Embedding of a division algebra into a matrix algebra over its centre Let $K$ be a number field and let $D$ be a central division algebra over $K$. Let $d$ be the index so that $[D:K]=d^2$. What is the minimal $n$ such that there exists an embedding of $D$ into $\mathrm{Mat}_{n \times n}(K)$?
Of course, we can always embed $D$ into $\mathrm{Mat}_{n \times n}(K)$ for $n=d^2$, but can we do better than this?
Here's an explicit example. Let $\mathbb{H}=\mathbb{Q}\oplus\mathbb{Q}i\oplus\mathbb{Q}j \oplus\mathbb{Q}k$ with $i^2=j^2=k^2=-1$ and $ij=k=-ji$ be the division algebra of rational quaternions.
Then $d=2$ and we can embed $\mathbb{H}$ into $\mathrm{Mat}_{4 \times 4}(\mathbb{Q})$. We can't embed $\mathbb{H}$ into $\mathrm{Mat}_{2 \times 2}(\mathbb{Q})$ because they have the same dimension. So in this case the question boils down to whether we can embed $\mathbb{H}$ into $\mathrm{Mat}_{3 \times 3}(\mathbb{Q})$. My intuition tells me that this isn't possible, but I can't pin down a proof. Edit:
user44191 has given a proof in the comments below. But I'd still be interested in an answer to the more general question.
Now $\mathbb{Q}(i)$ is maximal subfield of $\mathbb{H}$ and so we have an isomorphism $\mathbb{Q}(i) \otimes_\mathbb{Q} \mathbb{H} \cong \mathrm{Mat}_{2 \times 2}(\mathbb{Q}(i))$. Maybe something like this could help, but I don't quite see how. I've also thought about Brauer groups, but again I don't see how to take advantage of them for this problem.
Edit: I should have said that I don't require an embedding to preserve the multiplicative identity.
 A: Any embedding of $D$ into $M_n(K)$ defines a $D$-module structure on $K^n$. But $D$ is a simple algebra and we know all its modules: they are $D^m\cong K^{km}$ where $k=[D:K]$. Thus, $m=1$ is the best you can do.
A: $n = [D : K]$.
Assume $n < [D : K]$. Let $p: D \rightarrow \mathrm{Mat}_{n \times n}$ be an embedding, and let $\{\alpha_i\}_{1 \leq i \leq d^2}$ be a basis for $D$ over $K$, where $d^2=[D:K]$.
Choose any nonzero vector $v \in V \simeq K^n$. Then as there are $d^2$ elements of the set $\{p(\alpha_i)(v)\}$, they must be linearly dependent, by dimension. Therefore, there is some nontrivial linear relation $\sum_i c_i p(\alpha_i)(v) = 0$. As the embedding is linear, this implies that $p(\sum_i c_i \alpha_i)(v) = 0$. Then as the $\alpha_i$ form a basis, $\sum_i c_i \alpha_i$ is nonzero, and therefore is invertible with some inverse $\beta$. Then as this is an algebra embedding, $p(1)(v) = p(\beta (\sum_i c_i \alpha_i))(v) = p(\beta)(p(\sum_i c_i \alpha_i)(v)) = p(\beta)(0) = 0$. But remember - this is true for any vector, so $p(1) = 0$, which contradicts this being an embedding.
