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Math Help - Polynomial Rings Over Fields

  1. #1
    Super Member Bernhard's Avatar
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    Polynomial Rings Over Fields

    I am reading Dummit and Foote Section 9.5 Polynomial Rings Over Fields II and need some help and guidance with the proof of Proposition 15.

    Proposition 15 reads as follow:

    Proposition 15. The maximal ideals in F[x] are the ideals (f(x)) generated by irreducible polynomials. In particular, F[x]/(f(x)) is a field if and only if f(x) is irreducible.

    Dummit and Foote give the proof as follows:

    Proof: This follows from Proposition 7 of Section 8.2 applied to the Principal Ideal Domain F[x].


    My problem
    - can someone show me how Proposition 15 above follows from Proposition 7 of Section 8.2 (see below for Proposition 7 of Section 8.2)

    I would be grateful for any help or guidance in this matter.


    Peter


    Dummit and Foote - Section 8.2 - Proposition 7

    Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal.
    Last edited by Bernhard; May 31st 2013 at 02:57 AM.
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  2. #2
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    Re: Polynomial Rings Over Fields

    Quote Originally Posted by Bernhard View Post

    My problem
    - can someone show me how Proposition 15 above follows from Proposition 7 of Section 8.2 (see below for Proposition 7 of Section 8.2)
    Every field is a UFD, so F[x] is also a UFD. In light of Proposition 7 of Section 8.2, this means every non-zero prime ideal I of F[x] is maximal. I'm not sure if you've encountered this theorem, but it is immensely useful:

    Let R be any ring.
    An ideal I is maximal iff R/I is a field.
    An ideal I is prime iff R/I is an integral domain

    If you accept the theorem above (particularly the first one), all you'll need to show is that the irreducible polynomials are precisely the prime elements of F[x]. If you haven't read about the theorem above, or you're having trouble showing irreducible polynomials are prime, I can sketch the proof.
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Polynomial Rings Over Fields

    Thanks Gusbob, it would be very helpful if you sketched the proof

    Peter
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    Re: Polynomial Rings Over Fields

    Quote Originally Posted by Bernhard View Post
    Thanks Gusbob, it would be very helpful if you sketched the proof

    Peter
    Which one? Or both?
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  5. #5
    Super Member Bernhard's Avatar
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    Re: Polynomial Rings Over Fields

    Be great to see your proof of both

    Peter
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    Re: Polynomial Rings Over Fields

    Theorem: I is maximal iff R/I is a field

    Proof: Let J be a ideal such that I\subseteq J \subseteq R. By the correspondence theorem \overline{I}=(0)\subseteq\overline{J}\subseteq R/I, where \overline{I},\overline{J} denotes the image of I,J under the homomorphism r\mapsto r+I.

    If R/I is a field, there are only two possible ideals, so \overline{J} is either \overline{I} or R/I. This means J=I or J=R, so I is maximal.

    Conversely, if I is maximal, any ideal between I and R is I or R itself, so R/I only has two ideals, i.e. R/I is a field.




    Theorem: f(x) is irreducible iff (f(x)) is prime.

    Proof: Suppose (f(x)) is prime. If f(x) is constant and non-zero, then (f(x))=F[x], violating the primality of f(x). So f(x) cannot be constant. If f(x)=g(x)h(x), then g(x)h(x) \in (f(x)). Since (f(x)) is prime by assumption, without loss of generality assume g(x)\in (f(x)) (recall that an ideal I is prime if xy \in I \implies x\in I or y\in I). Then the degree of g(x) is at least that of the degree of f(x). But \deg f(x)=\deg g(x)+\deg h(x), so it must be that h(x) is a constant polynomial. Hence f(x) is irreducible.

    Now suppose f(x) is irreducible. Since we're in a UFD, it is enough to show that (f(x)) is maximal. Let I \subsetnoteq F[x] be an ideal containing f(x). We'll need to show that I=(f(x)). One has to note that F[x] is a principal ideal domain. So let I=(g(x)). Then (f(x))\subseteq(g(x))\subsetnoteq F[x]. The first inclusion means f(x)=g(x)h(x) for some polynomial h(x). The irreducibility of f(x) forces h(x) to be a unit (non-zero constant). Then f(x) divides g(x) as well, so (f(x))=(g(x))=I.
    Last edited by Gusbob; June 1st 2013 at 04:08 AM.
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    Re: Polynomial Rings Over Fields

    I made a mistake with my latex code, but now I can't edit it. In the last paragraph, the IF[x] should read I \subsetneq F[x] and the (g(x))F[x] should read (g(x))\subsetneq F[x].
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  8. #8
    Super Member Bernhard's Avatar
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    Re: Polynomial Rings Over Fields

    Thank you so much for your help ... Working through your posts carefully

    Thanks again!!! So very helpful!!!

    Peter
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