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

School of Mathematics and Statistics
MT4517 Rings & Fields
Solutions 3
Solution to Exercise 3.1. Recall that a subset A of a ring R is a subring if and only if a − b ∈ A
and a ∗ b ∈ A for all a, b ∈ A. Also recall that an ideal of a ring is a subring. Therefore: if a
subset of a ring is not a subring, it cannot be an ideal.
(i) N is not a subring (and hence not an ideal) in Z, as 1 ∈ N and 2 ∈ N while −1 = 1 − 2 /∈ N.
(ii) The set of all polynomials with integer constant can be characterised as
Let x ∗ q 1 + n 1 , x ∗ q 2 + n 2 ∈ A. Then
A = {x ∗ q + n : q ∈ Q[x], n ∈ Z}.
(x ∗ q 1 + n 1 ) − (x ∗ q 2 + n 2 ) = x ∗ (q 1 − q 2 ) + (n 1 − n 2 )
and
(x ∗ q 1 + n 1 ) ∗ (x ∗ q 2 + n 2 ) = (n 2 q 1 + n 1 q 2 + x ∗ q 1 ∗ q 2 )x + (n 1 ∗ n 2 ).
Since n 1 − n 2 ∈ Z and n 1 n 2 ∈ Z, we have that A is a subring in Q[x]. It is straightforward
to prove that A is not an ideal in Q[x] by taking the product of any element and any
number t ∈ Q \ Z.
(iii) By definition, the integers divisible by 3 are the elements of the principal ideal (and subring)
(3) in Z.
(iv) A = {q ∈ Q[x] : deg(q) ≥ 6} is a not a subring in Q[x] since x 6 − x 6 = 0 /∈ Q[x].
(v) {75a + 30b : a, b ∈ Z} = (30, 75) is an ideal (and so) subring in Z.
(vi) Let A be all the zero divisors of Z/(14). The elements of A are precisely those among
0, 1, . . . , 13 which are not comprime with 14. Hence A = {0, 2, 4, 6, 7, 8, 10, 12}. Consider
A as a subset of Z/(16). Then A is not a subring in Z/(16) as 7 − 6 = 1 /∈ A.
Solution to Exercise 3.2. Let A, B, C ⊆ S be arbitrary. It might be convenient to visualise these
sets using a Venn diagrams.
A1. It is possible to prove that A + (B + C) = (A + B) + C by direct computation.

A2. The zero of R is the empty set ∅ since
A + ∅ = (A ∪ ∅) \ (A ∩ ∅) = A \ ∅ = A
and
for all A ⊆ S.
∅ + A = (∅ ∪ A) \ (∅ ∩ A) = A \ ∅ = A
A3. Let A ⊆ S. Then A is the additive inverse for A:
for all A ⊆ S.
A + A = (A ∪ A) \ (A ∩ A) = A \ A = ∅
A4. A + B = B + A since A ∪ B = B ∪ A and A ∩ B = B ∩ A for all A, B ⊆ S.
M1. (A ∩ B) ∩ C = A ∩ (B ∩ C) holds by definition.
It is routine to check that D and M4 hold and so R is a commutative ring. The identity
element of R is the entire set S:
A ∗ S = A ∩ S = A
for all A ⊆ S.
If A ∈ R is a unit, then S = A ∗ A −1 = A ∩ A −1 ⊆ A. Hence A = S and so the only unit in R
is its one S.
With A + B = A ∪ B we do not get a ring. The axiom A3 fails: if A ∈ R, there exists B such
that A + B = A ∪ B = ∅ if and only if A = ∅.
Let A ∈ R be arbitrary. We will prove that the ideal (A) generated by A equals
I = { B ∈ R : B ⊆ A }.
Let B ⊆ S. Then B ∗ A = B ∩ A ∈ I and so (A) contains I. But I is a subring of R as:
(a) if B, C ∈ I, then B + B = ∅ and so −B = B. Hence B − C = B + C = (B ∪ C) \ (B ∩ C) ⊆
B ∪ C ⊆ A ∪ A = A yielding B − C ∈ I;
(b) if B, C ∈ I, then B ∗ C = B ∩ C ⊆ A and so B ∗ C ∈ I.
Hence since A ∈ I it follows that (A) ⊆ I and so (A) = I.
Solution to Exercise 3.3. We will prove directly that
I = {a 2 x 2 + · · · + a n x n : n ≥ 2} = { x 2 · f : f ∈ R[x] }
is an ideal in R[x]. Let f, g ∈ I. Then f = x 2 ∗ f 1 and g = x 2 ∗ g 1 for some f 1 , g 1 ∈ R[x]. Hence
f − g = x 2 ∗ (f 1 − g 1 ) ∈ I. If h ∈ I is arbitrary, then f ∗ h = h ∗ f = x 2 ∗ f 1 ∗ h ∈ I. Therefore I
is an ideal in R[x].
Solution to Exercise 3.4. The set of polynomials from the exercise can be given as
I = { f ∈ R[x] : f(1) = 0 }.
So, if f, g ∈ I, then (f − g)(1) = f(1) − g(1) = 0 and so f − g ∈ I. Also, f ∗ h(1) = h ∗ f(1) =
h(1) ∗ f(1) = h(1) ∗ 0 = 0 and so f ∗ h ∈ I for any h ∈ R[x]. Hence I is an ideal.
80

Solution to Exercise 3.5. It is straightforward to verify that R = { r + s √ 2 : r, s ∈ Q } is a
commutative ring with respect to the operations + and ∗ in R. It remains to prove that every
non-zero element of R is a unit. Let r + s √ 2 ∈ R. Then
1
r + s √ 2 = r − s √ 2
(r + s √ 2)(r − s √ 2) = r − s√ 2
r 2 − 2s 2 ∈ R
is the multiplicative inverse for r + s √ 2 in R. Thus R is a field.
Let Σ be the smallest subfield of R containing √ 2. We will prove that R = Σ. First note that,
since 1 ∈ Σ, n ∈ Σ for all n ∈ Z. Since Σ is a field, it follows immediately that Q ⊆ Σ. Since Σ
contains √ 2 and is closed under multplication and summation, we obtain that R = { r + s √ 2 :
r, s ∈ Q } ⊆ Σ. So, R ⊆ Σ. On the other hand, R is a subfield of R that contains √ 2. Hence, by
the choice of Σ, we have Σ ⊆ R. Therefore R = Σ, as required.
Solution to Exercise 3.6. Let I be the ideal in R generated by √ 2. Then
1 = √ 2 ·
1
√
2
∈ I
and so I = R.
Solution to Exercise 3.7. Let R be a commutative ring with identity whose only ideals are {0}
and R. In order to prove that R is a ring it suffices to show that every non-zero element in R is
a unit. So let r ∈ R \ {0}. Then the ideal generated by (r) = R. Hence there exists t ∈ R such
that rt = 1 and r is a unit.
Now, let R = Z/(6). Then R is a commutative ring with identity. All the non-invertible
elements of R are A = {0, 2, 3, 4}. Since 3 − 2 = 1 /∈ A, it follows that A is not an ideal in R.
Solution to Exercise 3.8. Obviously if A, B ∈ R, then A − B ∈ R. We are left to prove that
A ∗ B ∈ R for all A, B ∈ R. So, if
( ) ( )
a b
c d
A = and B = ,
2b a
2d c
then
( )
ac + 2bd ad + bc
A ∗ B =
∈ R.
2(ad + bc) ac + 2bd
Thus R is a subring of M 2 (R).
Let F be those matrices from R whose entries are from Q. It is straightforward to check that
F is a commutative subring of M 2 (R). Let A ∈ F \ {0}:
( ) a b
A = .
2b a
If det(A) = a 2 − 2b 2 = 0, then √ 2 ∈ Q, a contradiction. Hence det(A) ≠ 0 for all A ∈ F \ {0}
and so A −1 in M 2 (R). Since the entries of A −1 are obtained from those of A by adding and
multiplying it follows that A −1 ∈ F . Hence every non-zero element of F is a unit and F is a
field.
Let Σ be the set of matrices of the type
( ) a b
A =
2b a
81

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