Q as a module over Z

  • Thread starter sab47
  • Start date
  • #1
7
0
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.
 

Answers and Replies

  • #2
453
0
What would it mean to find a counterexample here?
 
  • #3
7
0
What would it mean to find a counterexample here?
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!
 
  • #4
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
for R over Q, merely the number of elements suffices.
 
  • #5
7
0
for R over Q, merely the number of elements suffices.
number of elements? how do you mean?
 
  • #6
AKG
Science Advisor
Homework Helper
2,565
4
How many elements can a finitely-generated module over Q possibly have? How many elements does R have?
 
  • #7
7
0
How many elements can a finitely-generated module over Q possibly have? How many elements does R have?
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!
 
  • #8
453
0
Should I somehow show that any finitely generated set over Q has finite number of elements?
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:
  • #9
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
do you know about countable, uncountable? this theory was introduced by cantor some 100 years ago.
 
  • #10
7
0
do you know about countable, uncountable? this theory was introduced by cantor some 100 years ago.
Not so much, no. I've just heard of it. But I'll look into it.
 
  • #11
mathwonk
Science Advisor
Homework Helper
2020 Award
11,100
1,302
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.
 
  • #12
453
0
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.
 

Related Threads on Q as a module over Z

  • Last Post
Replies
1
Views
2K
Replies
3
Views
3K
Replies
4
Views
702
Replies
1
Views
2K
Replies
10
Views
5K
Replies
2
Views
634
Replies
11
Views
1K
  • Last Post
Replies
2
Views
1K
Top