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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 $\varprojlim_n k[x_1,...,x_n]$ ?

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    Looks like $k[x_1,x_2,\ldots]$ to me...2012-09-30
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    I think you are mixing up projective limit and direct limit: $k[x_1,...]$ is the direct limit of $k[x_1,...,x_n] \hookrightarrow k[x_1,...,x_m]\;(n < m)$.2012-10-01
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    I also thought about this power series ring. But have have no idea on how to define the projections $k[[x_1,x_2,...]] \to k[x_1,...,x_n]$ ?2012-10-01
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    Ups, the previous comment concerns a comment of John Stalfos who suggested the projective limit in question is $k[[x_1,x_2,...]]$ (and was deleted while I was typing my comment).2012-10-01
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    What was the problem with $k[x_1,x_2,...],$ again? There *are* projections onto each element of your inverse system which commute with the maps in the system, though I haven't tried to check universality.2012-10-01
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    No element of $k[x_1, x_2, \cdots]$ (with the canonical projection) is capable of mapping to $\sum_{i=1}^n x_i \in k[x_1, \cdots, x_n]$ for every $n$... but there is a diagram of maps from $k[y]$ that does. (I assume the "canonical projection" mentioned in the question sends $x_n \to 0$, although that really isn't canonical)2012-10-01
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    @Kevin: The sequence $x=(x_1+...+x_n)_{n\ge 1}$ is an element in the projective limit (considered as subring of $\prod_n k[x_1,...,x_n]$) which I think has no counterpart in $k[x_1,x_2,...]$. In analogy with p-adic numbers I would interpret $x$ as $x_1 + x_2 + ...$ which is an element in $k[[x_1,x_2,...]]$. But I don't know if this a correct.2012-10-01
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    Ah, yes, ignore what I said, $k[x_1,\ldots]$ is a direct limit.2012-10-01

2 Answers 2

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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.

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    Dear Hurkyl, what you write seems correct but I think your answer would be clearer if you described your ring simply as a subring of $k[[x_1, x_2,x_3,...]]$. Meanwhile, +1.2012-10-02
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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 $\varprojlim_{n \to \infty} k[x_1,...,x_n]$.

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    Dear tj_, I don't think what you write is correct because you can't account for $x_1+x_2^2+x_3^3+...+x_r^r+...$, which *is* in the projective limit.2012-10-02