School of Mathematics and Statistics MT4517 Rings & Fields ...

where a, b ∈ Z/(3). Since Z/(3) is a field, using a similar argument to that used above, we can
prove that Σ is a field. It contains |Z/(3)| · |Z/(3)| = 9 elements.
Solution to Exercise 3.9. Let R be a commutative ring with identity, let I be an ideal of R, and
let r 1 , r 2 , . . . , r n ∈ I. We will prove that (r 1 , r 2 , . . . , r n ) ⊆ I.
Since r i ∈ I for all i, we have that λ i r i ∈ I for all λ i ∈ R. But I is closed under addition and
so λ 1 r 1 + λ 2 r 2 + · · · + λ n r n ∈ I for all λ 1 , λ 2 , . . . , λ n ∈ R, as required.
Solution to Exercise 3.10. We have already seen that every field F is an integral domain. So it
remains to prove that every ideal in F is principal. But there are only two ideals in F : (0) = {0}
and (1) = F .
Solution to Exercise 3.11. It is routine to prove the first part of the exercise from the definition
of an ideal.
Now, let i ∈ I and j ∈ J. Then ij ∈ I and ij ∈ J and so ij ∈ I ∩ J. Since I ∩ J is an ideal, it
is closed under addition and so every sum of elements of the form ij where i ∈ I and j ∈ J is
an element from I ∩ J. Hence IJ ⊆ I ∩ J.
Let R = { x ∗ f : f ∈ R[x] } and I = J = R. Then I and J are ideals in R. But
IJ = { x 2 ∗ f : f ∈ R[x] } is a proper subset of the ideal I ∩ J = R.
In Z[x] if I = (x, 2) and J = (x, 3), then { ij : i ∈ I, j ∈ J } is not an an ideal as x 2 ∈ I and
6 ∈ J but x 2 + 6 ∉ { ij : i ∈ I, j ∈ J }.
Solution to Exercise 3.12. Let
I = ⋃ I n
n∈N
and let a, b ∈ I. Then there exists n ∈ N such that a, b ∈ I n and so a − b ∈ I n ⊆ I. Moreover, if
x ∈ R, then ax ∈ I n ⊆ I for some n. Hence I is an ideal in R.
Now, {2m : m ∈ Z} ∪ {3n : n ∈ Z} is not a subring of Z since 3 − 2 = 1 /∈ {2m : m ∈
Z} ∪ {3n : n ∈ Z}.
Solution to Exercise 3.13. By Exercise 3.12, the union I of the ideals I n is an ideal in R.
By assumption, it is finitely generated and so there exist r 1 , r 2 , . . . , r k ∈ I such that I =
(r 1 , r 2 , . . . , r k ). Hence there exists N ∈ N such that r 1 , r 2 , . . . , r k ∈ I N and so s(r 1 , r 2 , . . . , r k ) ⊆
I N . Therefore
(r 1 , r 2 , . . . , r k ) ⊆ I N ⊆ I N+1 ⊆ I = (r 1 , r 2 , . . . , r k )
and so I N = I N+1 , as required.
82