Prime ideal $42 = (1+\sqrt{-41})(1-\sqrt{-41})$

Question

I found out that $\mathcal O_{-41} = 2 \cdot 3 \cdot 7 = (1+\sqrt{-41})(1-\sqrt{-41})$.

How can I show now that $(42) \subset (2, 1+\sqrt{-41})$ and that this is a prime ideal?

Answer 1

Observe that, $$2-(1+\sqrt{-41}) \in (2,1+\sqrt{-41}) \implies 1-\sqrt{-41} \in (2,1+\sqrt{-41}) \implies 42= (1+\sqrt{-41})(1-\sqrt{-41})\in (2,1+\sqrt{-41}) \implies (42) \subset (2,1+\sqrt{-41})$$.

Ideal $(42)$ is not prime because $42$ is not irreducible.