We know that $R[x]$ is not finitely generated as an $R$-module and has a basis of $\{1,x,x^2,\ldots\}$. I started thinking about whether or not $R[[x]]$ has a basis, and if it does have a basis, if it was countable. I've been thinking about this, but I didn't make too much progress. Any ideas or thoughts?
Does $R[[x]]$ have a basis and is it countable?
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0If $R$ is a field, then *of course* there is a basis, but it is "ugly" (requires Choice). For rings, this may become even "more ugly". – 2012-12-10
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0@HagenvonEitzen so it sounds like for a general ring $R$, there is a basis? Or you're saying the determination would be "more ugly"? – 2012-12-10
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0Rather the latter. For example, simply taking a maximal linear independent set (via Zorn's lemma) is not sufficient to obtain an basis. – 2012-12-10
2 Answers
You can easily construct uncountably many linearly independent elements of $R[[x]]$, e.g. $\sum_n a_n x^n$ for $a \in \{0,1\}^{\mathbb N}$.
EDIT: Oops, those are not all linearly independent, but there is a linearly independent uncountable subset. For example, for each $t \in (1/2,1)$, let $f(t) = \sum_{n=1}^\infty x^{b_n(t)}$ where $b_n(t) = \lfloor 2^n t \rfloor$. Note that for any nonempty finite set $\{t_1, \ldots, t_k\} \subset (1/2,1)$, at most finitely many terms can appear in more than one $f(t_j)$.
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0I was pretty confident that if there was a basis it would be uncountable, but I wasn't entirely sure of it. Your construction makes perfect sense, even! Thanks! – 2012-12-10
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2As I can see, people were very happy with this answer, but let me say a few words: do you think that the existence of uncountable many linear independent elements proves, if a basis there exists, that the basis is uncountable, too? Well, if $R$ is not a field but a commutative ring (as the notation suggests), then I expect from you to provide an argument. – 2012-12-10
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0Ah, good point. It's not as obvious as I thought that a countable basis couldn't produce uncountably many linearly independent elements over a commutative ring. – 2012-12-11
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1These aren't linearly independent, since $(x) + (x^2) = (x + x^2)$. There is an uncountable linearly independent subset, though. – 2012-12-11
$R[[X]]$ is isomorphic to $R^{\mathbb N}$ as $R$-modules.
If $R=\mathbb Z$, then $\mathbb Z^{\mathbb N}$ is not free (it's not trivial at all to prove this!).
In general, if $R^{\mathbb N}$ would have a countable basis then $R^{\mathbb N}\simeq R^{(\mathbb N)}$ (a direct sum of copies of $R$). Now let's count: $|R^{(\mathbb N)}|=\aleph_0|R|$ while $|R^{\mathbb N}|=|R|^{\aleph_0}$. What do you think now? Which one is greater?
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1Indeed this question asked for the general case and it is easy to see that this answer provides a counterexample when $|R|\leq\aleph_0$. However if $|R|=2^{\aleph_0}$ then the cardinality argument on its own fails because $\mathbb{|R^N|=|R^{(N)}|=|R|}$. – 2012-12-10
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0@AsafKaragila Agree! – 2012-12-10
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0@YACP: I'm trying to digest what you have written, and I think I get it, but I'm not totally certain. I am mostly considering the last part, and I don't quite see the argument yet... Could you possibly elaborate a little bit more? – 2012-12-10
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0Isomorphic objects have the same cardinal. Now try to read the Asaf's comment. (For instance, when $R=\mathbb Z$ one cardinal is strictly less than the other.) – 2012-12-11