Notation for a polynomial ring and formal polynomials 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?
 A: 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.)
A: 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.
