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Let $L/ K$ be a finite extension of number field. Let $\mathcal{P}$ be a prime ideal in $\mathbb{Z}_K= \{ \alpha \in K : \alpha \mbox{ is a algebraic integer}\}$. Prove that $\mathcal{P}\mathbb{Z}_L \neq \mathbb{Z}_L$

Any Hint?

Thanks!

3 Answers 3

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This is nonsense in this form. $\mathcal P$ is an ideal of $\Bbb Z_K$. Then $\mathcal P\Bbb Z_K = \mathcal P$, using $1\in\Bbb Z_K$. And, the whole ring is of course a prime ideal, so $\Bbb Z_K\Bbb Z_K=\Bbb Z_K$. So, $\mathcal P=\Bbb Z_K$ is a counterexample.

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    Berci's objection still applies, since if $L=K$ then $L$ is a finite extension of $K$. However, I think the usual convention is not to count the entire ring as a prime ideal, since that would destroy the theorem that says ideals have unique factorizations (up to order) as products of prime ideals.2012-10-29
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Let $I$ be a proper ideal of $\mathbf{Z}_K$. Suppose $I\mathbf{Z}_L=\mathbf{Z}_L$. Since $\mathbf{Z}_L$ is a finite $\mathbf{Z}_K$-module, Nakayama's lemma implies the existence of $a\in\mathbf{Z}_K$, $a\equiv 1\pmod{I}$, with $a\mathbf{Z}_L=0$. Since $1\in\mathbf{Z}_L$, this forces $a=0$, whence $1\in I$, a contradiction.

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    Thanks Keenan,$I$see that book.2012-10-29
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The way I understand the definitions, $\cal P$ isn't the whole ring, so it can't contain any units, so ${\cal P}{\cal O}_L$ can't contain any units, so it can't be all of ${\cal O}_L$

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    @Keenan, you have$a$point. I was thinking of elements of the form $ab$ with $a$ in $\cal P$ and $b$ is ${\cal O}_L$, but of course I need to consider finite sums of such elements.2012-10-29