***Reformulation of the question (see below for the original question):*** Let $K$ be an algebraic number field and $D$ a finite-dimensional $K$-division algebra. Is there a description of the field extensions $L\supseteq K$ such that $L\otimes\_K D$ is a division algebra?


I have encountered this question in the context of representation theory of finite groups, see below.




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*Original question:*


I have asked this question at math.stackexchange, but have not received an answer so far. Also, I'm not entirely sure that this is a suitable question for mathoverflow... See [this link](https://math.stackexchange.com/questions/1077589/over-which-fields-is-a-g-module-reducible) for the original question.


Let $K$ be an algebraic number field, $G$ a finite group and $V$ an irreducible $KG$-module with character $\chi$. Since V is irreducible, $\chi$ is of the form $\chi=m(\vartheta\_1+...+\vartheta\_s)$, where the $\vartheta\_i$ are the distinct conjugates of an absolutely irreducible character of G under the Galois group $\mathrm{Gal}(K[\vartheta]/K)$ of the character field $K[\vartheta]:=K[\vartheta(g) ~:~ g \in G]$ and m is the Schur index. We have $s=[K[\vartheta]:K]$.


My question is: Is there a characterisation of field extensions $L\supseteq K$ such that $V\_L := L\otimes\_K V$ is a reducible $LG$-module?


Since I asked this question at stackexchange I have found that for $m=1$ (and therefore $s\geq 2$, because otherwise we already have an absolutely irreducible representation) we have that $V\_L$ is reducible if and only if $L$ contains an intermediate field $K\subseteq F \subseteq K[\vartheta]$ such that $\mathrm{Gal}(K[\vartheta]/F)\leq \mathrm{Gal}(K[\vartheta]/K)$ has more than one orbit on $\{\vartheta\_1,...,\vartheta\_s\}$.


However, I am unsure about the case of $m\geq 2$, which I assume is more subtle. Is there a similarly succinct characterisation of field extensions $L|K$ such that $V\_L$ is reducible?


*Edit:* As Geoff Robinson points out, this problem may be restated as, "Which fields $L\supseteq K$ have the property that $\mathrm{End}\_{LG}(V\_L)$ is not a division algebra?" which is basically a question about division algebras over number fields.


Considering the (small) example of the quaternion group $Q\_8$ over the rationals and its irreducible four-dimensional representation $V$, a quick computation shows that $\mathrm{End}\_{\mathbb{Q}Q\_8}(V)$ becomes a matrix ring over $\mathbb{Q}(\sqrt{-1})$, $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{-5})$ but not over $\mathbb{Q}(\sqrt{-7})$.


I have, at present, no idea how to characterize these fields, especially in terms of some group theoretic or character theoretic property of $Q\_8$. Then again, this is probably not to be exprected, considering that $Q\_8$ and $D\_8$ have "equal" character tables, but the two-dimensional representations have distinct Schur indices...



Division algebras over extension fields / reducibility of $G$-modules
Let $K$ be a number field and $D$ a finite dimensional central division algebra over $K$. I believe that one can characterize the field extensions $L$ of $K$ for which $D\otimes\_K L$ is a division algebra. I am not sure how useful this particular characterization will be in the context of the representation theory of finite groups (which seems to be your primary interest), but perhaps if nothing else it will allow you to recast your problem in slightly different terms.


Anyway, I believe that the characterization you are after is a more or less straightforward consequence of the Albert-Brauer-Hasse-Noether theorem and follows from the ideas in Chapters 31 and 32 of Reiner's book *Maximal Orders*. 


Given a prime $\mathfrak p$ of $K$ denote by $m\_\frak{p}$ be the index of $D\otimes\_K K\_{\mathfrak p}$ (e.g., the degree of the division algebra part of $D\otimes\_K K\_{\mathfrak p}$). The following is Theorem 32.15 of Reiner.


**Theorem**. *Let $L$ be a finite extension of $K$ and suppose that $\dim\_K D=d^2$. Then $D\otimes\_K L\cong M\_d(L)$ if and only if for every prime $\mathfrak p$ of $K$ and prime $\mathfrak P$ of $L$ lying above $\mathfrak p$, we have:* 
\begin{equation} 
m\_{\mathfrak p}\mid [L\_\mathfrak P:K\_\mathfrak p].
\end{equation}


This theorem does not directly answer your question of course, but its proof contains all of the elements needed to do so. So it is worth saying something about how the theorem is proven. The idea is that the local Hasse invariant of $D$ (at a prime $\mathfrak p$ of $K$) is $\mathrm{inv}\_{\mathfrak p}(D)=\frac{s\_{\mathfrak p}}{m\_{\mathfrak p}}$ where $s\_{\mathfrak p}$ is an integer coprime to $m\_{\mathfrak p}$ satisfying $1\leq s\_{\mathfrak p}\leq m\_{\mathfrak p}$. The Hasse invariant of $D\otimes\_K L$ (at the prime $\mathfrak P$) on the other hand, is $[L\_\mathfrak P:K\_\mathfrak p]\cdot \frac{s\_{\mathfrak p}}{m\_{\mathfrak p}}$. The Albert-Brauer-Hasse-Noether theorem says that a central simple algebra is split globally if and only if it is split locally for all primes. Putting these facts together is how one proves the above theorem.


It is not too hard to see how all of this must be modified to characterize not when $D$ is split by $L$ but instead when $D\otimes\_K L$ is a division algebra. It goes without saying that you will need to ensure that $m\_{\mathfrak p}$ does not always divide $[L\_\mathfrak P:K\_\mathfrak p]$. It is possible for this condition to hold and for $D\otimes\_K L$ to still not be a division algebra though. It could be isomorphic to $M\_{d'}(D')$ for some division algebra $D'$ over $L$ and integer $d'>1$, for instance. To ensure that $D\otimes\_K L$ is a division algebra you will need to make use of the following result (which just cobbles together a bunch of results in Reiner and basically follows from the short exact sequence of Brauer groups that one gets in class field theory):


**Theorem**. *Let $S$ be a finite collection of primes of $L$ consisting of finite primes and real infinite places. Suppose that for each $\mathfrak P$ in $S$ we are given a reduced fraction $\frac{a\_{\mathfrak P}}{b\_{\mathfrak P}}$ such that:*


*1. $b\_{\mathfrak P}>1$ and $a\_{\mathfrak P}>0$;*


*2. $\frac{a\_{\mathfrak P}}{b\_{\mathfrak P}}=\frac{1}{2}$ whenever $\mathfrak P$ is real; and*


*3. $\sum\_{\mathfrak P \in S} \frac{a\_{\mathfrak P}}{b\_{\mathfrak P}}\in \mathbb Z.$*


*Then there exists a unique division algebra $D'$ over $L$ having $S$ as its set of ramified primes, Hasse invariant $\frac{a\_{\mathfrak P}}{b\_{\mathfrak P}}$ at the prime $\mathfrak P$, and degree $d'$ for $d'=\mathrm{lcm}[b\_{\mathfrak P}]$.*


So to make sure that $D\otimes\_K L$ is a division algebra you just need to ensure that the least common multiple of the denominators of its local invariants $[L\_\mathfrak P:K\_\mathfrak p]\cdot \frac{s\_{\mathfrak p}}{m\_{\mathfrak p}}$ is equal to the degree of $D$.


Finally, note that in certain cases one can recast this characterization in friendlier terms. Suppose for instance that $D$ is a quaternion algebra and that $L$ is a quadratic field extension of $K$. Then $m\_{\mathfrak p}$ is equal to $1$ or $2$ and $[L\_\mathfrak P:K\_\mathfrak p]=1$ precisely when $\mathfrak p$ splits in $L/K$. So in this case the first theorem I stated just says that $D\otimes\_K L$ is split if and only if no prime which ramifies in $D$ splits in $L/K$. You can come up with similar (though necessarily more complicated) characterizations in terms of the splitting behavior in $L/K$ of ramified primes in $D$ in the general case.


