Finding irreducible components of $\mathrm{Spec}(R/ \mathbb I)$

Question

I'm trying to solve the following exercise:

Let $R=\mathbb Z[\sqrt{-5}]$ and $\mathbb I=(1+\sqrt{-5})$. Find the irreducible components of $\mathrm{Spec}(R/\mathbb I)$.

For solving this problem I'm trying to use following exercise from Atiyah and Macdonald book:

Let $A$ be a commutative ring with unit, $X = \mathrm{Spec}(A) $ with the Zariski topology. Then the irreducible components of $X$ are $\lbrace V(p) : p\subset A \ \text{minimal prime ideal} \rbrace$ where $V(p) =\lbrace q \ \text{prime ideal } \mid p\subset q\rbrace$.

My Solution: As $R \cong \mathbb Z[x]/(x^2+5)$, hence $R/ \mathbb I \cong \mathbb Z/ 6 \mathbb Z$. Hence by above exercise for finding irreducible components of $\mathrm{Spec}(R/\mathbb I)$ we just need to find minimal prime ideals of $\mathbb Z/ 6 \mathbb Z$. Hence irreducible components of $\mathrm{Spec}(R/\mathbb I)$ are $V(2)$ and $V(3)$ where $(2)$ and $(3)$ denotes the prime ideals generated by $2$ and $3$ respectively. Is this solution correct?

Answer 1

This is correct. This is because under the isomorphism $\mathbb{Z} /6\mathbb{Z} \rightarrow \mathbb{Z}[\sqrt{-5}] / (1 + \sqrt{-5})$, 2 is sent to 2 and 3 is sent to 3.