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Math Help - Principal ideal and PID

  1. #1
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    Principal ideal and PID

    (i) Show that (2) +(x) ={ f2 + gx: f, g in Z[x]} is not a principal ideal. So Z[x] is not a PID even though Z is a PID

    (ii) Let r in R* where R is an integral domain. Show that (x) + (r) is a principal ideal <==> r in U(R). Hence show that R[x] is a PID <==> R is a field.

    This is my attempt for part (i)

    Let (a) = (2) + (x) for some a belongs to Z[x]
    Clearly, (a) is the least principal ideal which contains both (2) and (x). Hence (2,x) exists and (2,x) = (a)
    Since (a) = (2)+(x) we have (2) is the subset of (a) and (x) is a subset of (a). So, a|2 and a|x.
    But also, a is in (a) = (2) +(x) and so a = 2f + gx.
    Up to here, I dont know how to show it leads to contradiction so that (a) is not a principal ideal

    Thank you in advanced for helping me
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  2. #2
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    Quote Originally Posted by knguyen2005 View Post
    (i) Show that (2) +(x) ={ f2 + gx: f, g in Z[x]} is not a principal ideal. So Z[x] is not a PID even though Z is a PID

    (ii) Let r in R* where R is an integral domain. Show that (x) + (r) is a principal ideal <==> r in U(R). Hence show that R[x] is a PID <==> R is a field.

    This is my attempt for part (i)

    Let (a) = (2) + (x) for some a belongs to Z[x]
    Clearly, (a) is the least principal ideal which contains both (2) and (x). Hence (2,x) exists and (2,x) = (a)
    Since (a) = (2)+(x) we have (2) is the subset of (a) and (x) is a subset of (a). So, a|2 and a|x.
    But also, a is in (a) = (2) +(x) and so a = 2f + gx.
    Up to here, I dont know how to show it leads to contradiction so that (a) is not a principal ideal

    Thank you in advanced for helping me

    Hint: The ideal (2)+(x) is the set of all the pol's in \mathbb{Z}[x] with even free coefficient.

    Tonio
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