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Math Help - Prime Ideals

  1. #1
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    Prime Ideals

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

    It's easy to show that it's an ideal of \mathbb{Z} but I'm struggling to show that it must be prime; any help would be very useful!
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    Quote Originally Posted by Boysilver View Post
    If K is a number field with ring of integers \mathcal{O}_K, then prove that if P is a prime ideal of \mathcal{O}_K then P \cap \mathbb{Z} is a prime ideal of \mathbb{Z}.

    It's easy to show that it's an ideal of \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 \mathbb{Z}: if a,b \in \mathbb{Z} and ab \in P \cap \mathbb{Z}, then since ab \in P and P is prime in \mathcal{O}_K,we'll have a \in P or b \in P. so either a \in P \cap \mathbb{Z} or b \in P \cap \mathbb{Z}.

    note that P \cap \mathbb{Z} \neq \mathbb{Z} because 1 \notin P. a less trivial fact is this: if P \neq \{0\}, then P \cap \mathbb{Z} \neq \{0\}.
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    No doubt you're right, but I guess my issue is that I don't see how a \in P or b \in P implies a \in P \cap \mathbb{Z} or b \in P \cap \mathbb{Z}. Just because ab \in \mathbb{Z} this doesn't mean both a,b \in \mathbb{Z}. Am I missing something obvious?
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    Quote Originally Posted by Boysilver View Post
    No doubt you're right, but I guess my issue is that I don't see how a \in P or b \in P implies a \in P \cap \mathbb{Z} or b \in P \cap \mathbb{Z}. Just because ab \in \mathbb{Z} this doesn't mean both a,b \in \mathbb{Z}. Am I missing something obvious?
    Since our ideal is over  \mathbb{Z} we assume  a,b \in \mathbb{Z} .
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    Quote Originally Posted by Boysilver View Post
    No doubt you're right, but I guess my issue is that I don't see how a \in P or b \in P implies a \in P \cap \mathbb{Z} or b \in P \cap \mathbb{Z}. Just because ab \in \mathbb{Z} this doesn't mean both a,b \in \mathbb{Z}. Am I missing something obvious?
    don't forget that a and b are in \mathbb{Z} because you want to prove that P \cap \mathbb{Z} is a prime ideal of \mathbb{Z}.
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    Quote Originally Posted by NonCommAlg View Post
    a less trivial fact is this: if P \neq \{0\}, then P \cap \mathbb{Z} \neq \{0\}.
    Why is this true?
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    Quote Originally Posted by Jim63 View Post
    Why is this true?
    let 0 \neq a \in P and suppose that f(x)=x^n + c_1x^{n-1} + \cdots + c_n is the minimal polynomial of f(x) over \mathbb{Q}. it's a known fact that, since a \in \mathcal{O}_K, we have c_j \in \mathbb{Z}, for all j.

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

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


    Note: in the proof we didn't need P to be a prime ideal. so the above result holds for any non-zero ideal P of \mathcal{O}_K.
    Last edited by NonCommAlg; May 22nd 2010 at 04:34 PM.
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