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Prove that $\phi:R \longrightarrow R^{\prime}$ where $R=\lbrace a+b \sqrt{2}:a,b\in \mathbb{Z}\rbrace$ and $R^{\prime}= \lbrace \begin{pmatrix}{} a & 2b \\ b & a \end{pmatrix}:a,b\in \mathbb{Z} \rbrace$ given by $\phi(a+b \sqrt{2})$=$ \begin{pmatrix}{} a & 2b \\ b & a \end{pmatrix}$ is a ring isomorphism between $R$ and $R^{\prime}$. I tried to do it, and I concluded that $\phi$ is a group homomorphism with the $+$ operation, but it is not a group homomorphism with the product of two elements.

Note that the sum and product are the usual in $M_{2 \times 2}(\mathbb{Z})$, and the sum and the product in $R$ are the same as that in $\mathbb{Z}$. Can someone tell me if I had an error while doing the comparison between $\phi((a+b\sqrt{2})(c+d\sqrt{2}))$ and $\phi(a+b\sqrt{2}) \phi(c+d\sqrt{2})$?

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$$(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(bc+ad)\sqrt2$$

$$\pmatrix{a&2b\\b&a}\pmatrix{c&2d\\d&c}=\pmatrix{ac+2bd&2bc+2ad\\bc+ad&ac+2bd}$$

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    $\begingroup$ Thanks so much, I understand that my mistake be in multiply $(a+b\sqrt{2})(c+d\sqrt{2})$ $\endgroup$
    – user795628
    Nov 6, 2020 at 12:04

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