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Let $K$ be a quadratic extension of $\mathbb{Q}$. Then we know that $K$ is always a subfield of the algebra $\mathrm{M}_2(\mathbb{Q})$.

A "clean" way to produce embeddings is to observe that since $\dim_\mathbb{Q}(K)=2$ multiplication by elements in $K$ defines an embedding $$ K\longrightarrow\mathrm{End}_\mathbb{Q}(K)\simeq\mathrm{M}_2(\mathbb{Q}), $$ i.e. the choice of a $\mathbb{Q}$-basis of $K$ defines a said embedding.

Now let $D$ be a quaternion algebra over $\mathbb{Q}$. We know that embeddings $K\hookrightarrow D$ exists if and only if $D\otimes_\mathbb{Q}K\simeq\mathrm{M}_2(K)$, and actually one can characterize the situation in terms of local conditions.

The actual construction of embeddings seems a bit messier: if one writes $K=\mathbb{Q}(\sqrt d)$, one can analyze the reduced norm form in $D$ to deduce the existence of an element $x\in D$ such that $x^2=d$ and then embed $K$ in $D$ via $\sqrt d\mapsto x$.

I wonder if there are "cleaner" ways to produce embeddings $K\to D$, more akin to the split case $D=\mathrm{M}_2(\mathbb{Q})$.

It should be clear that I stated the question for quadratic extensions of $\mathbb{Q}$ and quaternion algebras over $\mathbb{Q}$ for the sake of simplicity. The question remains exactly the same replacing $\mathbb{Q}$ with any number field $E$.

Andrea Mori
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  • For a description of the possible subfields in a more general setting see this thread. Doesn't answer your question, I'm afraid. Just linking that. – Jyrki Lahtonen Jun 29 '25 at 10:39
  • @JyrkiLahtonen: thanks for the useful link, but in fact I know the general theory and that is of no help. I'd like to parametrize embedding $K\to D$ using something intrinsic to $K$ (like bases in the split case). Btw, I fixed the $M_2(K)$ typo. – Andrea Mori Jun 29 '25 at 11:27
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    I'm not sure I understand what you mean by "cleaner" ways: you can use Hilbert symbols $(a,b / \mathbb Q)$ to construct all $D$ containing $K= \mathbb Q(\sqrt a)$ inside $M_2(K)$, and I don't see how this is too different from the split case. (Yes, you need to choose $b$, and then apply a little theory to deal with the isomorphism question, but I don't know what more you're hoping for or why this isn't satisfactory.) – Kimball Jun 30 '25 at 02:09
  • @Kimball: I don't need to construct a $D$. The algebra $D$ is actually given and such that $D\otimes K\simeq M_2(K)$. How do you construct embeddings of $K$ in it? In the split case, you just pick a $\mathbb{Q}$-basis of $K$. I would like a "recipe" that involves just $K$. – Andrea Mori Jun 30 '25 at 08:06
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    To me, it looks like in the split case you're constructing $M_2(\mathbb Q)$ as $End_{\mathbb Q}(K)$. I don't see how you can construct an embedding of $K$ into $D$ without either constructing $D$ from $K$ (which requires auxiliary data to pick out your $D$) or finding a element in $D$ which generates $K$. – Kimball Jun 30 '25 at 09:04
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    One idea closer to the endomorphism approach in the split case is to start with a Hermitian form $\Phi$ on $K^2$, and then look at the subalgebra $D$ of $\alpha \in M_2(K)$ such that $\Phi(x \bar \alpha, y \alpha) = \Phi(x,y)$ for $x, y \in K^2$ and bar is the canonical involution. This gives you a quaternion algebra $D$ containing $K$. – Kimball Jun 30 '25 at 09:08
  • @Kimball: about your 1st comment: $K$ embeds into $\mathrm{End}_\mathbb{Q}(K)$ sending $x$ to the multiplication by $x$ map. The identification of the endomorphism ring is not canonical: changing identification chenges the embedding of $K$ in thematrix ring. – Andrea Mori Jun 30 '25 at 11:55
  • @Kimball: your 2nd comment is interesting. I will think about it. – Andrea Mori Jun 30 '25 at 11:57

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