Algebraic number theory.

• May 9th 2009, 11:04 AM
peteryellow
Algebraic number theory.
Let I be an ideal generated by $\displaystyle 2$ and $\displaystyle 1+\sqrt{-3}$ in the ring $\displaystyle \mathbb Z[{\sqrt{-3}}].$
Show that $\displaystyle I$ is different $\displaystyle (2).$
And $\displaystyle I^2 =2I.$
Show that ideals in $\displaystyle \mathbb Z[{\sqrt{-3}}].$ do not factor uniquely into prime ideals.
Show that I is a unique prime ideal contaning $\displaystyle (2).$
Conclude $\displaystyle (2)$ is not a product of prime ideals.
• May 9th 2009, 02:27 PM
NonCommAlg
Quote:

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

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

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