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In our notes I read that $R[[x]]$ (for ring $R$) is not generated by $\{1,x,x^2,\ldots\}$. I'm wondering if there exists a generating system and if yes, how it looks like?

Ben
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    Are you asking for generators of $R[[x]]$ as an $R$-module ? – Damien L Feb 04 '13 at 15:57
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    In any case, $R[[x]]$ is generated by all $f\in R[[x]]$. So you probably want more conditions on the generating system (as $R$-module or as $R$-algebra). –  Feb 04 '13 at 15:58
  • Is it "not generated only by" perhaps? I don't see how it could not be described naturally as generated by the cited set. – vonbrand Feb 04 '13 at 17:52

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If your question is to give a basis of $R[[x]]$ as a free $R$-module, this is equivalent to asking for a basis of an infinite direct product $\prod_{i=0}^\infty R$ of copies of $R$. The answer is that although such a basis must exist if one admits the Axiom of Choice (I think; I am sure this is true when $R$ is a field), there is no way to explicitly describe such a basis, exactly because its existence essentially depends on that Axiom. In any case, the direct product is much larger than the direct sum $\bigoplus_{i=0}^\infty R$ that can be embedded in it, so you correctly saw that a basis for the latter does not nearly generate the whole direct product.

(More precisely, the existence of a basis for all vector spaces implies the Axiom of Choice; the existence of a basis for this particular vector space might not imply the full power of the Axiom, but I am pretty sure that you cannot prove the existence even for $R=\Bbb F_2$ without using some form of the Axiom of Choice.)

Marc van Leeuwen
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  • Still, the space of formal series looks "easy" (because it is just an extension of the polynomial space), so one might hope to find some explicit basis. – tom_a2 Jun 08 '13 at 09:38
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    @tom_a2: More precisely it is a topological completion. In the same way $\Bbb R$ "looks easy" since it is a topological completion of$~\Bbb Q$. However constructing a basis of $\Bbb R$ as $\Bbb Q$-vector space is not possible (even though abstractly it exists), see http://math.stackexchange.com/questions/112859/basis-for-mathbbr-over-mathbbq for instance. The situation here is similar (although in some sense as a ring formal power series relate better to polynomials than $\Bbb R$ does to $\Bbb Q$). – Marc van Leeuwen Jun 08 '13 at 10:24