# Scalar extension of simple algebras

I have a question regarding quaternion algebras. Let $K$ be a number field and $L|K$ a quadratic field extension. Let $M|K$ be a field extension such that $M\otimes_K L$ is not a field, i.e. $M\otimes_K L \cong M\oplus M$. Now I consider a central simple algebra $\mathcal{D}$ over $L$, i.e. the center of $\mathcal{D}$ is $L$. If necessary, we can restrict ourselves to the case of $\mathcal{D}$ being a quaternion algebra over $L$. Is it true in general that $A:=\mathcal{D}\otimes _K M$ is a direct sum of two isomorphic central simple algebras over $M$ and if so, why is that true?

We certainly have $Z(A) \cong Z(\mathcal{D}) \otimes_K M \cong M\oplus M$ (where $Z(\phantom{x})$ denotes the center) and that $A$ is a semisimple $M$-algebra. So by Wedderburn's theorem it is a direct sum of two matrix rings over central $M$-division algebras $D_1$ and $D_2$. I.e. $A\cong D_1 ^{n_1\times n_1}\oplus D_2^{n_2\times n_2}$. Is it true that $D_1\cong D_2$ and $n_1=n_2$?

I came across this problem in a much more specific context, but my gut tells me that the claimed statement may be true in the context provided here...

The answer is yes. One has to check that the isomorphism $L\otimes_K M \cong M\oplus M$ is $L$-linear, so that it is an isomorphism of $L$-algebras. To do so, it is necessary to notice that the condition "$M\otimes_K L$ is not a field" implies that $L$ is a subfield of $M$.
Then one has $$D\otimes_K M \cong (D\otimes_L L) \otimes_K M \cong D\otimes _L (M\oplus M) \cong D\otimes_LM \oplus D\otimes_L M.$$ Since $D$ is a central simple $L$-algebra and $M$ is a simple $L$-algebra, we have that $D\otimes_L M$ is simple, so it is isomorphic to $\tilde{D}^{k\times k}$ for some $k\in\mathbb{N}$ and some central $M$-division algebra $\tilde{D}$.
$D$ being central simple over $L$ is the only necessary condition on $D$.
This generalizes to the situation where $L$ is an arbitrary finite extension of $K$ (not necessarily of degree $2$) such that $M\otimes_K L \cong \bigoplus_{i=1}^{[L:K]} M$.