>Suppose x is not in m for some maximal ideal m. Then x and m generate the unit ideal (1)... Why? I don't see how this is forced.

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Anonymous Mon Dec 31 20:23:30 2012 No.5400624 [Reply] [Original]

>Suppose x is not in m for some maximal ideal m. Then x and m generate the unit ideal (1)...

Why? I don't see how this is forced.

>>	Anonymous Mon Dec 31 20:24:00 2012 No.5400626 >>5400624 The setting is a commutative ring with identity of course*

>>	Anonymous Mon Dec 31 20:29:44 2012 No.5400634 m is a proper subset of the ideal (x)+m m is maximal, therefore (x)+m is the ring the ring equals the ideal (1)

>>	Anonymous Mon Dec 31 20:31:59 2012 No.5400637 >>5400634 Thanks, wasn't noticing that (x)+(m) is a proper superset of (m) for some reason. This commutative algebra makes me feel like an idiot.

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