By Thomas C. Craven

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**Example text**

An ideal M in R is maximal if M = R and whenever I is an ideal such that M ⊆ I ⊆ R, then M = I or M = R. 9 The same definition is used in noncommutative rings with 1. The examples (1), (2), and (4) above were maximal. And we saw that in each case, the quotient ring was a field. 15. Let M be an ideal in R. M is a maximal ideal iff R/M is a field. Proof. ( =⇒ ) Assume M is maximal. 9). We have R/M = 0 since M = R. Therefore 0 = 1 in R/M . Finally we check for inverses. Let a + M be a nonzero element of R/M .

But then gn = n1 g ∈ N g, so gN ⊆ N g. Replacing g by g −1 gives the other inclusion, so gN = N g. Why do we expect that normality is the condition we need to define the operation on cosets? For one thing, it makes things look closer to abelian. For another, we expect, from ring theory, that we should get quotient groups as sets of cosets, and they should be images of homomorphisms where the kernels are the subgroups we are factoring out. But the kernel of a homomorphism must be a normal subgroup: To see this, we first define kernel, just as we did for rings: for a group homomorphism f : G → H, ker f = { g ∈ G | f (g) = eH }.

Conversely, assume the set I of all nonunits is an ideal. Then it is clearly the unique maximal ideal of R since no other element can be in a maximal ideal other than a nonunit. Example: { m n ∈ Q | n is odd }. (2) is the unique maximal ideal. Math 412 First Midterm Oct. 6, 1998 Professor: Tom Craven Name Instructions: write your answers clearly and completely. The number in parentheses is the number of points the problem is worth. (20) 1. a. Complete the definition: An integer p is prime if b.