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Let $Q_1 = \left( \frac{a, b}{\mathbb{Q}} \right)$ and $Q_2 = \left( \frac{a, c}{\mathbb{Q}} \right)$ be quaternion algebras over $\mathbb{Q}$. Prove that $Q_1 \otimes_{\mathbb{Q}} Q_2$ is isomorphic to $\mathrm{M}_4(\mathbb{Q})$ or to $\mathrm{M}_2(Q_3)$, where $Q_3$ is a division algebra.

Definitions:

$(1)$ A central simple algebra (CSA) over a field K is a finite-dimensional associative K-algebra A which is simple, and for which the center is exactly K.

$(2)$ If $a$ and $b$ are nonzero elements of a field $K$ with characteristic different from 2, the quaternion algebra $\left( \frac{a, b}{K} \right)$ is defined as the $K$-algebra generated by elements $i$ and $j$ that satisfy the identities $i^2 = a$, $j^2 = b$, and $ij = -ji$.

Ideas:

$(1)$ $\left( \frac{a, b}{K} \right)$ is a central simple algebra.

$(2)$ The tensor product of central simple algebras is a central simple algebra.

$(3)$ The algebras $Q_1$ and $Q_2$ contain $\mathbb{Q}(\sqrt{a})$ as a subfield.

$(4)$ The function $\lambda: \left( \frac{a, m}{\mathbb{Q}} \right) \to \mathrm{M}_2(\mathbb{Q}(\sqrt{a}))$ given by:

$i, j \mapsto \begin{pmatrix} \sqrt{a} & 0 \\ 0 & -\sqrt{a} \end{pmatrix}, \begin{pmatrix} 0 & m \\ 1 & 0 \end{pmatrix}$

extends uniquely to an injective $\mathbb{Q}$-algebra homomorphism.

$(5)$ If $b = c$, I conjecture that $Q_1 \otimes_{\mathbb{Q}} Q_2$ is isomorphic to $\mathrm{M}_4(\mathbb{Q})$. However, I am unsure how to rigorously prove this.

$(6)$ If $b \neq c$, I conjecture that $Q_1 \otimes_{\mathbb{Q}} Q_2$ is isomorphic to $\mathrm{M}_2(Q_3)$, but i do not know how to describe $Q_3$, nor do I know how to establish this isomorphism.

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    What did you try? – Dominique Nov 20 '24 at 15:37
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    Sorry @Dominique, you are right. I have few ideas at the moment, but I will put them in the question right now. Thanks :) –  Nov 20 '24 at 19:42
  • I don't know what you know, but I'd suggest looking what happens over $\mathbb Q(\sqrt a)$. – Kimball Nov 21 '24 at 15:54
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    I think you're trying to be more specific than required. Let $A = $Q_1 \otimes_{\mathbb{Q}} Q_2$$. Since $A$ is simple, then $A \cong M_n(D)$ for some $n$ and some division $\mathbb{Q}$-algebra $D$ by Wedderburn's Theorem. Now $A$ has dimension 16 and $M_n(D)$ has dimension $n^2 \dim(D)$. The only possibilities are $n = 1,2,4$, and it looks like you're asked to rule out $n=1$ as a possibility, in which case $A$ itself is a division algebra. Can you show $A$ is not a division algebra? I think you can find explicit zero divisors. – Viktor Vaughn Nov 23 '24 at 00:54
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    @ViktorVaughn I suggest writing that as an answer. – coiso Nov 24 '24 at 14:25

1 Answers1

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I think you're trying to be more specific than required. Let $A=Q_1 \otimes_{\mathbb{Q}} Q_2$. Since is simple, then $A \cong M_n(D)$ for some $n$ and some division $\newcommand{\Q}{\mathbb{Q}}\Q$-algebra $D$ by Wedderburn's Theorem. Then $$ 16 = \dim(A) = \dim(M_n(D)) = n^2 \dim(D) \, . $$ so the only possibilities are $n\in \{1,2,4\}$. From the problem statment, you're asked to rule out $n=1$ as a possibility, in which case $A$ itself is a division algebra.

We show that $A$ is not a division algebra by finding explicit zero divisors. Let $i_1, j_1$ and $i_2, j_2$ be the generators of $Q_1$ and $Q_2$, respectively, so $$ i_1^2 = a, \ j_1^2= b, \quad i_2^2 = a, \ j_2^2 = c \, . $$ Then \begin{align*} (i_1 \otimes 1)^2 &= i_1^2 \otimes 1 = a \otimes 1\\ (1 \otimes i_2)^2 &= 1 \otimes i_2^2 = 1 \otimes a = a \otimes 1 \, , \end{align*} so \begin{align*} 0 &= a \otimes 1 - a \otimes 1 = (i_1 \otimes 1)^2 - (1 \otimes i_2)^2 = (i_1 \otimes 1 - 1 \otimes i_2)(i_1 \otimes 1 + 1 \otimes i_2) \, . \end{align*}

Viktor Vaughn
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