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Algebra question

  1. Nov 21, 2006 #1
    In my book it says that a Ring R can be imbedded in a ring R1 if there is an isomorphism of R onto R1 and we call R1 the over-ring. Some of the things that the author goes on to talk about makes me think that R can just be fit into R1 almost like it's a subset (which I know it's not) of R1. What I'm wondering is that if R is isomorphic to R1 then isn't R1 isomorphic to R so if R can be imbedded in R1 then R1 can be imbedded into R? Can anyone help to make this a little clearer to me?

    Another question I had has to do with polynomial rings. If you're trying to show that a certain polynomial is irreducible over a field F[x] is to show that the ideal A = (p(x)) in F[x] is a maximal ideal? What exactly does an ideal (p(x)) in F[x] look like? An example practice problem I have is show that x^3-9 is irreducible over the integers mod31. So I need to show that the ideal (x^3-9 ) is a maximal ideal. what does an ideal of this form look like. maybe not just this example but in general. Also, where does the -9 come in since the integers mod31 are {0,1,2,........,30} which are all positive.

    thanks for any help
     
  2. jcsd
  3. Nov 21, 2006 #2

    matt grime

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    All I can suggest is that you reread your book more closely since what you've written is obviously wrong: an 'isomorphism ... onto'? I don't think you meant to write that.

    -9 is just "minus 9", the additive inverse of 9, so that is obviously 22, mod 31, since 9+22=31 which is zero mod 31. The things mod 31 are not 'just' the numbers 0,...,30. Those labels are good choices of elements of equivalence classes of elements, with addition of the equivalence classes as the operation.
     
  4. Nov 21, 2006 #3

    HallsofIvy

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    Are you sure your book didn't say "homomorphism"? If R is isomorphic to R1 then R is not a subring of R1, it is essentially the same as R1.
     
  5. Nov 21, 2006 #4
    My book defines an isomorphism as a one-to-one homomorphism of R into R1. Then it says that two rings are isomorphic if there is an isomorphism of one onto the other. Then it says a ring R can be imbedded in a ring R1 if there is an isomrphism of R into R1.
     
  6. Nov 22, 2006 #5

    matt grime

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    That is an injective homomorphism. It is *not* an isomorphism.
     
  7. Dec 6, 2006 #6

    HallsofIvy

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    Notice that one says "onto" and the other "into".
     
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