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A question about submodules... |
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| Jan14-13, 05:05 PM | #1 |
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A question about submodules...
My textbook says that...
If M is a left R-module, then a submodule N of M...is an additive subgoup N of M closed under scalar multiplication: [tex]rn \in N[/tex] whenever [tex]n \in N[/tex] and [tex]r \in R[/tex]. So if we want to prove that something is a submodule, we need to show that... 1) It closed under scalar multiplication 2) The additive idenitity is in N 3) N is closed under additition 4) If x is in N, then so is its inverse Right? But, in the link that I attached, it only shows 1) and 3), right? Can anybody tell me why? Is the proof still considered complete? Thanks in advance |
| Jan14-13, 05:57 PM | #2 |
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Isn't it as simple as: If (1) holds, set r = 0 to get (2) and r = -1 to get (4)?
(OK, you might want to show that 0n is the additive identity for any n and that -n is the additive inverse of any n). |
| Jan14-13, 06:05 PM | #3 |
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| Jan15-13, 09:33 PM | #4 |
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Recognitions:
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A question about submodules...
1 does not imply 2, unless the subset considered is non empty. i.e. 1 implies that IF the subset contains anything, then it also contains 0.
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