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Math Help - Couple more quadratic field questions

  1. #1
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    Couple more quadratic field questions

    Why is 3 not a prime in Q(sqrt7) despite that there is no integer in Q(sqrt7) of norm 3?

    Like, if 3 can be expessed as a*b where a and b are non-unit integers in Q(sqrt7), then |N(a)|>=2, |N(b)|>=2 (they are not units), and |N(a)||N(b)| = N(3) = 9. Thus it must be that |N(a)|=|N(b)|=3. Contradictions, contradictions.. please help.

    Also, what exactly are the prime factors of 3 (in Q(sqrt7))?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    You seem to be confusing the notions of prime and irreducible. What you have shown is that 3 is irreducible in \mathbb{Q}(\sqrt{7}). A prime element is always irreducible, but an irreducible element is not necessarily prime. If the ring is a unique factorization domain, then the two notions coincide. In this case, \mathbb{Q}(\sqrt{7}) is a unique factorization domain and hence 3 is indeed prime.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    You seem to be confusing the notions of prime and irreducible. What you have shown is that 3 is irreducible in \mathbb{Q}(\sqrt{7}). A prime element is always irreducible, but an irreducible element is not necessarily prime. If the ring is a unique factorization domain, then the two notions coincide. In this case, \mathbb{Q}(\sqrt{7}) is a unique factorization domain and hence 3 is indeed prime.
    Then how do you tell if a number is prime? Also, the book i'm using says 3 is NOT a prime in Q(sqrt7)
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    I'm sorry, I made a mistake. Your book is correct; 3 is not prime, because it divides (1+\sqrt{7})(1-\sqrt{7}) but divides neither of the factors.

    In fact we do have an element of norm -3. Consider 2+\sqrt{7}; its norm is 2^2-7=-3. Think this through!
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