# rev2 - Math 5125 Friday October 28 Second Test Review...

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Math 5125 Friday, October 28 Second Test Review Sample Second Test no. 3. Answer All Problems. Please Give Explanations For Your Answers. 1. Let R be a commutative ring which has nonprincipal ideals. Prove that R has a max- imal nonprincipal ideal (that is, an ideal I of R such that I is not principal, but if I J ± R , then J is principal). (16 points) 2. Prove that 2, 3 ± 13 are irreducible in Z [ 13 ] . Also prove that 2 is not prime in Z [ 13 ] . (17 points) 3. Let R be a UFD and let P be a nonzero principal ideal of R . (a) Prove that there are only ﬁnitely many principal ideals of R containing P . (b) Is it true that there are only ﬁnitely many principal ideals of R / P ? (Prove or give counter example.) (17 points) Solution to Problem 2 on Sample Test no. 1 First note that S - 1 M ± S - 1 R . Obviously 0 S - 1 M . If m / s , n / t S - 1 M , then m / s + n / t = ( mt + ns ) / ( st ) S - 1 M . Also if r / u S - 1 R , then ( m / s )( r / u ) = ( mr ) / ( su ) S - 1 M . If S M 6 = /0, choose s S M . Then s / s = 1 S - 1 M , which shows that S - 1 M = S - 1 R and hence S - 1 M is not a maximal ideal. Now suppose S M = /0. If S - 1 M = S - 1 R , then 1 / 1 = m / s for some m M and s S . This tells us that ( m - s ) t = 0 for some t S and we deduce that st M which contradicts S M = /0. Therefore S - 1 M 6 = S - 1 R . Finally suppose S - 1 M J ± S - 1 R and J 6 = S - 1 R . Let θ : R S - 1 R denote the natural homomorphism r 7→ r / 1. Then M θ - 1 J ± R and θ - 1 J 6 = R . Therefore θ - 1 J = M by maximality of M . If x J , then x = y / s for some s S and y R , and y / 1 J θ ( R ) . Therefore y θ - 1 J and we deduce that y M . Then x = y / s S - 1 M , consequently J = S - 1 M and we conclude that

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rev2 - Math 5125 Friday October 28 Second Test Review...

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