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An Ideal in Z[x]

  1. Jul 14, 2008 #1
    The problem statement, all variables and given/known data
    Prove that the ideal I = [tex]\langle x^2 + 1 \rangle[/tex] is prime in Z[x] but not maximal.

    The attempt at a solution
    I'm having a hard time doing this because Z[x] is not a field. I know that x2 + 1 is irreducible in Z[x] so the proof must hinge on this fact.

    Let f(x) and g(x) belong to Z[x] and suppose f(x)g(x) is in I. Then there is some q(x) in Z[x] such that f(x)g(x) = (x2 + 1)q(x). How can I show that either f(x) or g(x) belongs to I? How does the irreducibility of x2 + 1 come into play here?
     
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  3. Jul 14, 2008 #2

    matt grime

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    A field has no ideals (other than 0 and itself) so I don't get your comment about it being harder (unless it was a joke).

    What are the definitions of maximal and prime ideal (in terms of quotients)?
     
    Last edited: Jul 14, 2008
  4. Jul 14, 2008 #3
    It's hard because most of the results I have concerning ideals involve fields.

    I know that if R is a commutative ring with unity and A is an ideal of R, then

    (1) A is prime iff R/A is an integral domain, and
    (2) A is maximal iff R/A is a field.

    I can try proving that Z[x]/I is an integral domain to show that I is prime but that's tedious. Is there a slicker way?
     
  5. Jul 14, 2008 #4

    matt grime

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    It's not tedious, it is simple and takes about one line.

    How can most of the results you know about ideals involve fields? As I pointed out fields don't have any non-trivial ideals.
     
  6. Jul 14, 2008 #5
    I forgot that Z[x]/I is already a ring since I is an ideal. So all I have to do is show that Z[x]/I contains unity and no zero divisors. I guess this all follows from the fact that Z[x] is an integral domain. Correct?

    I have a theorem that states: Let F be a field an let p(x) belong to F[x]. Then [tex]\langle p(x) \rangle[/tex] is a maximal ideal of F[x] if and only if p(x) is irreducible in F. Unfortunately I can't use this.

    I was also trying to use the division algorithm to show that I is a prime ideal but that only works for fields. The chapter where this problem is from also states a bunch of reducibility results but they all involve fields. This is why I wrote what I wrote.

    Thanks for your help.
     
  7. Jul 14, 2008 #6

    matt grime

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    What, independently of I? You need to prove something because x^2+1 is irreducible, remember. It is trivially true that the quotient isn't a field (just consider the polynomial 2 - it can't have a multiplicative inverse), and equally trivial that it contains a unit.
     
  8. Jul 14, 2008 #7
    You're right. I haven't used the fact that xx + 1 is irreducible. Is this used to show that Z[x]/I has no zero divisors?

    Consider the product (f(x) + I)(g(x) + I) = f(x)g(x) + I. If this product equals I, then f(x)g(x) = 0. It follows that the coefficients of f(x)g(x), which are just products of integers equals, equal 0. This means one of the integer factors of each of the coefficients is 0 and so one of f(x) or g(x) must be the 0 polynomial. Nowhere did I use the fact that xx + 1 is irreducible. Where did I go wrong?
     
  9. Jul 14, 2008 #8

    Hurkyl

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    Why?
     
  10. Jul 14, 2008 #9
    Hmm...I guess that isn't necessarily true. If f(x)g(x) + I = I, then all I can say is that f(x)g(x) is a member of I and so there is some q(x) in Z[x] such that f(x)g(x) = (xx + 1)q(x). Now what?
     
  11. Jul 14, 2008 #10

    Dick

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    Let r_f(x) and r_g(x) be the remainders after division of f(x) and g(x) by (x^2+1). If that relation is true then r_f(x)*r_g(x)=0. Now show Z[x] has no zero divisors.
     
  12. Jul 15, 2008 #11

    matt grime

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    Ugh, all this dividing and so on. Horrible. This is Z[x], after all, so 'division' isn't that nice to contemplate (we should really be thinking about the cosets f(x)+I as used above, anyway).

    I much prefer thinking of it as "x^2=-1".

    This makes it clear that we can think of Z[x]/(x^2+1) as

    {a+bx : a,b in Z and x^2=-1}

    and so zero divisors in Z[x]/(x^2+1) would imply that there are two deg 1 polys f(x) and g(x) such that f(x)g(x)=n(x^2+1) for some integer n.

    (Of course, you can always tensor with Q and use division algorithms there.)
     
  13. Jul 15, 2008 #12

    Dick

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    You are right. After rereading it, it looked so ugly I was going to delete it. Then the forum went offline.
     
  14. Jul 15, 2008 #13
    Why is it that you can treat xx as -1? I never understood this.
     
  15. Jul 15, 2008 #14
    Oh, and since f(x)g(x) = n(x2 + 1), how is it possible for f(x) or g(x) to be in I? I don't understand how I is a prime ideal yet.
     
  16. Jul 15, 2008 #15

    matt grime

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    You treat x^2 as -1 since you're treating x^2+1 as zero: that's what the notation Z[x]/(x^2+1) means.

    I don't understand your last post. The point was that if it is conceivable that two linear polys multiply together to be in I, then it must follow that f(x)g(x)=n(x^2+1). Polys of the form n(x^2+1) are the only polys of degree two in I. But x^2+1 is irreducible, so there you can't write (an integer multiple of) it as a product of two deg 1 polys.
     
  17. Jul 15, 2008 #16
    So in other words, f(x) and g(x) can't be linear. Right? But then what other choice do they have? How can I show that f(x) or g(x) is in I and thus prove that I is a prime ideal?
     
  18. Jul 15, 2008 #17

    matt grime

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    How many ways can I write n(x^2+1) as a product of two polynomials, recalling x^2+1 is irreducible over Z....? This is not hard algebra, this is stuff you were taught in high school.
     
  19. Jul 15, 2008 #18
    You could consider n as one polynomial and x2 + 1 as another polynomial.

    Now what? Are you suggesting that f(x) must be n (or x2 + 1 ) and g(x) must by x2 + 1 (or n)? But this will contradict the fact that f(x) and g(x) have degree at most 1.
     
  20. Jul 15, 2008 #19

    matt grime

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    Uh, what? You've lost me as to what it is that you think that you're doing.
     
  21. Jul 15, 2008 #20
    I'm trying to show that I is an ideal. I'm using the following definition.

    A prime ideal A of a commutative ring R is a proper ideal of R such that a, b in R and ab in A imply a in A or B in B.

    So I have to show that either f(x) or g(x) is in I.
     
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