Conversation started May 30, 2012 at 13:29.
Matt N.
Matt N.
May 30, 2012 13:29
Apparently, every $R$-module $M$ is the quotient of a free module. The proof has been skipped in my notes. So if $F(S)$ denotes the free module generated by the set $S$, we define a map $f: F(M) \to M$ as $$v=(0, \dots, 0, 1, 0, \dots, 0) \mapsto m$$ where the vector $v$ is one at coordinate $m$ and zero otherwise and $$(0, 0, \dots, 0) \mapsto 0$$
Then by the first isomorphism theorem, $M \cong F(M) / ker(f)$ since $f$ is surjective. So far so good.
What disturbs me is that I think the kernel of $f$ is $0$. Something is wrong here.
If $f$ really had kernel zero then every module $M$ would be isomorphic to $F(M)$.
David Wheeler
David Wheeler
i am confused...you are assuming M is finite?
Matt N.
Matt N.
That should've been: $$v=(0, \dots, 0, 1, 0, 0, \dots) \mapsto m$$ and $$(0, 0, \dots ) \mapsto 0 $$
David Wheeler
David Wheeler
ok, so we're taking M as a basis for F(M)
now, what happens if m is a torsion element in M?
because you're going to extend f R-linearly, yes?
i would call v, $v_m$
and then you have for a torsion element m, $f(rv_m) = rm = 0$
since F(M) is free, $rv_m \neq 0$, so the kernel isn't just the 0-element of F(M).
things that aren't 0 in F(M) can add up to 0 in M, see?
t.b.
t.b.
May 30, 2012 13:50
@MattN You have tons of relations in $M$ that are non-existent in $F(M)$. For example $\lambda m$ and $m$ are independent in $F(M)$. So $(\lambda m) - \lambda \cdot m$ is nonzero in $F(M)$ but is sent to zero in $M$ under $f$.
David Wheeler
David Wheeler
in F(M), you have R-linear combinations of the v's, all of which are non-zero, except the 0-combination
in M, these same linear combinations may "evaluate" to 0
Matt N.
Matt N.
@tb What's the difference between $(\lambda m)$ and $\lambda \cdot m$?
David Wheeler
David Wheeler
@tb i thought you didn't like algebra?
t.b.
t.b.
@MattN $(\lambda m)$ is a basis element in $F(M)$ and $\lambda \cdot m$ is $\lambda$ times the basis element $m$ of $F(M)$.
@DavidWheeler Right. Not particularly.
David Wheeler
David Wheeler
@MattN $\lambda m = v_{\lambda m}, \lambda \cdot m = \lambda v_m$
Matt N.
Matt N.
May 30, 2012 13:58
But isn't multiplication with elements $r$ in $R$ for $(m)$ in $F(M)$ defined as $r(m) := (rm)$?
David Wheeler
David Wheeler
one has 1 in the $\lambda m$-th place and 0's elsewhere, one has $\lambda$ in the m-th place, and 0's elsewhere
Matt N.
Matt N.
Oh. Wait.
David Wheeler
David Wheeler
two distinct elements of M give rise to 2 distinct basis elements of F(M)
F(M) is a LOT bigger than M
we give each element of M "it's own space" and then multiply everything in each space by every element of R
whereas in M, R just moves things around in M
Matt N.
Matt N.
So let's see what these things are: $m = (0, 0, \dots, 0, 1, 0, \dots, )$. Right?
David Wheeler
David Wheeler
for example, suppose M = Z, the integers. this is a free Z-module of rank 1. but F(Z) assigns a distinct element for each integer, so you have a free module of infinite rank
Matt N.
Matt N.
May 30, 2012 14:04
With one at position $m$.
Then $\lambda \cdot m = (0, 0, \dots, \lambda, 0, \dots )$
@tb Everything right so far?
t.b.
t.b.
yep (if that last thing has an entry in "m'th position".)
Matt N.
Matt N.
Yes.
Now. I want to figure out what $(\lambda m)$ is.
I think basis elements of $F(M)$ look like $(0, \dots, 1, 0, \dots)$.
t.b.
t.b.
Maybe you just use David's notation and write $v_{\lambda m} - \lambda v_m$. Where $v_x$ is the basis element in $F(M)$ corresponding to $x \in M$.
David Wheeler
David Wheeler
in my example, we might have 2 = (0,0,0,1,0,0,.....) (if we list the integers (0,1,-1,2,-2,...etc). but 2.1 = (0,2,0,.....)
but both of these elements will map to the integer 2 under f.
Matt N.
Matt N.
@tb Got it. Thank you.
Now I need to read your first comment again and see if I understand it now.
@tb Yay!! : )
@tb Thank you! : ) So the kernel of $f$ that seemed so obviously zero isn't zero.
David Wheeler
David Wheeler
May 30, 2012 14:10
the "sequence" notation is sort of bad, because R (the real numbers for example) is a module over Z, but you can't "list" the basis elements that way
t.b.
t.b.
I agree
David Wheeler
David Wheeler
it would be better to use an "index set"
t.b.
t.b.
@MattN far from it!
David Wheeler
David Wheeler
every "free thing" i've come across usually seems to be huge. except vector spaces. for some reason, they behave.
Matt N.
Matt N.
@tb See you later! And thanks. You saved me from a moment of panic there.
t.b.
t.b.
May 30, 2012 14:13
As David said, the free module $F(M)$ on $M = R$ has rank $\#R$. So it's huge in comparison with the free module $M$ of rank $1$.
@MattN later!
and no problem.
 
Conversation ended May 30, 2012 at 14:13.