What's the projective limit of these polynomial rings ?

Question

Define an inverse system of polynomial rings over a commutative ring $k$ by the canonical projection $k[x_1,...,x_n] \to k[x_1,...,x_m]\;(m< n)$.

Question: What is the projective limit $\varprojlim_n k[x_1,...,x_n]$ ?

Answer 1

Note that we can write, as sets:

$$ k[x_1, x_2, \cdots, x_n] \cong k[x_1] \times x_2 k[x_1, x_2] \times \cdots \times x_n k[x_1, x_2, \cdots, x_n]$$

(all but the first component of the product are ideals) An element of the product is interpreted as the sum of all of its components. Furthermore, the canonical projection (assuming you mean to send $x_n$ to $0$) is just the projections onto the first $m$ components.

Therefore,

$$ \lim_n k[x_1, x_2, \cdots, x_n] \cong k[x_1] \times x_2 k[x_1, x_2] \times x_3 k[x_1, x_2, x_3] \times \cdots $$

What addition is should be straightforward. Working with multiplication should be similar in flavor to working in power series rings.

The elements should probably be thought of as infinite sums; given a finite set of variables, each such sum should only have finitely many monomials that involve only those variables.

Answer 2

This question was recently asked again here, so I thought I'd give an alternative viewpoint, presenting the ring as a ring of formal sums.

A larger ring

Let $I$ range over sequences $(i_1,\ldots,i_n,\ldots)$ of natural numbers with only finitely many nonzero terms. Call this set of sequences $\newcommand\N{\Bbb{N}}\N^{(\N)}$. Consider formal sums over this set of sequences, $$ \sum_{I} a_I x^I, $$ with the $a_I\in R$, and where $x^I$ is a formal element we think of as representing the (finite) monomial $$\prod_{n=1}^\infty x_n^{i_n}.$$

With no restrictions on the $a_I$, the collection of all such formal sums form a ring, with the product the usual convolution product, $$ \newcommand\of[1]{\left({#1}\right)} \of{\sum_I a_I x^I} \of{\sum_J b_J x^J} = \sum_K c_K x^K, $$ where $$c_K = \sum_{I+J=K} a_Ib_J,$$ noting that the sum is finite, since $K$ has only finitely many nonzero terms.

Also note that the set of formal sums can be identified with the set of functions, $\mathbf{Set}(\N^{(\N)},R)$.

Some comments

Note that this ring is too big to be our limit, since it contains power series rings, things like $\sum_{n=0}^\infty x_1^n$.

In fact, this is variant (iii) of a formal power series ring in infinitely many variables $R[[x_1,\ldots,x_n,\ldots]]$ here.

The limit ring as a subring

Let $S$ be the ring of all formal sums defined above. We want to produce a subring $T$ which will be our limit ring. The key problem is that $S$ contains terms of infinite degree in a fixed variable, which is not allowed in our limit.

Let $\pi_n : S\to R[[x_1,\ldots,x_n]]$ be the projection onto this subset of the variables, defined by noticing that we have an inclusion $\iota_n: \N^n\hookrightarrow \N^{(\N)}$ as the first $n$ coordinates, so precomposition induces a map $$\iota_n^* : \mathbf{Set}(\N^{(\N)},R)\to \mathbf{Set}(\N^n,R)$$ that respects the ring structure.

We have that $$ \pi_n\of{\sum_I a_Ix^I}= \sum_J b_Jx^J, $$ where $J$ ranges over multi-indices $J=(j_1,\ldots,j_n)$ and $b_J = a_{\iota_n(J)}$.

Then define $$ T = \bigcap_n \pi_n^{-1}(R[x_1,\ldots,x_n]).$$

$T$ is our limit.

By definition of $T$, the maps $\pi_n: S\to R[[x_1,\ldots,x_n]]$ restrict to maps $\pi_n : T\to R[x_1,\ldots,x_n]$. It's not hard to check that such a definition gives us a cone from $T$ to our sequence in $R$-algebras.

On the other hand, given a cone from another $R$-algebra $T'$ to each polynomial ring, $\psi_n : T'\to R[x_1,\ldots,x_n]$, then define $\tilde{\psi} : T'\to T$ by $\tilde{\psi}(t) = \sum_I a_I x^I$, with $a_I$ the coefficient of $\pi_N(x^I)$ in $\psi_N(t)$, where $N$ is large enough that $i_m = 0$ for $m> N$, since $I$ has only finitely many nonzero terms. It's not hard to check that this is well defined and unique.

Edit note

Originally I defined $T$ in the following manner:

If $a=\sum_I a_I x^I$ is a formal sum, then define the degree of $x_n$ in $a$, denoted $[a]_n$ to be $$ \sup \{ k : \text{there exists $I$ with $a_I\ne 0$ and $i_n=k$}\}, $$ where the sup is taken in $\Bbb{N}\cup \{\infty\}$.

Then define $$T=\{a\in S : [a]_n <\infty\text{ for all $n$}.\}$$

But this is too strict, it excludes things like $$\sum_{n=0}^\infty x_1^nx_n.$$

Answer 3

I can answer your question. Call a linear combination of monomials $\prod_i x_i^{m_i}$ such that $m_i\ge 0, \sum_i m_i=m$ an $m$-form. Then as a subring $$\varprojlim_{n \to \infty} k[x_1,...,x_n] = \lbrace f_1 + \cdots + f_m \mid m \ge 0,\;\; f_i\; i\text{-form}\rbrace \le k[[x_1,x_2,...]]$$

For the proof note that there are canonical projections $p_n: k[[x_1,x_2,...]] \to k[[x_1,...,x_n]]$ and the restriction to $S = \lbrace f_1 + \cdots + f_m \mid m \ge 0,\;\; f_i\; i\text{-form}\rbrace$ has image in $k[x_1,...,x_n]$. Hence $p= \prod_n p_n: S \to \prod_n k[x_1,...,x_n]$ is an embedding of rings whose image is exactly $\varprojlim_{n \to \infty} k[x_1,...,x_n]$.