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
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