User:Michellepoliseno

HW 5: 7.4.37 <= Let R be a commutative ring with 1 and the set of all non-units be M. We want to show that M is a Unique Maximal in R. Assume that M is not maximal. Then $$ \exists \ $$ I ideal in R such that M $$ \subset \ $$ I $$ \subset \ $$ R. And M $$ \ne \ $$ I $$ \ne \ $$ R. We know that 1 does not exist in M, because 1 is a unit. So if M is not equal to I, then $$ \exists \ $$ u $$ \in \ $$ I, where u is a unit. Then 1 $$ \in \ $$ I. But that implies I=R. But if I=R, then I is not a maximal idea. If I is not maximal, then that implies that M is maximal. Now assume that M is not unique. Then $$ \exists \ $$ M' such that M' is a maximal ideal in R. Then M' $$ \subset \ $$ R and M $$ \ne \ $$ R. So 1 $$ \notin \ $$ M'. Thus M' is the set of all non-units. That implies that M'=M. Therefore M is a unique maximal ideal in R.