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 | | From: | Tonico | | Subject: | Quotient modules question | | Date: | 23 Jan 2005 03:17:23 -0800 |
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 | Hi all:
Last time I try to get an answer in this group about this, I
promise...at least in the next few months: Suppose A is a (commutative,
unitarian) ring, M, N ideals of A, and we look at all of them as
A-modules. If the quotient A-modules A/N, A/M are A-isomorphic (i.e.,
as A-modules), then M = N.
I still have the strong feeling that this isn't true, but can't come up
with a counterexample.
Any input on this will most sincerely be thanked.
Regards
Tonico
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| | From: | Tonico | | Subject: | Re: Quotient modules question | | Date: | 23 Jan 2005 05:35:54 -0800 |
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| Hi Jannick:
Now, I still have this doubt about the projections p, q: by definition,
for any x in A, we get tht p(x):= x + M. Now, for p to be an
A-homomorphism, it has to be that for any a in A we have p(ax) = ap(x)
<==> (ax) + M = a(x + M), from which we get that, apparently, the
A-module structure of A/M MUST be so that a(x + M):= (ax) + M....but
does it really must be so?! And likewise for q, of course...This is
weird to me since it seems that from the given data it follows the very
definition of A-module of A/M, and I'm not sure that it must be that.
I also have some trouble with the part "For any element x in M it holds
p(x)= 0, hence f(p(x))=0. This implies q(x)=0..." . Why is this true? I
know f is an A-isom. and thus f(p(x)) =0; nevertheless I can't see how
q joins the game...
Finally, what's the relevance of pointing out that f(p(1)) is a unit in
A/M? A/M is, for this question's sake, only a module, so its rings
stuff is irrelevant for us, isn't it? Furthermore, since f is ONLY an
A-homomorphism, couldn't it be that the unit 1 of A (or any other unit)
is NOT mapped to a unit in the ring A/M by the MODULE homomorphism p?
And even it p(1) is a unit in the ring A/M, why should it be that
f(p(1)) is a unit in A/N?
Thanx very much for your answers!
Best Regards
Tonico
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| | From: | Jannick Asmus | | Subject: | Re: Quotient modules question | | Date: | Sun, 23 Jan 2005 14:50:28 +0100 |
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| On 1/23/2005 2:35 PM, Tonico wrote:
> Now, I still have this doubt about the projections p, q: by definition,
> for any x in A, we get tht p(x):= x + M. Now, for p to be an
> A-homomorphism, it has to be that for any a in A we have p(ax) = ap(x)
> <==> (ax) + M = a(x + M), from which we get that, apparently, the
> A-module structure of A/M MUST be so that a(x + M):= (ax) + M.
It is like this by definition. If nothing else is said, then A/M carries
this A-module structure. This makes it fit into the world of commutative
algebra 'in a natural way'.
....but
> does it really must be so?! And likewise for q, of course...This is
> weird to me since it seems that from the given data it follows the very
> definition of A-module of A/M, and I'm not sure that it must be that.
>
> I also have some trouble with the part "For any element x in M it holds
> p(x)= 0, hence f(p(x))=0. This implies q(x)=0..." . Why is this true?
Use that f(p(1)) is a unit in A/N.
> nevertheless I can't see how
> q joins the game...
See above.
> Finally, what's the relevance of pointing out that f(p(1)) is a unit in
> A/M? A/M is, for this question's sake, only a module, so its rings
> stuff is irrelevant for us, isn't it?
It is ok to say that an element of A/M is a unit, since A/M *is* a ring.
> Furthermore, since f is ONLY an
> A-homomorphism, couldn't it be that the unit 1 of A (or any other unit)
> is NOT mapped to a unit in the ring A/M by the MODULE homomorphism p?
No, this cannot be.
> And even it p(1) is a unit in the ring A/M, why should it be that
> f(p(1)) is a unit in A/N?
This needs a proof. Just try it.
> Thanx very much for your answers!
> Best Regards
> Tonico
>
J.
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| | From: | Jannick Asmus | | Subject: | Re: Quotient modules question | | Date: | Sun, 23 Jan 2005 13:57:18 +0100 |
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| On 1/23/2005 12:17 PM, Tonico wrote:
> Suppose A is a (commutative,
> unitarian) ring, M, N ideals of A, and we look at all of them as
> A-modules. If the quotient A-modules A/N, A/M are A-isomorphic (i.e.,
> as A-modules), then M = N.
> I still have the strong feeling that this isn't true, but can't come up
> with a counterexample.
Tonico, the assertion is true.
Sketch of proof:
Let f: A/M -> A/N be an A-module isomorphism and denote the canonical
projections A -> A/M and A -> A/N by p and q, resp., which are A-module
homomorphisms.
f(p(1)) is a unit in A/N.
For any element x in M it holds p(x)= 0, hence f(p(x))=0. This implies
q(x)=0, i.e. x e N. Thus M is contained in N.
By symmetry you get the desired result.
My first shot hasn't got a close look on the module structures. So it
led to the wrong answer.
J.
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| | From: | Jannick Asmus | | Subject: | Re: Quotient modules question | | Date: | Sun, 23 Jan 2005 12:38:57 +0100 |
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| On 1/23/2005 12:17 PM, Tonico wrote:
> Suppose A is a (commutative,
> unitarian) ring, M, N ideals of A, and we look at all of them as
> A-modules. If the quotient A-modules A/N, A/M are A-isomorphic (i.e.,
> as A-modules), then M = N.
> I still have the strong feeling that this isn't true, but can't come up
> with a counterexample.
Your feeling is right: the assertion is wrong.
Try to construct a counterexample in A = Z x Z (Z denotes the ring of
integers) with the ideal M generated by (1,0) and N ...
Cheers,
J.
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| | From: | Jannick Asmus | | Subject: | Re: Quotient modules question | | Date: | Sun, 23 Jan 2005 13:05:35 +0100 |
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| On 1/23/2005 12:38 PM, Jannick Asmus wrote:
> Your feeling is right: the assertion is wrong.
>
> Try to construct a counterexample in A = Z x Z (Z denotes the ring of
> integers) with the ideal M generated by (1,0) and N ...
I just crosschecked. This example does _not_ turn out to be a
counterexample.
Let M be as above and N generated by (0,1). Then the quotients are both
Z, but _not_ isomorhic as A-modules.
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