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:
- It feels suspiciously like a function ring, but in general the map taking formal polynomials to polynomial functions is neither injective nor surjective.
- 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 $\mathbb{A}^n$?)
- 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.)
- 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?
Best Answer
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.)