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The title couldn't quite contain the question, so I didn't attempt to make it precise. I should note that this is the third or fourth question I've asked these past two days about problems I've been having, and I want to thank you all for being really helpful.

The question is this: Say that you start with a number field $K$ that doesn't contain $\zeta_p$. Let $\Delta$ be the automorphism group of $K(\zeta_p)/K$. I will view $K(\zeta_p)^{\times} \otimes \mathbb{F}_p$ as a $\mathbb{F}_p[\Delta]$-module. Let $\omega$ be the cyclotomic character (meaning that for any $\delta \in \Delta$ and $\zeta_p$, $\delta(\zeta_p)=(\zeta_p)^{\omega(\delta)}$). Is there an easy description of $(K(\zeta_p)^{\times} \otimes \mathbb{F}_p)^{\omega}$?

Note that I mean the $\omega$-isotypic component, as is explained in one of my previous questions: Basic Representation Theory

Obviously $\zeta_p^n$ is in $(K(\zeta_p)^{\times} \otimes \mathbb{F}_p)^{\omega}$ for all integer $n$. What else is in there? Is there a nice description?

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You should register and ask the moderators to merge your several accounts into one. You should then also start accepting answers that you are satisfied with, so that people don't think that the questions have not been answered. Otherwise, this will continue deflecting peoples' attention from other questions that still need answering. –  Alex B. Apr 7 '11 at 4:31
    
Okay, Alex. Will do. –  Nicole Apr 7 '11 at 14:55

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up vote 4 down vote accepted

Write $L=K(\zeta_p)$, and let $G_K$ (resp. $G_L$) be the abolute Galois group of $K$ (resp. $L$), so that $G_K/G_L = \Delta$. Inflation-restriction gives an isomorphism $H^1(G_K,\mathbb Z/p) = H^1(G_L,\mathbb Z/p)^{\Delta},$ where the superscript $\Delta$ denotes the submodule of $\Delta$-invariants. (More precisely, there is a map from the left-hand side to the right-hand side, which in this case is an isomorphism because the flanking terms in the inflation-restriction exact sequence involve cohomology of $\Delta$ with coefficients in $\mathbb Z/p$, and this cohomology vanishes, since $|\Delta| \mid p-1$, which is coprime to $p$.)

Now if we choose an isomorphism $\mathbb Z/p \cong \mu_p$ (i.e. fix a choice $\zeta_p$ of primitive $p$th root of $1$), then we get an isomorphism $H^1(G_L,\mathbb Z/p)^{\Delta} = H^1(G_L,\mu_p)^{\omega}.$ (The superscript $\omega$ here I am using in your sense, namely it is the submodule on which $\Delta$ acts via $\omega$.)

[Added: Note that $\Delta$ acts on $\mu_p$ through the character $\omega$. So a cohomology class $c$ in $H^1(G_L,\mathbb Z/p)^{\Delta}$ is fixed by $\Delta$ if and only if, when it is regarded as a class in $H^1(G_L,\mu_p)^{\omega}$, the action of $\Delta$ is given by the character $\omega$.]

Finally, Hilbert's Theorem 90 gives an isomorphism $H^1(G_L,\mu_p) = L^{\times}\otimes \mathbb F_p.$

So in conclusion, there is an isomorphism $$(L^{\times} \otimes \mathbb F_p)^{\omega} \cong H^1(G_K,\mathbb Z/p) = Hom(G_K^{ab},\mathbb Z/p).$$


The preceding isomorphism has a concrete interpretation: namely, a (non-trivial) map from $G_K^{ab}$ to $\mathbb Z/p$ corresponds to a degree $p$ abelian exension of $K$, say $F$. If we form the compositum of $F$ and $L$, we obtain a degree $p$ extension of $L$. (Because the degree of $L$ over $K$ divides $p-1$, this compositum is still genuinely of degree $p$ over $L$.) Kummer theory allows us to describe this extension by extracting the $p$th root of some element $l \in L^{\times}$. The image of $l$ in $L^{\times}\otimes \mathbb F_p$ is then an element of $(L^{\times}\otimes \mathbb F_p)^{\omega}.$

Conversely, if we have an element of $(L^{\times}\otimes \mathbb F_p)^{\omega},$ say the image of some element $l \in L^{\times}$, then adjoining the $p$th root of $L$ to $L$ gives a cyclic extension of $L$ of degree $p$. The assumption that $\Delta$ acts on the image of $l$ in $L^{\times}\otimes\mathbb F_p$ via $\omega$ shows (with a little calculation) that $L(l^{1/p})$ is in fact abelian (of degree $(p-1)p$) over $K$, and so it contains a degree $p$ subextension $F$ of $K$, so that $L(l^{1/p}) = F L.$

The two constructions just described give an explicit description of each direction of the isomorphism obtained by cohomological methods above.


For example, the element $\zeta_p$ that you mention in your question corresponds to the degree $p$ subextension of $K$ contained in $K(\zeta_{p^2})$. But any number field admits (infinitely) many other degree $p$ abelian extensions, and these give rise to (infinitely) many other elements of $(L^{\times}\otimes \mathbb F_p)^{\omega}$.

Let me finally explain why it is reasonable that there should be many such elements: the space $L^{\times}\otimes \mathbb{F}_p$ is an infinite dimensional vector space over $\mathbb F_p$, which (since $\Delta$ has order prime-to-$p$) breaks up into a sum of eigenspaces under the $\Delta$-action. But $\Delta$ has only finitely many characters, so just by counting, we see that at least one eigenspace is going to have to be infinite dimensional. In fact, all the eigenspaces will be infinite dimensional, and the preceding discussion proves this for the $\omega$-eigenspace.

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Thanks! That was very helpful! –  Nicole Apr 7 '11 at 15:08
    
Would you mind expatiating a little bit on $H^1(G_L,\mathbb{Z}/p)^{\Delta}=H^1(G_L,\mu_p)^{\omega}$? I'm still a little vague on this. –  Nicole Apr 7 '11 at 15:55
    
@Nicole: Dear Nicole, I added a line of explanation as you asked. Regards, –  Matt E Apr 8 '11 at 0:36

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