## Exhibit a quadratic field as a field of matrices

Let $K = \mathbb{Q}(\sqrt{D})$, where $D$ is a squarefree integer. Let $\alpha = a+b\sqrt{D}$ be in $K$, and consider the basis $B = \{1,\sqrt{D}\}$ of $K$ over $\mathbb{Q}$. Compute the matrix of the $\mathbb{Q}$-linear transformation ‘multiplication by $\alpha$‘ (described previously) with respect to $B$. Give an explicit embedding of $\mathbb{Q}(\sqrt{D})$ in the ring $\mathsf{Mat}_2(\mathbb{Q})$.

We have $\varphi_\alpha(1) = a+b\sqrt{D}$ and $\varphi(\alpha)(\sqrt{D}) = bD + a\sqrt{D}$. Making these the columns of a matrix $M_\alpha$, we have $M_\alpha = \begin{bmatrix} a & bD \\ b & a \end{bmatrix}$, and this is the matrix of $\varphi_\alpha$ with respect to $B$. As we showed in the exercise linked above, $\alpha \mapsto M_\alpha$ is an embedding of $K$ in $\mathsf{Mat}_2(\mathbb{Q})$.

Compare to this previous exercise about $\mathbb{Z}[\sqrt{D}]$.

Advertisements
Post a comment or leave a trackback: Trackback URL.