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Q as a module over Z

  1. Mar 14, 2007 #1
    Hey guys,

    I'm self-teaching maths to preper myself for the next term of uni, so I'm reading this book on abstract algebra, and somewhere it says that R (the set of real numbers) is not finitely generated as a module over Q (set of rational numbers). Now, I can see that it's not, but i can't think of a rigorous proof for it. I thought maybe i hould just find a countr example like i did in a different case (Q is not finitely ggenerated over Z) but i'm prety bad at these counter examples! lol. Can anyone help me make sense of this? cause i prefer to understand everything before i continue to the next part.
     
  2. jcsd
  3. Mar 14, 2007 #2
    What would it mean to find a counterexample here?
     
  4. Mar 14, 2007 #3
    there was a similar statement which said Q is not finitely generated over Z. So what I did with that was i said if we assume q1,...,qn generate Q. Then take a z in Z which is coprime with the denominator of all members of the generatind set (i.e coprime with all qi) then 1/z cannot be generated by this set q1,...,qn . so Q is not finitely generated.

    So this is what I meant by a counter example, finding something like 1/z above, which can't be generated by the generating set. Pehaps counter example isn't the best way to put it!
     
  5. Mar 14, 2007 #4

    mathwonk

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    for R over Q, merely the number of elements suffices.
     
  6. Mar 14, 2007 #5
    number of elements? how do you mean?
     
  7. Mar 14, 2007 #6

    AKG

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    How many elements can a finitely-generated module over Q possibly have? How many elements does R have?
     
  8. Mar 14, 2007 #7
    Should I somehow show that any finitely generated set over Q has finite number of elements? Sorry to be so slow, like I said I'm self teaching these things. There must be a theorem or something about number of elements of finitely generated modules which I've forgotten!
     
  9. Mar 14, 2007 #8
    No, of course not. Q is finitely generated over itself, how many elements does it have? What if F is a finitely generated free module over Q? How many elements does it have? Can this help us solve the more general problem?
     
    Last edited: Mar 14, 2007
  10. Mar 14, 2007 #9

    mathwonk

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    do you know about countable, uncountable? this theory was introduced by cantor some 100 years ago.
     
  11. Mar 14, 2007 #10
    Not so much, no. I've just heard of it. But I'll look into it.
     
  12. Mar 14, 2007 #11

    mathwonk

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    the point is a finitely generated module iover Q has the same number of elements as Q, while that is less than the number of elements of R.
     
  13. Mar 14, 2007 #12
    I suggest you learn some basic set theory before going too deep into algebra. Knowledge of cardinalities, the schroder-bernstein theorem and Zorn's lemma are all pretty important prerequisites for studying modules and rings properly.
     
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