By Thomas C. Craven
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Extra info for Abstract algebra [Lecture notes]
Since 1 = 0 in R/M , we have M = R. Assume there is an ideal I such that M ⊆ I ⊆ R. If I = M , let a ∈ I, a ∈ / M . Then a + M has an inverse u + M in R/M , so au + M = 1 + M . In particular, au = 1 + m for some m ∈ M . Since m ∈ M ⊆ I, we have 1 = au − m ∈ I and so I = R. Therefore M is maximal. 16. Every maximal ideal is prime. Exercises, pages 157–159. 5. In Z6 , the maximal ideals are (2) and (3). 17. The inverse image of a prime ideal is prime. Let f : R → S be a homomorphism of rings and let Q be a prime ideal in S.
The cancellation results can be used to provide other ways to define a group (see exercises 37–39 on page 181). Facts about groups: (1) The identity element is unique. ] (2) Cancellation laws hold on both sides. ] (3) Inverses are unique. [If x and y are both inverses for g, then x = x(gy) = (xg)y = y. ] (4) (ab)−1 = b−1 a−1 and (a−1 )−1 = a. ] (5) am an = am+n , (am )n = amn for all m, n ∈ Z. We now define a0 = e, the identity element of the group. ] Sometimes it happens that ak = e for some k > 0.
Cn and denote it by (c1 , c2 , . . , cn ). Comments. 1. If R does not have an identity, there is a complication in the definition since one wants the elements ci ∈ I (see exercise 31, p. 143). 2. If R is not commutative, one needs multiplication on both sides in the definition of I. 3. Ideals with finitely many generators are called finitely generated ideals. One has to work a bit to find ideals which are not finitely generated and we will avoid them in this course. One example is the ideal generated by all the indeterminates in the polynomial ring R[x1 , x2 , x3 , .