# showing $\psi: R\to \mathbb C$ is ring isomorphism.

Below is an example from I.N. Herstein:

Let $R=\Bigg\{\left( \begin{array}{ccc} a & b \\ -b & a \end{array} \right)\Bigg|a,b\in \mathbb R\Bigg\}$ and let $\mathbb C$ be the field of complex numbers. Define $\psi :R\to \mathbb C$ by $\psi\left( \begin{array}{ccc} a & b \\ -b & a \end{array} \right)=a+bi.$

It is asked to prove that $\psi$ is a ring isomorphism of $R$ onto $\mathbb C$ .

can anyone help me with some hint how to prove the ring isomorphism.

• First show it's a homomorphism, then show it's bijective. Where are you having trouble? – user98602 Oct 26 '14 at 7:53
• By restricting $R$ via the projection $r \rightarrow (1\quad 0) r$, $\psi$ simply becomes the natural isomorphism from $\mathbb R^2$ to $\mathbb C$. – user139000 Oct 26 '14 at 7:57
• @kittuu Please note the discussion under Sami Ben Romdhane's answer below. You should probably indicate what kind of isomorphism you want: one of rings, of vector spaces over $\mathbb{R}$, or of $\mathbb{R}$-algebras. – Dan Shved Oct 26 '14 at 8:07

Prove that

• $\psi$ is a morphism of rings i.e.

$$\psi( M_z+M_{z'})= \psi(M_z)+\psi(M_{z'})$$ and $$\psi(M_zM_{z'})=\psi(M_z)\psi(M_{z'})$$ and $$\psi(M_1)=1$$ where $$z=a+ib\quad;\quad M_z=\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$

• $\psi$ is injective i.e. $\ker \psi=\{M_0\}$
• $\psi$ is surjective i.e. $\operatorname{Im}\psi=\Bbb C$.
• What the introduction of a scalar and mention of a linear transformation? Not really necessary to prove $\psi$ is a ring isomorphism. – manthanomen Oct 26 '14 at 8:01
• One should also check that multiplication is preserved ! – Dan Shved Oct 26 '14 at 8:04
• Yes you're right I didn't see the ring-theory tag and my answer is for the isomorphism of vector spaces. – user63181 Oct 26 '14 at 8:05
• @manthanomen True, but the OP didn't clearly state that he wants a ring isomorphism. Maybe he wants an $\mathbb{R}$-algebra isomorphism, in which case one indeed needs to check that $\psi$ preserves multiplication by elements of $\mathbb{R}$. – Dan Shved Oct 26 '14 at 8:05
• @DanShved My question deals with ring-isomorphism.. – kittuu Oct 26 '14 at 8:06