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Math Help - Maximal Ideal Problem within a Field

  1. #1
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    Maximal Ideal Problem within a Field

    Let M be a maximal ideal of a commutative ring R with identity. Prove that for each  a \in R, MR[X]+(X+a)R[X] is a meximal ideal of R.

    Proof so far.

    Suppose that MR[X]+(X+a)R[X] \subseteq J \subseteq R , and I want to show that J = R.

    Or should I process another way, prove that J = MR[X]+(X+a)R[X] \subseteq J \subseteq R ?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let M be a maximal ideal of a commutative ring R with identity. Prove that for each  a \in R, MR[X]+(X+a)R[X] is a meximal ideal of R[X] .
    I=MR[X] + (X+a)R[X] is an ideal of R[X] not R. suppose J is an ideal of R[X] and I \subsetneq J \subseteq R[X]. choose f(X) \in J-I. then f(X)=(X+a)g(X) + c, for some c \in R and

    g(X) \in R[X]. why? now X+a \in I \subset J and f(X) \in J. thus c=f(X)-(X+a)g(X) \in J. but c \notin M because otherwise c \in I and thus f(X) \in I, which is not possible. so since M

    is maximal, we must have M+Rc=R. hence m + rc = 1, for some r \in R, \ m \in M. but then: 1=rf(X)-r(X+a)g(X)+m \in J. why? therefore: J=R[X]. \ \ \Box
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  3. #3
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    Thanks for your help!

    So now, I want to show that MR[X]+(X+a)R[X]=MR[X]+XR[X] if and only if a \in M

    Proof so far.

    Suppose that  a \in M , MR[X]+(X+a)R[X]=MR[X]+XR[X]+aR[X], now, does  aR[X] get absorbed by  MR[X] since a is in M?

    The converse would be the same, isn't it?

    Thanks
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    Thanks for your help!

    So now, I want to show that MR[X]+(X+a)R[X]=MR[X]+XR[X] if and only if a \in M

    Proof so far.

    Suppose that  a \in M , MR[X]+(X+a)R[X]=MR[X]+XR[X]+aR[X], now, does  aR[X] get absorbed by  MR[X] since a is in M?

    The converse would be the same, isn't it?

    Thanks
    your proof is not correct, because in general: (X+a)R[X] \neq XR[X]+aR[X]. the reason is that an element of the LHS is in the form (X+a)f(X) but an element of the RHS is in the form

    Xf(X) + ag(X). so in general we only have: (X+a)R[X] \subseteq XR[X]+aR[x]. here's how to solve your problem: first if a \in M, then we'll have:

    MR[X]+(X+a)R[X] \subseteq MR[X] + XR[X]+aR[X] = MR[X]+XR[X], because a \in M. on the other hand we have: (-a)R[X] \subseteq MR[X]. thus:

    MR[X]+XR[X] \subseteq MR[X]+(X+a)R[X]+(-a)R[X] = MR[X]+(X+a)R[X]. this proves half of your problem. now suppose MR[X]+(X+a)R[X]=MR[X]+XR[X]. then since we

    have X+a \in MR[X]+(X+a)R[X], we must also have X+a \in MR[X]+XR[X]. hence: X+a=\sum_{i=0}^k m_i X^i + X\sum_{i=0}^{\ell}q_iX^i, for some q_i \in R, \ m_i \in M. so the constant coefficients in

    both sides must be equal. thus: a=m_0 \in M. \ \ \Box
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