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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.

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  • $\begingroup$ 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
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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. $\operatorname{Im}\psi=\Bbb C$.
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  • $\begingroup$ 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

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