Math 330, Abstract Algebra I Solutions to Homework 8 Problems ...

Math 330, Abstract Algebra I
Solutions to Homework 8 Problems
that if b k ∈ A, so is (−b) k since for even k, (−b) k = b k and for odd k, (−b) k = −b k , and A
is closed under additive inverses. Now, consider (a − b) h+k :
(a − b) h+k =
=
∑h+k
( ) h + k
a h+k−i (−b) i
i
i=0
k∑
i=0
( h + k
i
)
a h+k−i (−b) i +
∑h+k
i=k+1
( ) h + k
a h+k−i (−b) i
i
Notice that every term in the first summation is the product of a h with some other ring
element and every term in the second summation is the product of (−b) k with some other
ring element. Since A is an ideal of R and we know a h ∈ A and (−b) k ∈ A, we know that
every term in each summation is contained in A, hence the sum itself is contained in A.
That is, (a − b) h+k ∈ A, hence a − b ∈ N(A).
Now, for a ∈ N(A) and some arbitrary element r ∈ R, we know that there is some
positive integer h with a h ∈ A. Since R is a commutative ring, we can write (ar) h = a h r h =
r h a h = (ra) h , which we know is an element of A, since a h ∈ A and A is an ideal. So, by the
ideal test N(A) is an ideal of R.
Chapter 14. Problem 56. Let R be a commutative ring with unity and let I be a proper
ideal with the property that every element of R that is not in I is a unit of R. Prove that I
is the unique maximal ideal of R.
Solution: Let R and I be given as above. Suppose A is an ideal of R with I ⊂ A ⊆ R (so
that I is properly contained in A). Then there is some element a ∈ A, a ∉ I. Since a ∉ I, a
must be a unit. So there is some a −1 ∈ R, and the product of this ring element with a must
be contained in the ideal A: a · a −1 = 1 ∈ A. Since 1 is an element of A, A = R, hence I is
maximal.
Now, suppose B ≠ I is a maximal ideal of R. As B is maximal and B ≠ I, B is not a
subset of I (for then B I R implies that B is not maximal). Thus there must be an
element b ∈ B so that b ∉ I, hence b is a unit. By the above argument, we see that B must
in fact be all of R, as B contains 1. Hence, B is not actually a maximal ideal, and I is, in
fact, the unique maximal ideal of R.