Example of two field extensions such that their tensor product is not a field Example of two fields $K$ and $L$, which are extensions over $k$, such that $K\otimes_k L$ is not a field. 
Here is what I did. But I am a little bit unsure. Can someone suggest anything, or perhaps an easier example? 
The example that I came up with is $k=\mathbb{Q}$, $K=\mathbb{Q}(\sqrt{2})$, and $L=\mathbb{Q}(i)$. Then $K$ and $L$, viewed as modules over $k$, will have tensor product equal to $\mathbb{Q}^4$, under the universal map, 
$$ (a+b\sqrt{2})\otimes (c+di) = (a,b,c,d) $$
The ring structure on $\mathbb{Q}^4$ will then be given by, 
$$ (a,b,c,d)\cdot (x,y,z,w) = (ax + 2by, ay + bx, az - bw, aw + bz) $$
Then we see that $(1,1,0,0)\cdot (0,0,1,1) = (0,0,0,0)$. Thus, $\mathbb{Q}^4$ is not even an integral domain. 
 A: $\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(\sqrt{2})\simeq\mathbb Q(i)\otimes_{\mathbb Q} \mathbb Q[X]/(X^2-2)\simeq\mathbb Q(i)[X]/(X^2-2)$ which is a field since $X^2-2$ is irreducible over $\mathbb Q(i)$. (This field is in fact $\mathbb Q(i,\sqrt 2)$.)
However, if you consider $\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(i)\simeq\mathbb Q(i)\otimes_{\mathbb Q} \mathbb Q[X]/(X^2+1)\simeq\mathbb Q(i)[X]/(X^2+1)$ this is not a field since $X^2+1$ is reducible over $\mathbb Q(i)$.
A: Recall that the tensor product is bilinear (over $\mathbb{Q}$ in your example). Therefore,
$$(a + b\sqrt{2})\otimes(c + di) = ac(1\otimes 1) + ad(1\otimes i) + bc(\sqrt{2}\otimes 1) + bd(\sqrt{2}\otimes i).$$
Note that $\{1\otimes 1, 1\otimes i, \sqrt{2}\otimes 1, \sqrt{2}\otimes i\}$ is a basis for $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$ as a vector space over $\mathbb{Q}$. Using this basis, the isomorphism $\varphi : \mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i) \to \mathbb{Q}^4$ that you obtain is generated by
$$\varphi((a + b\sqrt{2})\otimes(c + di)) = (ab, ac, bc, bd).$$
I said generated by because not every element of $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$ can be written in the form $(a + b\sqrt{2})\otimes(c + di)$, but can be written as a sum of such elements. The inverse isomorphism is given by
$$\varphi^{-1}(e, f, g, h) = e(1\otimes 1) + f(1\otimes i) + g(\sqrt{2}\otimes 1) + h(\sqrt{2}\otimes i).$$
Now note that if $\alpha_1\otimes\beta_1, \alpha_2\otimes\beta_2 \in \mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$, then $$(\alpha_1\otimes\beta_1)\cdot(\alpha_2\otimes\beta_2) = (\alpha_1\alpha_2)\otimes(\beta_1\beta_2).$$ Extending linearly gives the product structure on $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$. More explicitly, we have
\begin{align*}
=&\ ax(1\otimes 1) + ay(1\otimes i) + az(\sqrt{2}\otimes 1) + aw(\sqrt{2}\otimes i) + bx(1\otimes i) - by(1\otimes 1)\\
&+ bz(\sqrt{2}\otimes i) - bw(\sqrt{2}\otimes 1) + cx(\sqrt{2}\otimes 1) + cy(\sqrt{2}\otimes i) + 2cz(1\otimes 1) + 2cw(1\otimes i)\\
&+ dx(\sqrt{2}\otimes i) -dy(\sqrt{2}\otimes 1) + 2dz(1\otimes i) -2dw(1\otimes 1)\\
&\\
=&\ (ax - by + 2cz - 2dw)(1\otimes 1) + (ay + bx + 2cw + 2dz)(1\otimes i)\\ 
&+ (az - bw + cx - dy)(\sqrt{2}\otimes 1) + (aw + bz + cy + dx)(\sqrt{2}\otimes i).
\end{align*}
Therefore, the induced product structure on $\mathbb{Q}^4$ is actually
\begin{align*}
&(a, b, c, d)\cdot(x, y, z, w)\\ 
=&\ (ax - by + 2cz - 2dw, ay + bx + 2cw + 2dz, az - bw + cx - dy, aw + bz + cy + dx).
\end{align*}

It turns out that the example you picked is a field as pointed out by KCd in the comments; it is $\mathbb{Q}(\sqrt{2}, i)$.
The approach of picking field extensions $K$ and $L$ of $k$ such that $K\otimes_kL$ is not an integral domain (and hence not a field) is a good one. In fact, it is known precisely when $K\otimes_kL$ is an integral domain: when $K$ and $L$ are somewhere $k$-linearly disjoint; see Proposition $128$ of these notes by Pete Clark. Later in that document (Proposition $130$), there is a condition on $K$ and $L$ which determines when $K\otimes_kL$ is in fact a field.
As suggested by KCd in the comments, you should try the case $K = L$. For any non-trivial field extension $K$ of $k$, $K\otimes_kK$ is not an integral domain (in the language above, $K$ and $K$ are not somewhere $k$-linearly disjoint). For example, can you find two non-zero elements of $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})$ with product zero? What about $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$?
