# Ring homomorphism in the quaternions

I been trying to find a ring homomorphism but without luck, for example in the complex numbers we have the conjugation but this not work in the quaternions, since $f(xy) = f(y)f(x)$ supposing $f$, is the conjugation, and we don't have commutativity in the quaternions.

So, is there some ring homomorphism distinct from the identity or the zero ones?

• A homomorphism from where to where? For one from $\Bbb H$ to itself, how about $a+bi+cj+dk\mapsto a+bj+ck+di$? – Lord Shark the Unknown Sep 21 '17 at 1:59

If you mean a ring homomorphism from $\mathbb H$ to itself, you can take any noncentral element $x$ and use the ring isomorphism $q\mapsto xqx^{-1}$.

Actually, it can be proven that these (with the identity automorphism) give you all possible automorphisms.

To show that, we can refer to the Skolem-Noether theorem on central simple algebras. Since $\mathbb H$ is a central simple $\mathbb R$ algebra, we know already that all $\mathbb R$-linear automorphisms are inner.

But if $\phi$ is any automorphism $\mathbb H\to \mathbb H$, then it's easy to see that $r$ is central in $\mathbb H$ iff $\phi(r)$ is central in $\mathbb H$, so that $\phi$ restricted to $\mathbb R$ is an automorphism of $\mathbb R$. But the only automorphism of $\mathbb R$ is the identity, so $\phi$ is already $\mathbb R$-linear. Thus all automorphisms of $\mathbb H$ have been found.

Distinct $x$'s are not guaranteed to yield distinct maps. Since $G=\mathbb H\setminus\{0\}$ is a group, we can refer to the basic group theory result about the conjugation action of $G$ on itself. It's easy to prove that the kernel of the action is the center of $G$ (which corresponds to $\mathbb R\setminus\{0\}$ in our $G$) so that $x$ and $y$ produce the same inner automorphism on $\mathbb H$ iff $x=ry$ for some $r\in\mathbb R$.

• Doesn’t that get them all? – Lubin Sep 21 '17 at 2:48
• That work but why takes them all? – Hector Gurle Sep 21 '17 at 3:19
• @EdgarM Lubin is right: the Skolem-Noether theorem says all automorphisms of a central simple algebra are inner, so this (along with the identity) catches all automorphisms of $\mathbb H$ – rschwieb Sep 21 '17 at 10:39
• @Kimball But if $\phi$ is any automorphism of $\mathbb H$, it's easy to see that $r$ is central iff $\phi(r)$ is central, and so $\phi$ restricted to $\mathbb R$ is an automorphism of $\mathbb R$. But the identity is the only automorphism of $\mathbb R$, so in fact $\phi$ is $\mathbb R$ linear already. Only possible because of this nice property of $\mathbb R$, of course. – rschwieb Sep 22 '17 at 0:24
• While we’re on the topic, perhaps it should be pointed that $\Bbb Q_p$ has this property too: the only automorphism is the identity. – Lubin Sep 22 '17 at 13:00