Recall that the field $\mathbb{Q}[\sqrt{k}\,]$, as a set, is given by $\{a+b\sqrt{k}:a,b \in \mathbb{Q}\}$.

For each $k$ such that $\sqrt{k} \notin \mathbb{Q}$, there is an isomorphic matrix field with isomorphism $$ \varphi : a+b\sqrt{k} \ \longmapsto \left(\begin{array}{cc} a & b \\ kb & a \end{array}\right) $$

When $k=-1$ we have $\mathbb{Q}[\operatorname{i}]$ and we have the beautiful property that complex conjugation corresponds to matrix transposition, i.e. $\varphi(\overline{z}) = \varphi(z)^{\top}$ for all $z \in \mathbb{Q}[\operatorname{i}]$.

For any field extension, there is the idea of a conjugate. In the case of $\mathbb{Q}[\sqrt{k}\,]$ we say that the conjugate of $a+b\sqrt{k}$ is $a-b\sqrt{k}$. If $p \in \mathbb{Q}[\sqrt{k}\,]$ and $p^*$ is its conjugate then $$\varphi(p^*) = \varphi(p)^{\top} \ \text{for all} \ \ p \in \mathbb{Q}[\sqrt{k}\,] \iff k=-1$$

Why is $k=-1$ so special? There must be some reason deep-down for conjugation and transposition corresponding if and only if $k=-1$ and we're in $\mathbb{Q}[\operatorname{i}]$.


1 Answer 1


I'm not sure that the following is the kind of explanation you are looking for. But I want to discuss a few related things, so here comes.

1) The set of matrices $$ F_k=\left\{\left(\begin{array}{cc}a&b\\kb& a\end{array}\right)\mid a,b\in\Bbb{Q}\right\}\subset M_2(\Bbb{Q}) $$ is stable under taking the transpose if and only if $k^2=1$. In other words if $A\in F_k$, then also $A^T\in F_k$. So taking the transpose is a mapping from the subring $F_k$ to itself only for those two values of $k$.

2) In the realm of matrix representations of skewfields as well as fields taking the transpose is useful, but one has to keep in mind that the transpose is inherently an $anti$automorphism: $(AB)^T=B^TA^T$ for all square matrices $A$ and $B$. The possibly best known instance of this is the representation of Hamiltonian quaternions by 4x4 real matrices: $$ q=a+bi+cj+dk\mapsto M(q)=\left( \begin{array}{rrrr} a&-b&-c&-d\\ b&a&-d&c\\ c&d&a&-b\\ d&-c&b&a \end{array}\right). $$ Note that conjugation $q\mapsto \overline{q}=a-bi-cj-dk$ of quaternions corresponds to taking the transpose, $M(\overline{q})=M(q)^T$. Therefore conjugation is an antiautomorphism: $\overline{q_1\cdot q_2}=\overline{q_2}\cdot\overline{q_1}.$ In your case the subrings $F_k$ are all commutative, so in the stable cases $k=\pm1$ the order of multiplication is irrelevant, and you do get an automorphism of the ring (or the field $\Bbb{Q}(i)$ in the case $k=-1$).

3) A general related fact is the following consequence of Skolem-Noether Theorem. The matrix rings $M_n(\Bbb{Q})$ are all central simple algebras over the rationals. If $F$ is an $n$-dimensional extension field of $\Bbb{Q}$, and $f,g:F\to M_n(\Bbb{Q})$ are any two homomorphism of $\Bbb{Q}$-algebras, then Skolem-Noether says that there exists an invertible matrix $B\in M_n(\Bbb{Q})$ such that $$ g(x)=B f(x) B^{-1} $$ for all $x\in F$. In particular, if $F$ is a subring of $M_n(\Bbb{Q})$ that also happens to be a field, then we can let $f$ be the inclusion mapping (=identity), and let $g=\sigma$ be any automorphism of $F$. Skolem-Noether then tells us that we can realize the automorphism $\sigma$ by matrix conjugation. IOW there exists a matrix $B_\sigma$ such that $$ \sigma(x)=B_\sigma x B_\sigma^{-1} $$ for all $x\in F$.

The reason why I bring up Skolem-Noether is that it applies to all the fields $F_k, k\notin \Bbb{Q}^2$ in your question - not just to $F_{-1}$. The conjugation $\sigma:a+b\sqrt k\mapsto a-b\sqrt k$ can be realized as conjugation by $$ B_\sigma=\left(\begin{array}{rr}1&0\\0&-1\end{array}\right), $$ as (the above $B_\sigma$ is its own inverse) $$ \left(\begin{array}{rr}1&0\\0&-1\end{array}\right)\left(\begin{array}{cr}a&b\\kb& a\end{array}\right)\left(\begin{array}{rr}1&0\\0&-1\end{array}\right)= \left(\begin{array}{cr}a&-b\\-kb& a\end{array}\right). $$

Note: Conjugation, unlike taking the transpose, is inherently order preserving: $$ B(A_1A_2)B^{-1}=(BA_1B^{-1})(BA_2B^{-1}). $$ This is why I feel that conjugation is a more natural place to look for automorphism than transpose. In the case of fields the order of multiplication is immaterial, but in a more general setting you still face the problem of finding subfields of $M_n(\Bbb{Q})$ that are stable under transpose. As we just saw, Skolem-Noether makes that problem go away in the sense that it promises the existence of a suitable matrix to conjugate with.

  • $\begingroup$ This answer looks great. Could you please add some links and definitions. For example, what does it mean for a set to be stable under an operation? $\endgroup$ Nov 7, 2013 at 20:02
  • $\begingroup$ Ahhh, I see. So stability in this sense is just closure. This is a great, great answer Jyrki; the best I've ever had here on this site. I wish I could give you +20. Thank you so very much! $\endgroup$ Nov 7, 2013 at 20:07

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