If $$R = \mathbb Z[\sqrt2] = \{a+b\sqrt2\mid a,b \in \mathbb Z\}\\I = \{a+b\sqrt2\mid a,b \in 3\mathbb Z\}$$
Is ${R /I} \cong {{\mathbb Z} /{9\mathbb Z}}$?
2026-04-05 14:43:38.1775400218
Is ${R /I} \cong {{\mathbb Z} /{9\mathbb Z}}$?
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No. Note that $1 + 1 + 1 = 0$ in $R/I$ but not in $\mathbb{Z}/9\mathbb{Z}$. (In fact, $R/I$ is the field with 9 elements).