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Let the Ring, $R = \{ a+b \sqrt2 \vert a \in \mathbb{Z}, b \in \mathbb{Z}\}$(Euclidean domain)

Say $I = \langle 4+\sqrt2 \rangle \lhd R$

Find the number of the idempotent(I.e. $a^2 = a$ ) in $R/I$.

All the $\vert R/I \vert$ and $char(R)$ would be $14$. But I can't find the next step to find the isomorphic ring with the $R/I$. Any help would be appreciated.

p.s) My guess is $R/I \simeq \mathbb{Z}_{14}$. So the number of the idempotenet is $4$ (By $C.R.T$)

se-hyuck yang
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2 Answers2

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Note that, in $R/I$, $\overline{4+\sqrt{2}}=\overline{0}$. That is, $-\overline{4}=\overline{\sqrt{2}}$ iff $\overline{16}=\overline{2}$ iff $\overline{15}=\overline{1}$ and $\overline{14}=\overline{0}$ .

Thus if $\pi:R\to R/I$ is the natural map, $\pi(a+b\sqrt{2})=\pi(a)+\pi(b)\pi(\sqrt{2}))=\overline{a}+\overline{b}\cdot\overline{\sqrt{2}}=\overline{a}+\overline{b}\cdot\overline{(-4)}=\overline{a}-\overline{4b}$. This relation suggest that $f:R\to\mathbb{Z}_{14}$ given by $$f(a+b\sqrt{2})=\overline{a}-\overline{4b}$$ is a surjective homomorphism. (It's not so difficult to prove it. Convince yourself!) And its kernel is $I$. Therefore $R/I\simeq\mathbb{Z}_{14}$.

Jaca
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For $\left(4+\sqrt2\right)\left(a+b\sqrt2\right)$ to be an integer, we must have $a+4b=0$. Thus this integer is $4a+2b=-14b$, a multiple of $14$. Thus the integers from $0$ to $13$ are in different equivalence classes and can be taken as a set of representatives of $R/I$. Since $14\in I$, addition and multiplication on these representatives works as in $\mathbb Z_{14}$. It follows that $R/I\simeq\mathbb Z_{14}$.

joriki
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