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Here is the function I am speaking about:

Recall that the field $\mathbb{Q}(\sqrt{3})$ has basis $\{1 ,\sqrt 3\}$ as a vector space over $\mathbb{Q}.$ Let $\mathbb{Z}[\sqrt{3}] \subset \mathbb{Q}(\sqrt{3})$ be the subring $\{ a_{1} + a_{2}\sqrt{3}| a_{1},a_{2} \in \mathbb{Z}\}.$

Show that the function $$\phi:\mathbb{Q}(\sqrt{3}) \rightarrow M_{2}(\mathbb{Q}), \phi (a_{1} + a_{2}\sqrt{3}) = \begin{pmatrix} a_{1} & 3a_{2} \\ a_{2} & a_{1} \end{pmatrix} $$ is an injective ring homomorphism.

My question is:

I have no problem in showing the following:

1- Showing that it is injective.

2- Showing that it preserves addition.

3- Showing that it preserves multiplication.

But I have a problem in

4- Showing that it maps the identity element of $\mathbb{Q}(\sqrt{3})$ to the identity element of $M_{2}(\mathbb{Q}).$

I know the identity element of $M_{2}(\mathbb{Q})$ which is $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ but it is not clear for me what is the identity element of $\mathbb{Q}(\sqrt{3}),$ could anyone clarify this for me please?

Servaes
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    The identity is obviously $1$. This is represented by $1 \cdot 1 + 0 \cdot \sqrt{3}$ (i.e. $a_1=1, a_2=0$). – Crostul Nov 05 '20 at 08:28
  • It is realty not that difficult to tell which element of $\Bbb Q(\sqrt3)$ is mapped to that matrix. You already have the formula for $\phi$. – Arthur Nov 05 '20 at 08:29
  • @Crostul is not it the additive identity not the multiplicative identity ? in proving the ring homomorphism? So the additive identity in this ring is 1? –  Nov 05 '20 at 08:33
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    The additive identity is obviously $0$... – Derek Luna Nov 05 '20 at 08:39
  • See also https://math.stackexchange.com/a/3844191/589 – lhf Nov 05 '20 at 11:45

1 Answers1

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The ring $\Bbb{Q}(\sqrt{3})$ contains $\Bbb{Q}$ as a subring. Hence the identity (additive or multiplicative) of $\Bbb{Q}(\sqrt{3})$ is that of $\Bbb{Q}$.

Servaes
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