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An old qual problem reads

Let $D$ be a 9-dimensional central division algebra over $\mathbb{Q}$ and $K \subset D$ be a field extension of $\mathbb{Q}$ of degree $>1$. Show that $K \otimes_\mathbb{Q} K$ is not a field and deduce that $D \otimes_\mathbb{Q} K$ is no longer a division algebra.

I'm not sure the appropriate tool to use to approach this. Clearly $K$ is of degree $3$. Can someone help out? Perhaps you could also add a reference where this material is covered in more generality.

user26857
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Eric Auld
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2 Answers2

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By the primitive element theorem, we can write $K = \mathbb{Q}[\alpha]/f(\alpha)$ for some irreducible polynomial $f$. Then

$$K \otimes K \cong K[\alpha]/f(\alpha)$$

which is not a field since $K$, by construction, contains a root of $f$, hence $K[\alpha]/f(\alpha)$ is a product of at least two fields (one for every irreducible factor of $f$ over $K$), one of which is $K$. (If $K$ were Galois then we'd be able to say something stronger: namely, $K \otimes K$ would be a product of $\deg f$ copies of $K$.)

It follows that $D \otimes K$ contains a subring $K \otimes K$ which isn't a domain, so isn't a division ring. I'm a little worried here because I haven't used the hypothesis that $D$ is $9$-dimensional, but as far as I can see this isn't necessary.

Qiaochu Yuan
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    I hazard a guess that the exact dimension of $D$ plays a role in a follow-up exercise. We can conclude that $\dim_{\Bbb{Q}}K$ must be $3$. Therefore $K$ is a maximal subfield of $D$, and hence splits it. So it follows that $D\otimes K\cong M_{3\times3}(K)$. – Jyrki Lahtonen Jan 13 '16 at 08:20
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I came across the same problem while qual-prepping and, despite the high quality of Qiaochu's answer, I prefer to think of this problem in a different (but hopefully still correct!) way. We have a $\mathbb{Q}$-algebra homomorphism $\mu\colon K\otimes_\mathbb{Q}K\to K$ defined by \begin{equation} x\otimes y\mapsto xy. \end{equation} Because $\dim_\mathbb{Q}(K) > 1$, \begin{equation} \dim_\mathbb{Q}(K\otimes_\mathbb{Q} K) = \dim_\mathbb{Q}(K)^2 > \dim_\mathbb{Q}(K), \end{equation} so $\mu$ has a nontrivial kernel, but certainly $\mu$ is not the zero map. Then $\ker\mu\subset K\otimes_\mathbb{Q}K$ is a nontrivial, proper ideal, meaning that $K\otimes_\mathbb{Q}K$ is neither a field nor a domain. As Qiaochu says, the ring $D\otimes_\mathbb{Q} K$ then has a subring which is not a domain, and is thus not a division ring.

Keenan Kidwell uses the map $\mu$ in asnwers here and here; I hope I've adapted it appropriately.

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