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Q as a module over Z |
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| Mar14-07, 08:44 AM | #1 |
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Q as a module over Z
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. |
| Mar14-07, 09:30 AM | #2 |
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What would it mean to find a counterexample here?
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| Mar14-07, 09:47 AM | #3 |
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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! |
| Mar14-07, 10:25 AM | #4 |
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Recognitions:
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Q as a module over Z
for R over Q, merely the number of elements suffices.
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| Mar14-07, 03:37 PM | #5 |
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| Mar14-07, 04:17 PM | #6 |
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Recognitions:
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How many elements can a finitely-generated module over Q possibly have? How many elements does R have?
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| Mar14-07, 05:27 PM | #7 |
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| Mar14-07, 07:44 PM | #8 |
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| Mar14-07, 09:46 PM | #9 |
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Recognitions:
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do you know about countable, uncountable? this theory was introduced by cantor some 100 years ago.
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| Mar14-07, 09:50 PM | #10 |
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| Mar14-07, 09:56 PM | #11 |
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Recognitions:
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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.
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| Mar14-07, 10:51 PM | #12 |
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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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