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Thread: Algebraic number theory.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by peteryellow View Post
    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! 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

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