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Math Help - Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

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    Super Member Bernhard's Avatar
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    Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

    In Dummit and Foote Ch 9 Polynomial Rings, Proposition 2 (page 296 - see attached) reads as follows:

    ================================================== =========================================

    Proposition 2: Let I be an ideal of the ring R and let (I) = I[x] denote the ideal of R[x] generated by I (the set of polynomials with co-efficients in I)

    Then  R[x]/I[x] \cong (R/I)[x]

    In particular, if I is a prime ideal of R then I[x] is a prime ideal of R[x]

    ================================================== ==========================================

    I am concerned with the proof that  R[x]/I[x] \cong (R/I)[x].

    In this proof (i think) we need the First Isomorphism Theorem. Dmmit and Foote give this theorem as part of Theorem 7 in Ch 7 on page 243 (see attached)

    They state the First Isomorphism Theorem as follows: (see attached)


    ================================================== ==========================================

    If  \phi : R \rightarrow S is a homomorphism of rings, then the kernel of  \phi is an ideal of R, the image of  \phi is a subring of S, and  R / ker \phi is isomorphic as a ring to  \phi (R)

    ================================================== ==========================================

    I follow the proof of down to the following statement (see attached page 296 D&F)

    ... ... , which is to say that ker  \phi = I[x] = (I), proving the first part of the proposition.


    BUT surely we need to now use the First Isomorphism Theorem to show that we have an isomorphism ( I suspect D&F are assuming we know this bit ... )

    However, by the FIT, given that ker  \phi = I[x] we have that


     R[x]/ ker \phi  =  R[x] / I[x][ \cong  \phi [ R[x] ]


    So we require  \phi [R[x]] = (R/I) [x]


    But how do we know that the image of R[x] under  \phi is (R/I)[x] and not a subset of (R/I)[x]


    Would appreciate some help in clarifying this matter.

    Peter
    Last edited by Bernhard; February 8th 2013 at 06:03 PM.
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    Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

    Who is \phi?

    Presumably \phi(a_kx^k+a_{k-1}x^{k-1}+...+a_0)=(a_k+I)x^k+(a_{k-1}+I)x^{k-1}+...+a_0+I

    With this definition of \phi, it's "easy" to show \phi is a homomorphism with image (R/I)[x] and kernel I[x].
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    Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

    yes, indeed, to apply the FIT, we need to know that φ is onto.

    but suppose that f(x) is in (R/I)[x], that is f(x) has coefficients in R/I. recall that the NATURAL homomorphism R-->R/I which sends r-->r+I for each r in R IS onto. so every element b in R/I is a+I for SOME a in R.

    so given f(x) = (a0+I) + (a1+I)x +...+ (an+I)xn in (R/I)[x], we clearly have for

    g(x) = a0 + a1x +...+ anxn, φ(g(x)) = f(x), which shows that φ is surjective.
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    Super Member Bernhard's Avatar
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    Re: Polynomial Rings _ Dummit and Foote Chapter 9 - Proposition 2

    Thanks Deveno!
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