Algebraic number theory.

Quote:

Originally Posted by peteryellow

Let I be an ideal generated by $2$ and $1+\sqrt{-3}$ in the ring $\mathbb Z[{\sqrt{-3}}].$
Show that $I$ is different $(2).$
And $I^2 =2I.$
Show that ideals in $\mathbb Z[{\sqrt{-3}}].$ do not factor uniquely into prime ideals.
Show that I is a unique prime ideal contaning $(2).$
Conclude $(2)$ is not a product of prime ideals.

i normally don't answer questions with so many parts and no work done by the OP, because that would be a crime! (Nod) to get you started:

most parts of the problem are easily proved. what you need to know here is that $I$ is actually a maximal ideal of your ring. also $2I \subseteq I^2$ is

obvious because $2 \in I.$ to prove that $I^2 \subseteq 2I,$ you only need to show that $ab \in 2I,$ for all $a,b \in I.$ for that you will need this fact that

$(1+\sqrt{-3})^2=-2 + 2 \sqrt{-3}=2(1+\sqrt{-3} - 2) \in 2I.$