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let $F$ be a field. for which ring $R$, $F$ is an $R$-module. i know already that as an abelian group $F$ is a $\mathbb Z$- module, what else can we say for a general field $F$.

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Small comment: note that this question is equivalent to the question: for which rings $R$ there is an homomorphism $R \to F$. – the L Jun 7 '11 at 10:07
$R$ can be any subring of $F$. – wxu Jun 7 '11 at 10:11
@anonymous: why there is such equivalence – palio Jun 7 '11 at 11:18
actually there is no. My mistake. Such homomorphisms will give you all possible structures of $F$ as an $R$-algebra, but I see you want less. – the L Jun 7 '11 at 11:38
up vote 4 down vote accepted

I'm afraid there isn't an awful lot we can say in general. The way the question was posed, the field multiplication is not tied to the $R$-module action at all. Therefore any kind of a module action on the additive group will work. For example, if $F=\mathbf{Q}(i,\sqrt{5})$, then additively $F$ is isomorphic to the Cartesian power $\mathbf{Q}^4$, and therefore $F$ can be viewed as a module over any ring that has a 4-dimensional rational representation, e.g. the ring of rational Hamiltonian quaternions. Similarly $F$ can be turned into a $\mathbf{Q}[G]$-module for a group $G$ that is isomorphic to any $\mathbf{Q}[G]$-module $V$ with $\dim_{\mathbf{Q}}V=4$ simply by defining the module action in such a way that a linear isomorphism between $V$ and $F$ becomes a module isomorphism.

The characteristic of the field will place some restrictions. If $F$ has characteristic zero, then it has no additive torsion, and the ring $R$ cannot have any torsion either. If $F$ has characteristic $p$, then the additive order of $1_R$ must be either infinite or a multiple of $p$. In other words $F$ will always be either a $\mathbf{Q}$-module (when $char F=0$) or a $\mathbf{Z}/p\mathbf{Z}$-module (when $char F=p$).

Similarly, as you have seen in a field theory course, $F$ is always a module over any of its subfields.

The question would become quite a bit more interesting, if you tie the $R$-action to the $F$-multiplication. A natural way of doing that would be to insist that for all $x\in F$ the mapping $\rho_x:F\rightarrow F, a\mapsto ax$ would also be a homomorphism of $R$-modules. IOW, $F$ would be what is known as a $(R,F)$-bimodule. This places more severe constraints on the ring $R$.

But before we attempt anything like that I ask you to comment. Is this anything like what you expect? I am uncertain as to exactly what you want to learn? I understood the question to mean that given a field $F$, what kind of rings $R$ act on it in a way that turn the additive group $(F,+)$ into an $R$-module. If you wanted to ask something else, please be more specific :-)

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thanks jykri. my question is that given a field $F$ what are the most known ways to turn $F$ into module for example i know that it is a $\mathbb Z$-module as an abelian group and also an $F$-module... is there other known ways? – palio Jun 7 '11 at 17:28
@palio: I listed a few such ways. Most of the ones that I can think of right away depend on our ability to view $F$ as a vector space over any of its subfields, and it is not difficult to define module actions on a more or less arbitrary vector space. To add one more to the list: We can view $F$ an $F[x]$ module by letting $x$ act as multiplication by a constant $\lambda\in F$. This gives a module action in which a polynomial $p(x)\in F[x]$ acts a multiplication by the constant $p(\lambda)$. – Jyrki Lahtonen Jun 7 '11 at 17:47
I agree with your interpretation as asking for which R can (F,+) be an R-module. I think though one can give a succinct characterization of such rings: those with a ring homomorphism into the endomorphism ring of the additive group of F. This is tautological, so probably useless. Anonymous's interpretation as looking for R in which F is an R-algebra is also reasonable (R with a homomorphism into F). Your bimodule interpretation may be a nice middle ground. – Jack Schmidt Jun 7 '11 at 18:43

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