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$R=\left\lbrace\left.\begin{bmatrix}a & -b\\b&a\end{bmatrix}\,\right| a, b\in \mathbb{R}\right\rbrace$

$(R,+,\cdot)$ is ring with binary operation $+$ and $\cdot$ ,

Give me a hint to show that $R$ is isomorphic to $\mathbb{C}$ where $(\mathbb{C},+,\cdot)$ is the ring of complex numbers with binary operation $+$ and $\cdot$.

nonuser
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Beginner
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  • I know $a+bi=\begin{bmatrix}a & -b\b&a\end{bmatrix}$ – Beginner Feb 10 '18 at 12:42
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    The thing you state above is not true. A complex number is not "equal to" a matrix. There is an isomorphism that connects the two, but I feel like that is the thing that you are asking about? I'm actually not entirely clear on what you are asking, particularly in light of your comment? Are you looking for any old homomorphism? Because you could just send everything to zero... – Xander Henderson Feb 10 '18 at 12:48
  • Please see my edit and confirm that an isomorphism is what you want. – Arnaud Mortier Feb 10 '18 at 12:49
  • You clearly already know what the isomorphism is supposed to be (although you can't write it exactly like you did in your comment; either use $\mapsto$, or introduce a name for the function and use function notation). It is trivial to show that it is bijective and straight-forward to show that it is a homomorphism. Where are you stuck, exactly? – Arthur Feb 10 '18 at 12:50
  • yes. i think isomorphic not homomorphisms. – Beginner Feb 10 '18 at 12:52
  • i try $f:R\to C$ isomorphic if and only if $f(x+y)=f(x)+f(y)$ and $f(x\cdot y)=f(x)\cdot f(y)$ – Beginner Feb 10 '18 at 12:54

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You are almost there. It is not $$\begin{bmatrix}a & -b\\b&a\end{bmatrix}=a+bi$$ but

$$\begin{bmatrix}a & -b\\b&a\end{bmatrix}\mapsto a+bi$$ is a homomorphism and it is bijective.

nonuser
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