$\mathbb{Q}(\sqrt2)\otimes_\mathbb{Q} \mathbb{Q}(\sqrt2)$ isn't a field.

Question

I have the following problem:

In our script there is an example stating: $\mathbb{Q}(\sqrt2)\otimes_\mathbb{Q} \mathbb{Q}(\sqrt2)$ isn't a field. But it isn't proven and I don't know how to do it. Can someone explain that to me? Thanks in advance.

Answer 1

Let us make some distinction, denote $$ \begin{aligned} x &= \sqrt 2\otimes 1\ ,\\ y &= 1\otimes \sqrt 2\ . \end{aligned} $$ Then the given ring is a vector space of dimension four over $\Bbb Q$, with basis $1$, $x$, $y$, $xy$. Now let us look at: $$ (x+y)(x-y)=x^2-y^2=2\otimes 1-1\otimes 2=0\ .$$ (The elements $x\pm y$ are not zero.)