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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!

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3 Answers

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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Sorry, was $\mathcal{P}\mathbb{Z}_L \neq \mathbb{Z}_L$ –  P. M. O. Oct 29 '12 at 2:00
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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. –  Gerry Myerson Oct 29 '12 at 2:07
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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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Evidently, the assumption that $\cal P$ be a prime ideal is unnecessary --- all that's needed is that it be a proper ideal. –  Gerry Myerson Oct 29 '12 at 2:43
    
I can use the lemma 4.5 part iii) Page 389 Algebra-Hungerford to reach the contradiction? –  P. M. O. Oct 29 '12 at 2:53
    
Dear @P.M.O., No, that particular version of Nakayama's lemma doesn't apply because the ideal $I$ need not be contained in the Jacobson radical of $\mathbf{Z}_K$ (in fact the Jacobson radical of $\mathbf{Z}_K$ is $0$). I'm using a more general version of Nakayama's lemma (which implies Hungerford's). It can be found in Theorem 2.2 on page 8 of Matsumura's Commutative Ring Theory. –  Keenan Kidwell Oct 29 '12 at 10:48
    
Thanks Keenan, I see that book. –  P. M. O. Oct 29 '12 at 20:02
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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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Dear @Gerry, I don't think it's obvious that $\mathcal{P}\mathscr{O}_L$ doesn't contain any units just because $\mathcal{P}$ doesn't. If $A$ is any domain which is not a field, $a\in A\setminus A^\times$, $a\neq 0$, then $(a)\neq A$ but $aK=K$, where $K$ is the field of fractions of $A$. I realize this is not the same setup as in the OP's question, but it shows that some fact about the extension of rings $\mathcal{O}_K\subseteq\mathcal{O}_L$ has to be used (like finiteness). –  Keenan Kidwell Oct 29 '12 at 2:28
    
@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. –  Gerry Myerson Oct 29 '12 at 2:40
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