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.

  • $\begingroup$ First show it's a homomorphism, then show it's bijective. Where are you having trouble? $\endgroup$ – user98602 Oct 26 '14 at 7:53
  • $\begingroup$ By restricting $R$ via the projection $r \rightarrow (1\quad 0) r$, $\psi$ simply becomes the natural isomorphism from $\mathbb R^2$ to $\mathbb C$. $\endgroup$ – user139000 Oct 26 '14 at 7:57
  • $\begingroup$ @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. $\endgroup$ – 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. $\operatorname{Im}\psi=\Bbb C$.
  • $\begingroup$ What the introduction of a scalar and mention of a linear transformation? Not really necessary to prove $\psi$ is a ring isomorphism. $\endgroup$ – manthanomen Oct 26 '14 at 8:01
  • $\begingroup$ One should also check that multiplication is preserved ! $\endgroup$ – Dan Shved Oct 26 '14 at 8:04
  • $\begingroup$ Yes you're right I didn't see the ring-theory tag and my answer is for the isomorphism of vector spaces. $\endgroup$ – user63181 Oct 26 '14 at 8:05
  • $\begingroup$ @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 $\mathbb{R}$. $\endgroup$ – Dan Shved Oct 26 '14 at 8:05
  • $\begingroup$ @DanShved My question deals with ring-isomorphism.. $\endgroup$ – kittuu Oct 26 '14 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.