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While doing some exercises about rings and prime ideals i got stuck with the following:

Having a ring R: {$a + b \sqrt7 | a,b \in \mathbb{Z}$}, being a subring of $\mathbb{R}$, and knowing that $2-\sqrt7$ divides 3 in R, is the following ideal maximal:

I = $\langle 3 \rangle$ generated by 3 in R

I dont really know how to show this, but was trying to show that it is not a prime ideal($\Rightarrow$ not a maximal ideal)...

Any help would be appriciated

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Your approach is correct The ring $ R$ is a commutative ring with identity.So an ideal $P$ in $R$ is prime iff $ab\in P$ implies either $a\in P$ or $b\in P$.Observe that $(2-\sqrt{7})(2+\sqrt{7})=4-7=-3\in \langle3\rangle$ but neither $(2-\sqrt{7})$ nor $(2+\sqrt{7}) \in P$.hence not prime and hence not maximal

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  • $\begingroup$ Perfect, just what I needed! Was trying to get somewhere with the definition of prime ideals, but just got stuck all the time.. Thanks a lot $\endgroup$ – Nohr Nov 17 '14 at 15:57
  • $\begingroup$ glad to help you $\endgroup$ – Learnmore Nov 17 '14 at 15:58

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