *expression*. (In Mathematica, for example, $p$ would be an object like `#^3 - # &` whereas $p(X)$ would be an object like `X^3 - X`.) Personally, I think of evaluation as a homomorphism from the polynomial ring into any ring, written on the right (how weird), so $p(X) \in k[X]$ induces a family of functions $p: R \to R$, one for each homomorphism $k \to R$.2011-01-23
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    @Zhen: okay, that's a problem. On the one hand you want to refer to polynomial rings without referring to their generators. On the other hand you want to be able to plug in _specific elements of rings_ into these polynomials. These desires are incompatible. The latter requires a choice of generators to do; if you aren't making such a choice, then the objects you're plugging in aren't elements of other rings, they are points in the corresponding affine variety.2011-01-23
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    @Qiaochu: Hmmm. Now I'm lost. Is there an even more abstract notion of polynomials than as a collection of coefficients? I suppose I should state what I mean formally. A polynomial $p$ in $n$ variables over a ring $A$ is a finite collection of coefficients $(a_{j_1, \ldots, j_n} \in A : (j_1, \ldots, j_n) \in J)$. Given $n$ elements $b_1, \ldots, b_n$ of the ring $B$ containing $A$, define $\displaystyle p(b_1, \ldots, b_n) = \sum_{(j_1, \ldots, j_n) \in J} a_{j_1, \ldots, j_n} {b_1}^{j_1} \cdots {b_n}^{j^n}$. Then, we can legitimately say e.g. $p(X_1, \ldots, X_n) \in A[X_1, \ldots, X_n]$....2011-01-23
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    [...] and this mapping $p \mapsto p(X_1, \ldots, X_n)$ is even bijective. But I guess you're thinking of $A[X_1, \ldots, X_n]$ as a ring in its own right, in which case I suppose this would be akin to taking a basis...?2011-01-23
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    @Zhen: I am thinking of a polynomial as an element of a polynomial ring. If you don't want this polynomial ring to have a fixed set of generators, you can't refer to a polynomial by its coefficients. Right now your definition requires generators, but I thought the point of this question was to not do this. (Yes, it is exactly like taking a basis.) The blog post I linked to describes a way to refer to elements of a ring abstractly. (What are you doing if you aren't "thinking of A[X_1, ... X_n] as a ring in its own right"?)2011-01-23
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    @Qiaochu: I think of $A[X_1, \ldots, X_n]$ as the free associative commutative unital $A$-algebra on $\{ X_1, \ldots, X_n \}$ (when $A$ is commutative), so automatically it comes with evaluation maps. How else could one recognise a polynomial ring?2011-01-23
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    @Zhen: okay, I officially don't know what we're talking about here. (Perhaps we should conduct this conversation in person.) Do you want to fix a choice of generators or not? Have I answered your original questions to your satisfaction or not, and if not, what have I missed?2011-01-23
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    That's always nice when people accidentally meet online people from their class :)2011-01-23
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    @Zhen: in case your actual question is about how to define a polynomial ring without fixing a choice of generators, there's nothing wrong with "a ring which is isomorphic to the free ring on some number of generators" because that definition doesn't specify a choice of isomorphism. This is analogous to defining F_n to be a group isomorphic to the free group on n elements, or defining R^n to be a vector space isomorphic to the free vector space on n elements.2011-01-23
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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 )2011-01-24
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    @Mariano: oddly enough, here at Cambridge almost nobody else has their laptops open in class; I'm almost the only one. People here still take notes on paper for some reason...2011-01-24
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    @Qiaochu: I suspect most of us aren't quite practised enough in LaTeX to take notes live! (I might be able to take down the text, but the diagrams would kill me.) Anyway, I thought of an analogy to clarify my question — see the edited first paragraph.2011-01-24
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    @Zhen: I added a fresh answer. I assume this answer addresses at least the last question, since you haven't brought that up in this discussion.2011-01-24