Ring of infinite sequences.

Question

Let $(R,+,\cdot)$ be a commutative ring with unity and $R^{\infty}$ be the ring of sequences $(a_0,a_1,a_2,\ldots)$ with elements in $R$. The binary operations in $R^{\infty}$ is defined as:

$(a_0,a_1,a_2,\ldots)+(b_0,b_1,b_2,\ldots)=(a_0+b_0,a_1+b_1,a_2+b_2,\ldots)$ and

$(a_0,a_1,a_2,\ldots)\cdot (b_0,b_1,b_2,\ldots)=(a_0\cdot b_0,a_1\cdot b_1,a_2\cdot b_2,\ldots)$.

Then I have the following question:

(1) Prove that $(R^{\infty},+,\cdot)$ is a commutative ring with unity.

(2) How the ideals of $R^{\infty}$ will look like.

(3) $(a_0,a_1,a_2,\ldots)$ is unit in $R^{\infty}$ if and only if $a_i$ is unit in $R$ for every $i$.

(4) If $R$ is unique factorization domain then what can we say about $R^{\infty}$?

(5) If $R=\mathbb{Z}$, then what will be the structure of $\mathbb{Z}^{\infty}$?

I have done the (1) and (3) but not able to proceed further. Kindly give some idea to solve this.

Answer 1

1. Explicitly, all of the axioms hold because they hold componentwise.

2. This is the hardest question and I'll save it for last.

3. Explicitly, an inverse is a componentwise inverse.

4. $R^{\infty}$ won't be a UFD even if $R$ is; the problem turns out not to be the uniqueness of factorizations but existence. An element of $R^{\infty}$ is irreducible iff it has the form $\prod r_i$ where exactly one $r_i \in R$ is irreducible and the others are units. So the elements which are products of irreducibles have the form $\prod r_i$ where all but finitely many of the $r_i$ are units. For example, $\prod p_i \in \mathbb{Z}^{\infty}$, where the $p_i$ are the primes, is not a product of irreducibles. However, we do have an "infinite irreducible factorization" given by the componentwise irreducible factorization.

5. I don't know what you mean by "structure" here.

Now for some words on the ideals. Already the prime ideals are quite complicated. First, let's observe that there are some obvious ones, namely: if $P$ is any prime ideal of $R$, then there's a projection $\pi_i : R^{\infty} \to R$ given by taking the $i^{th}$ coordinate, and then a quotient map $R^{\infty} \to R \to R/P$. These are the "obvious" prime ideals, one copy of $\text{Spec } R$ for every coordinate. Geometrically, finite products of commutative rings correspond to disjoint unions of affine schemes.

But infinite products are more complicated. As is done in this answer by Eric Wofsey and this blog post, IMO the cleanest way to proceed is as follows. It turns out that there is a natural continuous map

$$\text{Spec } R^{\mathbb{N}} \to \beta \mathbb{N}$$

which sends a prime ideal $P$ of $R^{\mathbb{N}}$ to an ultrafilter $U$ on $\mathbb{N}$. This ultrafilter has the property that $P$ comes from a prime ideal of the quotient $R^{\mathbb{N}}/U$, which is an ultrapower of $R$. This ultrapower is elementarily equivalent to $R$, and so for example if $R$ is a field then so is any ultrapower, so in that case the space $\beta \mathbb{N}$ of ultrafilters is a complete description of the spectrum.

In general the ultrapower $R^{\mathbb{N}}/U$ has "nonstandard prime ideals" given by the kernels of quotient maps $R^{\mathbb{N}}/U \to (\prod R/P_i)/U$ to ultraproducts, where $P_i$ is a sequence of prime ideals; see Eric Wofsey's answer linked above for a worked example of this for $R = \mathbb{Z}$. There are even more complicated examples that I don't understand, but again see Eric's answer.