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Suppose we have the Galois field extensions $\mathbb{F} \leq \mathbb{L} \leq \mathbb {K}$ with Galois groups $\Gamma = Gal(\mathbb{K} / \mathbb{F})$, $N=Gal (\mathbb{K} / \mathbb{L})$. The Galois group of $\mathbb{L}$ over $\mathbb{F}$ is ${\Gamma} / {N}$. Let $D$ be a central simple algebra over $\mathbb{L}$ that is split by $\mathbb{K}$. Is there any natural way to define a $\Gamma / N$ Galois action on $D$?

One obvious example is when $D=M_n (\mathbb{L})$. My first thought was to think of $D$ as some subalgebra of $M_n (\mathbb{K})$, but except for the case above, D will contain elements that are in $\mathbb{K} - \mathbb{L}$ and the regular action of $\Gamma / N$ doesn't act on such elements. Since we are dealing with Galois extensions, then for each $\tau \in Gal(\mathbb{L}/\mathbb{F})$ there is a $\sigma \in\Gamma$ that its restriction to $\mathbb{L}$ is $\tau$ and we can use these function to define the action, but then how can we choose them so they will be compatible with each other?


another example is if $D'$ is central simple over $\mathbb{F}$ and is split by $\mathbb{K}$ then $D'\otimes_{\mathbb{F}}\mathbb{L}$ is central simple over $\mathbb{L}$ and has the natural action $\sigma(a\otimes \lambda)=a\otimes \sigma(\lambda)$

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Duh. Obviously the best known example of the kind of Galois group action my answer is trying to describe is the extension of the complex conjugation to Hamilton's quaternions. If we write $\mathbf{H}$ as the set of $2\times2$ complex matrices of the form $$\left(\begin{array}{rr}z_1&z_2^*\\-z_2&z_1^*\end{array}\right),$$ where $^*$ denotes the usual complex conjugation, then it is immediately clear that entrywise complex conjugation of such matrices is an automorphism of $\mathbf{R}$-algebras. I am not quite convinced, whether you would call this action 'natural', though. –  Jyrki Lahtonen Jul 7 '11 at 4:10
    
@Jyrki Lahtonen : I might have misused the word "natural" a bit. But when you look on an algebra as an extension by scalars of another algebra, it is "natural" to define the action on the scalar part of the tensor product, and then you get that the invariants of the action is the algebra you started with. –  Prometheus Jul 7 '11 at 5:21
    
Now I recalled a result stating that (perhaps under some conditions) any automorphism of a subfield of $D$ can be realized as an inner automorphism of $D$, i.e. as a conjugation. Need to dig out the exact formulation. Stay tuned. –  Jyrki Lahtonen Jul 7 '11 at 5:57
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I don't know too much about this, but I will make a couple points.

Let us assume first that $K$ is a maximal subfield of $D$. Then $D$ will be of dimension $n^2$ over $L$ and $[K:L]=n$. So viewing $D$ as a vector space over $K$ acting from the right, the left regular representation of $D$ consists of $K$-linear mappings, and we can thus realize $D$ as a subspace of $M_n(K)$. The question then becomes, will $D$ be stable under the entrywise action of $\Gamma$. In general the answer is no, but I think that it will be stable under the action of $Gal(K/L)$.

To make this a bit concrete take a look at my answer here: An example of a division ring $D$ that is **not** isomorphic to its opposite ring There $E$ has the same role as $K$ here and $F$ has the role of $L$. The extension is cyclic (and also cubic, but in general the extension of the degree does not change my points here), and you see immediately that acting by any element of $Gal(K/L)$ maps that matrix to another one of the same form, i.e. to another element of $D$. However, if you apply an element $\tau$ of $Gal(L/F)$ to all the matrix entries, you will not get the same set, if $\tau(\gamma)\neq\gamma$, where $\gamma$ is that special non-norm element. What happens in that case is that $D$ is mapped to another division algebra $D^\tau$. In other words $\tau$ acts on the Brauer group $Br(K/L)\le Br(L)$. I'm sure this action can be described also at the level of Hasse invariants (if $K,L,F$ are number fields).

The case of cyclic division algebra is nothing special here. That element $\gamma$ there just defines a corresponding 2-cocycle of $Gal(K/L)$ acting on $L^*$. In the case of a non-cyclic extension $K/L$ the desription of a 2-cocycle is more complicated, and my copy of Jacobson's Basic Algebra II is in my office, so I will pass that. Similar phenomenons occur, though. The 2-cocycle may be stable or may not, if we let an element of $Gal(K/F)$ act on it, and again $D^\tau$ may be a different element of $Br(K/L)$.

[Edit] If $K$ is a subfield of $D$ but not maximal then, IIRC we can find a maximal subfield of $D$ containing $K$ that is Galois over $L$. Similar tricks work then, I think. Two related results that I found are (chapter 4 of Jacobson's Basic Algebra II):

Theorem: Let $A$ be a simple subalgebra of a finite dimensional central simple algebra $B$. Then any homomorphism of $A$ into $B$ can be extended to an inner automorphism of $B$.

Theorem: (Skolem - Noether) Any automorphism of a finite dimensional central simple algebra is inner.

I think that both these theorems assume the automorphism to be linear over the center, i.e. homomorphisms of $L$-algebras. Therefore these results do not say anything about mappings extending $Gal(L/F)$.[/edit]

On the other hand, if $K$ is very large, then I don't think we have much hope of defining an action of $Gal(K/L)$ on $D$.

I wish I could have made my answer independent of that particular representation of $D$. Hopefully some more knowledgable poster can add to this, and correct eventual errors.

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