# Conjugacy classes of maximal tori in PGL2: only 2?

If $K$ is a field, and $F$, $L$ are two distinct quadratic extensions of $K$, then must the subgroups of $\operatorname{PGL}(2, K)$ defined by $F$ and $L$ be conjugate?

To define the subgroup, identify the field $F$ with a two-dimensional $K$-vector space. The subgroup is the subgroup induced by the action of the nonzero elements of $F$ on this vector space. Different identifications yield conjugate subgroups.

Everything makes good sense when the absolute Galois group of the field is locally cyclic, so that extension fields are extremely unique. However, I'm not sure this makes much sense for a field like $\mathbf{Q}$. What goes wrong with (at least one of) the following arguments:

They should not be conjugate: Surely conjugacy would lift to an isomorphism of $K$-algebras between $F$ and $L$? Surely it simply defines an isomorphism of the groups of units of $F$ and $L$? However, I think not all quadratic fields over $\mathbf{Q}$ have the same group of units (the torsion parts differ at least in $\mathbf{Q}[i]$ and $\mathbf{Q}[ω]$).

They should be conjugate: I believe conjugacy classes of maximal tori correspond in some way to elements of the Weyl group. The Weyl group has two elements, and the identity element corresponds to the subgroup defined by $K$ acting on $K ⊕ 0$.

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By "groups of units" do you mean the units in the corresponding ring of integers? Of course those groups are very different in different quadratic extensions of $\mathbb{Q}$, for instance $\mathbb{Q}[i]$ has four units and $\mathbb{Q}[\sqrt{2}]$ has infintely many. – Alon Amit Jun 8 '11 at 18:15
@Alon: sorry, I mean it fairly literally: the nonzero elements of the field. The idea is to view the field as an algebra of 2×2 matrices over the field, but of course for GL we only take the invertible (nonzero) ones. – Jack Schmidt Jun 8 '11 at 18:21
Oh I see, that makes sense. So now, as you point out, the torsion parts may be different - isn't that enough to rule out the option that the groups are conjugate? $\omega$ has order 3 and so does it's action on the 2D vector space over $\mathbb{Q}$ spanned by $\{1, \omega\}$. There's no element of order 3 in $\mathbb{Q}[i]$, so does this not imply that the corresponding subgroups of $\mbox{PGL}(2,\mathbb{Q})$ must be different? – Alon Amit Jun 8 '11 at 20:28
@Alon: yup, but then I've also learned the conjugacy classes of maximal tori are in 1–1 correspondence with conjugacy classes of the Weyl group. The two answers disagree, and that leaves me confused. – Jack Schmidt Jun 8 '11 at 22:57

Let $L,F$ be quadratic extensions of a field $K.$ Upon choosing a bases for $L$ and $F$ over $K$ we obtain monomorphisms $i_L$ and $i_F$ from $L$ and $F,$ respectively, into the ring $\mathfrak{M}(2,K)$ of 2-by-2 matrices over $K.$ The multiplicative groups $L^{\times}$ and $F^{\times}$ are mapped under this homomorphism into $GL(2,K).$ Let $\pi:GL(2,K) \rightarrow PGL(2,K)$ be the projection map. We claim that if $\pi(i_L(L^{\times}))$ and $\pi(i_F(F^{\times}))$ are conjugate in $PGL(2,K)$ by some element $\overline{A},$ then $L\cong F$.
Let $A$ be a lift of $\overline{A}$ to $GL(2,K).$ Conjugation by $A$ is an isomorphism from $i_L(L)$ to $i_L(L)^A$ corresponding to a change in the basis chosen for $L/K.$ Thus, applying $i_L$ and acting by $A$ is a monomorphism from $L$ into $GL(2,K)$. Subsequently, we may replace $i_L$ by our original embedding followed by conjugation, and restrict our attention to the case where $\pi(i_L(L^{\times})) = \pi(i_F(F^{\times})).$
Let $\alpha$ be a primitive element of $L/K$ and $\beta \in F$ such that $\pi(i_L(\alpha)) = \pi(i_F(\beta)).$ Then $i_L(\alpha) = a i_F(\beta)$ for some $a\in K.$ It follows that if $p_{\alpha}(X) = X^2 + b_1X + b_0$ is the characteristic polynomial for $i_L(\alpha),$ then $p_{\beta}(X) = X^2 + a b_1X + a^2 b_0 = a^2p_{\alpha}(\frac{1}{a}X)$ is the characteristic polynomial for $i_F(\beta).$ By the Cayley-Hamilton theorem $p_{\alpha}(i_L(\alpha)) = i_L(p_{\alpha}(\alpha))$ and $p_{\beta}(i_F(\alpha)) = i_F(p_{\alpha}(\beta))$ are zero transformations. But $i_L$ and $i_F$ are embeddings, so $p_{\alpha}(\alpha) = 0$ and $p_{\beta}(\beta) = 0.$ It follows, as the degree of $p_{\alpha}$ is equal to [K(\alpha):K], the polynomial $p_{\alpha} = irr(\alpha, K)$ and $irr(\alpha, K)(\frac{\beta}{a^2}) = 0$. We conclude $L = K(\alpha) = K(\alpha/a^2)$ embeds into $F,$ and as the two fields have the same dimension over $K,$ they are in fact isomorphic.