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Prime Ideals of direct sum of Z and Z

  1. Feb 3, 2012 #1
    I am trying to find nonzero prime ideals of [itex]\mathbb{Z} \oplus \mathbb {Z}[/itex], specifically those which are not also maximal.

    If I try to do direct sums of prime ideals, the resulting set is not a prime ideal. (e.g., [itex]2 \mathbb{Z} \oplus 3 \mathbb{Z}[/itex] is not prime since [itex](3,3) \cdot (2,2) = (6,6)\in 2 \mathbb{Z} \oplus 3 \mathbb{Z}[/itex], but [itex] (2,2),(3,3) \notin 2 \mathbb{Z} \oplus 3\mathbb{Z} [/itex].)

    In fact, I don't think a prime ideal could be constructed in this way since I can always take a product of the form (1,x)(y,1) and obtain (y,x) and since 1 isn't a multiple of any integer other than 1, neither of the factors would have come from the ideal.

    Can somebody please help me find the prime ideals? Thanks.
     
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  3. Feb 3, 2012 #2

    mathwonk

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    modding out a prime ideal is supposed to give a domain. does that help?

    notice the zero ideal is not prime, why not?
     
  4. Feb 3, 2012 #3
    0 is not prime because [itex](a,0)(0,b)=0[/itex], but [itex] (a,0),(0,b)\notin 0 [/itex].

    Okay, so would [itex]\langle (2,3) \rangle [/itex] be a prime ideal and [itex]\langle (4,6) \rangle [/itex] be a prime ideal that is not maximal?

    If you mod out by both, I believe you get an ID, but [itex]\langle (4,6) \rangle \subset \langle (2,3) \rangle [/itex] .
     
  5. Feb 3, 2012 #4

    micromass

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    We have

    [tex](2,5)*(5,3)\in <(2,3)>[/tex]

    but neither (2,5), not (5,3) is in the ideal.

    What about [itex](\alpha\mathbb{Z})\times \mathbb{Z}[/itex]
     
  6. Feb 4, 2012 #5
    Okay, then [itex]p \mathbb{Z} \oplus \mathbb{Z} [/itex] is a prime ideal if p is prime, but it would also be maximal... or would it? If I mod out by it, I think I get a field, so it would have to be maximal.

    If we have [itex] I= rs \mathbb{Z} \oplus \mathbb{Z} [/itex], then I can always write [itex](r,1)\cdot (s,1)=(rs,1)\in I [/itex], but [itex](r,1),(s,1)\notin I[/itex]. Thus, composites don't work.
     
  7. Feb 9, 2012 #6
    Okay, I have proven that for any rings R and S, [itex]R \oplus S [/itex] can only have a prime ideal of the form [itex] I \oplus S [/itex] where I is a prime ideal of R or [itex]R \oplus J[/itex] where J is a prime ideal of S. Since the prime ideals of Z are pZ for prime p, the prime ideals of [itex] \mathbb{Z}\oplus \mathbb{Z} [/itex] are [itex]p \mathbb{Z} \oplus \mathbb{Z} [/itex].

    BUT, we have [itex](\mathbb{Z} \oplus \mathbb{Z})/(p \mathbb{Z} \oplus \mathbb{Z})\cong (\mathbb{Z}/p\mathbb{Z}) \oplus (\mathbb{Z}/\mathbb{Z})\cong \mathbb{Z}/p \mathbb{Z}[/itex] which is a field.

    Either I'm missing something simple, or there are no primes that are not also maximal.
     
  8. Feb 12, 2012 #7

    morphism

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    Don't forget [itex]\mathbb Z \oplus p\mathbb Z[/itex] and the two nonmaximal primes [itex]0 \oplus \mathbb Z[/itex] and [itex]\mathbb Z \oplus 0[/itex].
     
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