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 | | From: | Michael Barr | | Subject: | Quotient modules question | | Date: | 23 Jan 2005 17:43:57 -0800 |
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 | I may be repeating a lot of what has been said, but here is how I see
it. If f:A/M --> A/N is an A-isomorphism, then let x in N. f([x1]) =
xf([1]) = 0 where [z] is the residue class containing z. But then [x]
= 0 which is true iff x in M. Hence N is included in M and the
reverse inclusion follows by the reverse ismorphism
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| | From: | Jannick Asmus | | Subject: | Re: Quotient modules question | | Date: | Mon, 24 Jan 2005 02:54:05 +0100 |
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| On 1/24/2005 2:43 AM, Michael Barr wrote:
> I may be repeating a lot of what has been said, but here is how I see
> it. If f:A/M --> A/N is an A-isomorphism, then let x in N. f([x1]) =
> xf([1]) = 0 where [z] is the residue class containing z. But then [x]
> = 0 which is true iff x in M. Hence N is included in M and the
> reverse inclusion follows by the reverse ismorphism
It's done.
J.
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