Tensor product of quadratic number field with itself

Question

If $F=\mathbb{Q}(\sqrt{D})$, is there a nice structure to $F\otimes_{\mathbb{Q}} F$?

A spanning set of that tensor product is $1\otimes 1$, $1\otimes \sqrt{D}$, $\sqrt{D}\otimes 1$, and $\sqrt{D}\otimes\sqrt{D}$. I think these are linearly independent, since $\sqrt{D}\notin\mathbb{Q}$. In fact, if I write them as matrices (with respect to the previous basis, in order), I get that

$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}, \begin{pmatrix} 0 & D & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & D\\ 0 & 0 & 1 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 & D & 0 \\ 0 & 0 & 0 & D\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 & D^2 \\ 0 & 0 & D & 0\\ 0 & D & 0 & 0\\ 1 & 0 & 0 & 0\end{pmatrix}$$

All of these are full-rank, actually. I tried checking whether an arbitrary linear combination of these can have determinant 0, but I couldn't solve the resulting equation.

Based on another question, I can map $F\otimes_{\mathbb{Q}} F\rightarrow F$ by $x\otimes y\mapsto xy$, and by dimensionality this has a kernel of dimension 2. In fact I can find a basis of this kernel: $1\otimes\sqrt{D}-\sqrt{D}\otimes 1$ and $\sqrt{D}\otimes\sqrt{D} - D\otimes 1$. So this has a non-trivial ideal, so it is not a field, but it seems to be an integral domain. Is there much more I can say about it?

Answer 1

Yes, there is a very nice description of $F\otimes_{\mathbb Q} F$, and, more generally, $E\otimes_k F$ where $k$ is a field, $E$ is a finite separable extension, and $F$ is an arbitrary field extension.

In the simple case of the question, $$ {\mathbb Q}(\sqrt{D})\otimes_{\mathbb Q} {\mathbb Q}(\sqrt{D} \;\approx\; \mathbb Q[x]/\langle x^2-D\rangle \otimes_{\mathbb Q} \mathbb Q(\sqrt{D}) \;\approx\; \mathbb Q(\sqrt{D})[x]/\langle x^2-D\rangle $$ $$ \approx\; \mathbb Q(\sqrt{D})[x]/\langle x-\sqrt{D}\rangle \oplus \mathbb Q(\sqrt{D})[x]/\langle x+\sqrt{D}\rangle \;\approx\; \mathbb Q(\sqrt{D}) \oplus \mathbb Q(\sqrt{D}) $$ The more general version uses the obvious analogous argument.

Answer 2

If $V$ and $W$ are two finite dimensional $k$-vector spaces with bases $B_V = \{v_1, \dots, v_n\}$ and $B_W = \{w_1, \dots, w_k\}$ then a basis of $V\otimes_k W$ is

$$ B_{V\otimes_k W} = \{v_i\otimes w_j : 1\le i\le n, 1\le j\le k\}$$

So since quadratic number fields are $2$ dimensional $\mathbb{Q}$-vector spaces and $F = \mathbb{Q}(\sqrt{D})$ has basis $\{1, \sqrt{D}\}$ then you immediately get that $\{1\otimes 1, 1\otimes\sqrt{D}, \sqrt{D}\otimes 1, \sqrt{D}\otimes \sqrt{D}\}$ is a basis of $F\otimes_{\mathbb{Q}} F$ as a $4$-dimensional $\mathbb{Q}$-vector space. No need to solve any annoying $4\times 4$ matrices.

As for further structure I'm not $100\%$ sure, but I hope this appropriately answers at least the first part of your question so as to be posted here rather than as a comment.