Let $K$ be a field of characteristic zero, or if this is too general, an algebraic number field. Let $G$ be a finite group and $V$ an irreducible and finite-dimensional $KG$-module.

Let $\chi$ be the character of $V$. 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 absolute Galois group of $K$ and $m$ is the Schur index.

Here is my question: Is there a succinct way of characterising those field extensions $L\supseteq K$ such that $V_L:=L\otimes_K V$ is reducible and how does this relate to the character field $K[\vartheta_i]:=K[\vartheta_i(g) ~:~ g \in G]$?

Here's what I have figured out so far: 1) $V_L$ is reducible if $K[\vartheta_i]\subseteq L$ and $s>1$. The condition that $s>1$ is necessary, as may be seen by studying the irreducible $4$-dimensional representation of the quaternion group $Q_8$ over $\mathbb{Q}$.

2) The statement of 1) cannot be reversed, i.e. $$V_L \text{ reducible} \Longrightarrow L\supseteq K[\vartheta_i]$$ is false. A counter-example is provided by the irreducible four-dimensional representation $V:=\mathbb{Q}(\zeta_5)$ of the cyclic group with five elements, which becomes reducible when the scalars are extended from $\mathbb{Q}$ to $\mathbb{Q}(\sqrt{5})$, a proper subfield of the character field, which is the fifth cyclotomic field.

So, I wonder if there is a characterisation of the form "$V_L$ is reducible if and only if $L$ fulfills the property [...]"?

Edit: I have taken the liberty of crossposting this to mathoverflow: https://mathoverflow.net/questions/191465/over-which-fields-is-a-g-module-reducible

Also, I have noticed that in the case of $m=1$, $s\geq 2$ we have that $V_L$ is reducible if and only if $L$ contains an intermediate field $F$ of $K[\vartheta]/K$ such that $\mathrm{Gal}(K[\vartheta]/F)$ has more than one orbit on $\{\vartheta_1,...,\vartheta_s\}$.


This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Ben Linowitz below.

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.


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