Fall 2016 - Problem 2

Solution.

By Artin-Wedderburn, $K$ is a product of at most two matrix algebras: $K \simeq M_{n_1}\p{D_1} \times M_{n_2}\p{D_2}$. Since $K$ is a two-dimensional $\Q$-algebra, we have $n_1^2 \dim{D_1} + n_2^2 \dim{D_2} = 2$.

Without loss of generality, suppose $n_1 = 1$ so either $\dim{D_1} = 2$ and $n_2 = 0$, or $\dim{D_1} = \dim{D_2} = n_2 = 1$. In the first case, $K$ is a field since $K$ is a two-dimensional algebra and $\Q$ is commutative, so $K$ is a two-dimensional field extension, i.e., $K \simeq \Q\p{\sqrt{d}}$ for some non-square $d \in \Q$. In the second case, we simply have $K \simeq \Q \times \Q$.

Case 1: $K \simeq \Q\p{\sqrt{d}}$

With $\set{1, \sqrt{d}}$ as our $\Q$-basis, we see that $a + b\sqrt{d}$ acts via

Note that $\rho\p{K^\times}$ is self-centralizing. If $A = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$ commutes with $\rho\p{\sqrt{d}}$, then

which gives $p = s$ and $q = dr$, so $A \in \rho\p{K^\times}$. Notice also that $\rho\p{K^\times}$ is not self-normalizing. For example,

In particular, the centralizer $C$ is a proper (normal) subgroup of the normalizer $N$. Finally, an equivalence class $AC \in N/C$ acts on $\rho\p{K^\times}$ via conjugation: let $AB$ be a representative of $AC$ so that

where the first equality comes from centrality and the second comes from normality. Note that if $b = 0$, then $\rho\p{a} = aI_2$, so this action fixes $\Q$. Thus, we may identify $N/C \leq \Gal\p{K/\Q}$, but $\abs{\Gal\p{K/\Q}} = 2$, and we have shown that $\abs{N/C} > 1$, so $\abs{N/C} = \br{N : \rho\p{K^\times}} = 2$.

Case 2: $K \simeq \Q \times \Q$

Using the canonical $\Q$-basis, pointwise multiplication gives the representation

The normalizer of invertible $2 \times 2$ matrices is the subgroup $S_2 \rtimes \diag_2$, where $S_2$ acts on invertible diagonal matrices by permuting the columns. Hence, because $\rho\p{K^\times} \simeq \diag_2$, it follows that the index of $\rho\p{K^\times}$ in its normalizer is $\abs{S_2} = 2$.