Solutions of Practice Test 3 MA407H
1. Let R =
a b
c d
a 0
| a, b, c, d ∈ Z and let S =
|a∈Z .
0 a
(a) Prove or disprove that S is a subring of R.
(b) Prove or disprove that S is an ideal of R.
0
a 0
a 0
Solution: (a) Let
and
be arbitrary elements of S. Then
0 a
0 a0
and
a 0
0 a
−
a 0
0 a
a0 0
0 a0
=
a − a0
0
0
a − a0
∈S
0
aa0 0
a 0
∈ S,
=
·
0 aa0
0 a0
so S is a subring.
ra rb
r 0
a b
6∈ S, S is not an ideal of R.
=
·
(b) Since
rc rd
0 r
c d
his is true, since (a−1 )p = (ap )−1 = a−1 .
2. (a) Show that the characteristic of Z[i]/ha + bii divides a2 + b2 .
(b) Give the characteristic of Z[i]/h2 − ii.
Solution:
(a)Since Z[i]/ha + bii is a ring with unity, its characteristic is the order of
1 under addition. Let I = ha + bii Then a + I = −bi + I, so after taking
squares we get that a2 + I = −b2 + I, which implies that a2 + b2 ∈ I. The
characteristic of a ring is the additive order of its unity, which is 1 + I here.
Thus (a2 + b2 )(1 + I) = 0 + I in Z[i]/ha + bii. So the additive order of 1
divides a2 + b2 , which was the claim.
(b) since 22 + 11 = 5, by part (a) the characteristic divides 5, so it is 1 or
5. If it were 1, then 1 + h2 − ii = 0 + h2 − ii, i.e. 1 ∈ h2 − ii. Thus there
exists a + bi ∈ Z[i] such that (a + bi)(2 − i) = 1. The inverse of 2 − i is
1
2/5 + i/5, but it has rational coefficients, so it is not a Gaussian integer.
Thus 1 6∈ h2 − ii. This proves that the characteristic of Z[i]/h2 − ii is 5.
3.
(a)(15 points) Prove that Z3 [x]/hx2 + 1i is a field.
(b)(15 points) Prove that Z3 [x]/hx2 + x + 1i is not a field.
Solution:
(a) Let I = hx2 + 1i ideal in Z3 [x], and ax + b + I ∈ Z3 [x]/I for some
a, b, ∈ Z3 . Then if a2 + b2 6= 0 then
(ax + b)−1 =
a − bx
.
a2 + b2
We need to show that a2 + b2 = 0, i.e. a2 = −b2 , implies that a = b = 0.
This is true, since otherwise, if b 6= 0 then (a/b)2 mod 4 = −1, but we can
check that none of the elements of Z3 has a square which is -1. Thus b is
zero, which implies that a = b = 0. Thus Z3 [x]/hx2 + 1i is a field.
(b) Since 1 is a root of x2 +x+1 modulo 3, we have that x2 +x+1 ≡ (x−1)2
mod 3, thus (x − 1) + hx2 + x + 1i is a zero divisor in Z3 [x]/hx2 + x + 1i,
and it is not a field.
4. Let F be a field. Find all maximal ideals in the ring F ⊕ F .
Solution:
Since the only ideals of a field are F and {0}, the only maximal ideals of
F ⊕ F are {0} ⊕ F and F ⊕ {0}.
5. In Z, let A = h2i and B = h8i. Show that the group A/B is isomorphic
to Z4 , but that the ring A/B is not ring-isomorphic to the ring Z4 .
Solution:
Let φ : A → Z4 given by φ(2k) = k mod 4. Then φ is a group homomorphism since
φ(2k + 2l) = φ(2(k + l)) = (k + l)
2
mod 4 = φ(2k) + φ(2l).
(Note that φ is not a ring homomorphism, since e.g. φ(2 · 2) = 2 mod 4
but φ(2)φ(2) = 1 mod 4.) Also, φ is surjective and ker φ = B. Thus A/B
is group isomorphic to Z4 .
However, since Z4 has a unity element, but A/B has none (because only
0 is idempotent in A/B: 2 · 2 = 4 mod 8, 4 · 4 = 0 mod 8 and 6 · 6 = 4
mod 8), A/B is not ring-isomorphic to Z4 .
6. In Z ⊕ Z let I = {(a, 0)|a ∈ Z}. Show that I is a prime ideal but not a
maximal ideal.
Solution:
The projection π : Z ⊕ Z → Z defined by π(a, b) = b is a surjective homomorphism with ker π = I. Thus Z ⊕ Z/I ∼
= Z. Since Z is an integral domain
but not a field, this shows that I is a prime ideal but not a maximal ideal.
Second Solution:
To see that I is not maximal, take J = {(a, 2b)|a, b ∈ Z Then I ⊂ J ⊂ Z⊕Z,
and both inclusion is proper.
To see that I is an integral domain, suppose that (a, b)(c, d) ∈ I. But
(a, b)(c, d) = (ac, bd) ∈ I, so bd =. This implies that b = 0 or d = 0, so
either (a, b) ∈ I or (c, d) ∈ I.
3