Field extension whose tensor product with itself over $\mathbb{Q}$ is not a field 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.
 A: 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. 
A: 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.
