Here is the question I want to answer:
Let $R$ be the quadratic integer ring $\mathbb Z[\sqrt{-5}]$ and define the ideals $I_2 = (2, 1 + \sqrt{-5}), I_3 = (3, 2 + \sqrt{-5}),$ and $I_3^{'} = (3, 2 - \sqrt{-5}).$
$(a)$ Prove that $2, 3, 1 + \sqrt{-5}$ and $1 - \sqrt{-5}$ are irreducibles in $R,$ no two of which are associate in $R,$ and that $6 = 2 . 3 = (1 + \sqrt{-5}).(1 - \sqrt{-5})$ are two distinct factorizations of $6$ into irreducibles in $R.$
$(b)$ Prove that $I_2, I_3,$ and $I_3^{'}$ are prime ideals in $R.$ [One approach: for $I_3,$ observe that $R/I_3 \cong (R/(3))/(I_3/(3))$ by the Third Isomorphism Theorem for Rings. Show that $R/(3)$ has $9$ elements, $I_3/(3)$ has $3$ elements, and that $R/I_3 \cong \mathbb Z/ 3 \mathbb Z$ as an additive abelian group. Conclude that $I_3$ is a maximal (hence prime) ideal and that $R/I_3 \cong \mathbb Z/ 3 \mathbb Z$ as rings.]
$(c)$ Show that the factorization in $(a)$ imply the equality of the ideal $(6) = (2)(3)$ and $(6) = (1 + \sqrt{-5})(1 - \sqrt{-5})$. Show that these two ideal factorizations give the same factorization of the ideal $(6)$ as the product of prime ideals (cf. Exercise $5$ in section $2$)
My question is:
What is the difference between answering the last part in $(a)$ and $(c)$? could anyone explain this to me please?
Prove that $6 = 2 . 3 = (1 + \sqrt{-5}).(1 - \sqrt{-5})$ are two distinct factorizations of $6$ into irreducibles in $R.$
$(c)$ Show that the factorization in $(a)$ imply the equality of the ideal $(6) = (2)(3)$ and $(6) = (1 + \sqrt{-5})(1 - \sqrt{-5})$. Show that these two ideal factorizations give the same factorization of the ideal $(6)$ as the product of prime ideals (cf. Exercise $5$ in section $2$).