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I'm trying to show that a subring of $M_2(\mathbb{R})$ is isomorphic to $\mathbb{C}$. I know it works because I already have matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$. I thought I could prove that the isomorphism works before and after addition and multiplication. Is that sufficient? Here's my math.

$\phi(M_1+M_2) $$= \phi(\begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix}+ \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix})$ $$=\phi(\begin{pmatrix} a_1+a_2 & b_1+b_2 \\ -b_1-b_2 & a_1+a_2 \end{pmatrix})$$ $$=a_1+a_2+(b_1+b_2)i$$

$\phi(M_1)+\phi(M_2)$ $$=\phi(\begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix})+\phi(\begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix})$$ $=a_1+b_1i+a_2+b_2i=a_1+a_2+(b_1+b_2)i$

$\phi(M_1M_2)$ $=\phi(\begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix}\begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix})$ $=a_1a_2-b_1b_2+(a_1b_2+a_2b_1)i$

$\phi(M_1)\phi(M_2)=\phi(\begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{pmatrix})\phi(\begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{pmatrix})$ $=(a_1+b_1i)(a_2+b_2i)$ $=a_1a_2+a_1b_2i+a_2b_1i+b_1b_2i^2$ $=a_1a_2-b_1b_2+(a_1b_2+a_2b_1)i$

I'm not sure why my lines aren't working. Those are supposed to be 2x2 matrices. Is this sufficient, or must I mention the kernel and whatnot?

Afntu
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Antonio Baez
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  • "The order is irrelevant" proves the homomorphism part, and you have yet to prove the isomorphism part. – Trebor May 04 '24 at 04:04
  • @Trebor and showing it from the other side (complex numbers to matrices) doesn't work either? Like both together, order with regard to addition and multiplication and order with regard to the rings. I don't understand why that wouldn't be enough – Antonio Baez May 04 '24 at 04:10
  • Well you didn't show it from the other side: I can only see one function $\phi$. And it is not enough. You need to show that the two functions compose to the identity. That's simply the definition of isomorphisms. – Trebor May 04 '24 at 04:27
  • Yes, I have not shown it from the other side yet. But If I did, is that enough? Or do I have to involve the kernel in some way. – Antonio Baez May 04 '24 at 04:29
  • As I said, no it is not. – Trebor May 04 '24 at 04:54
  • Well telling me I can't do something doesn't really help. Maybe suggest something that could actually help me? – Antonio Baez May 04 '24 at 05:14
  • I suggested it. "You need to show that the two functions compose to the identity." – Trebor May 04 '24 at 05:17
  • Alright, alright. So, how would I go about doing that? Isn't that just showing a second mapping the other way and showing they don't affect each other? – Antonio Baez May 04 '24 at 05:22

1 Answers1

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Showing $\phi$ respects addition and multiplication only shows that it is a homomorphism. To show it is an isomorphism, you need to show that it is a bijection. One way to do that is to define a $\psi$ the other way ($\Bbb C\to A\subseteq \mathsf{GL}_2(\Bbb R)$, where $A$ is the set of the matrices of those kind), and show that $\phi(\psi(a+bi))=a+bi$ and $\psi(\phi(M))=M$.

Hope this helps. :)

ultralegend5385
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