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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.

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    $\begingroup$ There isn't even an algebra embedding of $\mathbb{Q}(i)$ into $\mathrm{Mat}_{3}(\mathbb{Q})$ (assuming the embedding preserves the identity, and if it doesn't, you're essentially embedding into $\mathrm{Mat}_{2}$ anyway), so in that case, $n = 4$. $\endgroup$
    – user44191
    Commented Apr 6, 2021 at 21:03
  • $\begingroup$ Ah, I was a bit too hasty. I don't require an embedding to preserve the identity. Thanks for pointing this out! $\endgroup$ Commented Apr 6, 2021 at 21:22
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    $\begingroup$ Not preserving the identity doesn't allow an embedding of $\mathbb{H}$ into $\mathrm{Mat}_3$, though - just of $\mathbb{Q}(i)$. The image of the identity has to satisfy $A^2 = A$, so it is a projection; the embedding can then be interpreted as an embedding into the endomorphism of the image of the projection. As such, you're still getting an embedding into $\mathrm{Mat}_2$, which (as you've said) is impossible. $\endgroup$
    – user44191
    Commented Apr 6, 2021 at 21:33
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    $\begingroup$ The above can be generalized; not preserving the identity just means the identity becomes a projection, which means it doesn't allow you to decrease $n$ - just to "fill in" between dimensions where maps that do preserve the identity are possible. e.g. there are no identity-preserving embeddings of $\mathbb{Q}(i)$ in $\mathrm{Mat}_n$ for $n$ odd, but there are obvious embeddings that don't preserve the identity. $\endgroup$
    – user44191
    Commented Apr 6, 2021 at 21:37
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    $\begingroup$ There is. If there is such an embedding, consider the image $A$ of $i$. It must satisfy $A^2 + I = 0$, so its eigenvalues can only be $\pm i$, but its characteristic polynomial is a cubic, which must have a real root. $\endgroup$
    – user44191
    Commented Apr 7, 2021 at 11:51

2 Answers 2

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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.

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  • $\begingroup$ Thank you for this very clear answer! With hindsight, I take your point in the comments, and do feel a bit silly. I guess this is a nice example of the idea that sometimes the best way to understand a ring is through its modules. $\endgroup$ Commented Apr 7, 2021 at 20:42
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$n = [D : K]$.

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.

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  • $\begingroup$ Thanks for this nice answer and for all your help in the comments above. $\endgroup$ Commented Apr 7, 2021 at 20:45

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