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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.
  2. jcsd
  3. Feb 3, 2012 #2


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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
    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


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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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