Prime Ideals

Oct 2009
40
1
If \(\displaystyle K\) is a number field with ring of integers \(\displaystyle \mathcal{O}_K\), then prove that if \(\displaystyle P\) is a prime ideal of \(\displaystyle \mathcal{O}_K\) then \(\displaystyle P \cap \mathbb{Z}\) is a prime ideal of \(\displaystyle \mathbb{Z}\).

It's easy to show that it's an ideal of \(\displaystyle \mathbb{Z}\) but I'm struggling to show that it must be prime; any help would be very useful!
 

NonCommAlg

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If \(\displaystyle K\) is a number field with ring of integers \(\displaystyle \mathcal{O}_K\), then prove that if \(\displaystyle P\) is a prime ideal of \(\displaystyle \mathcal{O}_K\) then \(\displaystyle P \cap \mathbb{Z}\) is a prime ideal of \(\displaystyle \mathbb{Z}\).

It's easy to show that it's an ideal of \(\displaystyle \mathbb{Z}\) but I'm struggling to show that it must be prime; any help would be very useful!
it's even easier to show that it's prime in \(\displaystyle \mathbb{Z}\): if \(\displaystyle a,b \in \mathbb{Z}\) and \(\displaystyle ab \in P \cap \mathbb{Z},\) then since \(\displaystyle ab \in P\) and \(\displaystyle P\) is prime in \(\displaystyle \mathcal{O}_K,\)we'll have \(\displaystyle a \in P\) or \(\displaystyle b \in P.\) so either \(\displaystyle a \in P \cap \mathbb{Z}\) or \(\displaystyle b \in P \cap \mathbb{Z}.\)

note that \(\displaystyle P \cap \mathbb{Z} \neq \mathbb{Z}\) because \(\displaystyle 1 \notin P.\) a less trivial fact is this: if \(\displaystyle P \neq \{0\},\) then \(\displaystyle P \cap \mathbb{Z} \neq \{0\}.\) (Nod)
 
Oct 2009
40
1
No doubt you're right, but I guess my issue is that I don't see how \(\displaystyle a \in P\) or \(\displaystyle b \in P\) implies \(\displaystyle a \in P \cap \mathbb{Z}\) or \(\displaystyle b \in P \cap \mathbb{Z}\). Just because \(\displaystyle ab \in \mathbb{Z}\) this doesn't mean both \(\displaystyle a,b \in \mathbb{Z}\). Am I missing something obvious?
 

chiph588@

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Sep 2008
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Champaign, Illinois
No doubt you're right, but I guess my issue is that I don't see how \(\displaystyle a \in P\) or \(\displaystyle b \in P\) implies \(\displaystyle a \in P \cap \mathbb{Z}\) or \(\displaystyle b \in P \cap \mathbb{Z}\). Just because \(\displaystyle ab \in \mathbb{Z}\) this doesn't mean both \(\displaystyle a,b \in \mathbb{Z}\). Am I missing something obvious?
Since our ideal is over \(\displaystyle \mathbb{Z} \) we assume \(\displaystyle a,b \in \mathbb{Z} \).
 
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NonCommAlg

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No doubt you're right, but I guess my issue is that I don't see how \(\displaystyle a \in P\) or \(\displaystyle b \in P\) implies \(\displaystyle a \in P \cap \mathbb{Z}\) or \(\displaystyle b \in P \cap \mathbb{Z}\). Just because \(\displaystyle ab \in \mathbb{Z}\) this doesn't mean both \(\displaystyle a,b \in \mathbb{Z}\). Am I missing something obvious?
don't forget that \(\displaystyle a\) and \(\displaystyle b\) are in \(\displaystyle \mathbb{Z}\) because you want to prove that \(\displaystyle P \cap \mathbb{Z}\) is a prime ideal of \(\displaystyle \mathbb{Z}.\)
 
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NonCommAlg

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Why is this true?
let \(\displaystyle 0 \neq a \in P\) and suppose that \(\displaystyle f(x)=x^n + c_1x^{n-1} + \cdots + c_n\) is the minimal polynomial of \(\displaystyle f(x)\) over \(\displaystyle \mathbb{Q}.\) it's a known fact that, since \(\displaystyle a \in \mathcal{O}_K,\) we have \(\displaystyle c_j \in \mathbb{Z},\) for all \(\displaystyle j.\)

now if \(\displaystyle n=1,\) then since \(\displaystyle 0=f(a)=a+c_1,\) we have \(\displaystyle a=-c_1 \in \mathbb{Z}\) and so \(\displaystyle 0 \neq a \in P \cap \mathbb{Z}.\) if \(\displaystyle n \geq 2,\) then \(\displaystyle c_n \neq 0\) because of minimality of \(\displaystyle f(x).\) thus \(\displaystyle f(a)=0\) gives us

\(\displaystyle c_n=-(a^n + c_1a^{n-1} + \cdots + c_{n-1}a) \in P\) and so \(\displaystyle 0 \neq c_n \in P \cap \mathbb{Z}.\)


Note: in the proof we didn't need \(\displaystyle P\) to be a prime ideal. so the above result holds for any non-zero ideal \(\displaystyle P\) of \(\displaystyle \mathcal{O}_K.\)
 
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