Do quaternion algebra isomorphisms conserve norms?

Question

The original exercise I'm trying to solve is, if $H(a,b) \cong H(c,d)$ then $ax^2+by^2-abz^2=0$ is isomorphic to $cx^2+dy^2-cdz^2=0$ (the algebras are over some field $k$ if it matters)

In the following, $1,i,j,ij$ is the canonical $k$-base of $H(a,b)$ (i.e. such that $i^2=a, j^2=b, ij=-ji)$, and $1,i',j',i'j'$ is the basis of $H(c,d)$ over $k$.

To do this, I thought that $ax^2+by^2-abz^2=N(xi+yj+zij)$, so,

suppose $\psi$ is an isomorphism from $H(a,b)$ to $H(c,d)$

then, if norms are conserved, $\psi(N(xi+yj+zij))=N(\psi(xi+yj+zij))=N(Xi'+Yj'+Zi'j')=cX^2+dY^2-cdZ^2$

where in the second equality, I have used that if isomorphisms conserve norms, then $\psi(q\bar q)=\psi(q)\bar \psi(q)=\psi(q)\psi(\bar q)$ which would imply $\psi(\bar q)=\bar \psi(q)$ (assuming $q\neq0$, if it is $0$ then it is obviously true)

Hence my question,

$\bullet$ do such isomorphisms conserve norms? i.e. $\psi(N(q))=N(\psi(q))$?

$\bullet$ Alternatively, do isomorphisms commute with conjugation? i.e. $\psi(\bar q) = \bar \psi(q)$?

$\bullet$ Or, do isomorphisms send pure quaternion on pure quaternion? i.e. $\bar q = -q \Leftrightarrow \bar \psi(q) = -\psi(q)$?

I think these three questions are equivalent to the same property

In all of the previous exercises I saw on quaternion algebras, it seems like every isomorphism I saw had this property, but I have a feeling this is not true in general

Answer 1

Let $z=x+p$ with $p$ the imaginary part of $z$

Let $\psi$ be the isomorphism and $\bar .$ be the conjugation

Then $z=x+p$ and $\bar z=x-p$

then we have $\psi(z)=x+\psi(p)$ and $\psi(\bar z)=x- \psi(p)$

Adding together we get $2x =\psi(z)+\psi(\bar z)=2\Re(\psi(z))$

So we have $\Re(\psi(z)+\psi(\bar z))=2\Re(\psi(z))$ and $\Im(\psi(z)+\psi(\bar z))=0$

Therefore it must be that $\bar\psi(z)=\psi(\bar z)$

So $N(\psi(z))=\psi(z)\bar \psi(z)=\psi(z)\psi(\bar z)=\psi(z \bar z) = \psi(N(z))$ as desired

(and $\psi$ can be just a $k$-linear morphism, not necessarily an isomorphism)

In particular, pure quaternions are always sent on pure quaternions: $\bar \psi(p)=\psi(\bar p)=-\psi(p)$