MATH 251 Homework 7 Solutions 1. Let R be a ring and {S j | j ∈ J ...

MATH 251 Homework 7 Solutions
1. Let R be a ring and {S j | j ∈ J} a collection of subrings (resp. ideals) of R. Prove that ∩ j∈J S j is a
subring (resp. ideal) of R.
Proof : Put S = ∩ j∈J S j .
Suppose that S j is a subring of R for all j ∈ J. Pick a, b ∈ S. Then a, b ∈ S j for all j ∈ J. Since S j is a
subring of R, we have that a − b, ab ∈ S j for all j ∈ J. Hence a − b, ab ∈ S. So S is a subring of R.
Suppose next that S j is an ideal for all j ∈ J. Pick a ∈ S and r ∈ R. Then a ∈ S j for all j ∈ J. Since S j
is an ideal of R, we have that ra, ar ∈ S j for all j ∈ J. Hence ra, ar ∈ S. So S is an ideal of R. ✷
2. Which of the following subsets of Q are subrings of (Q, +, ·)? Prove your answer!
(a) The set of all rational numbers with odd denominators (when written in lowest terms).
(b) The set of all rational numbers with even denominators (when written in lowest terms).
(c) The set of all squares of rational numbers.
(d) The set of all rational numbers with odd numerators (when written in lowest terms).
(e) The set of all rational numbers with even numerators (when written in lowest terms).
Solution : (a) SUBRING. Pick x, y ∈ S. Then x = a s and b = b where a, b, s, t ∈ Z, (a, s) = (b, t) = 1
t
and s, t are odd. Then
at − bs
x − y = and xy = ab
st
st
Since st is odd, we get that x − y, xy ∈ S (although it is possible that (at − bs, st) ≠ 1 ≠ (ab, st)). So S
is subring of Q.
(b) NOT A SUBRING : 1 6 ∈ S but 1 6 + 1 6 = 1 3
/∈ S. So S is not a subring of Q.
(c) NOT A SUBRING : 1 ∈ S but 1 + 1 = 2 /∈ S since ± √ 2 /∈ Q. So S is not a subring of Q.
(d) NOT A SUBRING : 1 3 ∈ S but 1 3 + 1 3 = 2 3
/∈ S. So S is not a subring of Q.
(e) SUBRING : Pick x, y ∈ S. Then x = a s and b = b where a, b, s, t ∈ Z, (a, s) = (b, t) = 1 and a, b are
t
even. Since (a, s) = (b, t) = 1 we have that s, t are odd. Then
x − y =
at − bs
st
and
xy = ab
st
Note that at − bs, ab are even and st is odd. So x − y, xy ∈ S (although it is possible that (at − bs, st) ≠
1 ≠ (ab, st)). So S is subring of Q. ✷
1

3. The center of a ring R is the set {r ∈ R | ar = ra for all a ∈ R}. Find the center of the quaternions
H = {a + b i + c j + d k | a, b, c, d ∈ R}.
Solution : Let a + b i + c j + d k ∈ Z(H). Then
−b + a i + d j + (−c) k = (a + b i + c j + d k) i = i (a + b i + c j + d k) = −b + a i + (−d) j + c k
Hence d = −d and −c = c. So c = d = 0. Also
a j + b k = (a + b i) j = j (a + b i) = a j + (−b) k
So b = −b. Hence b = 0. So Z(H) ⊆ R. Clearly, R ⊆ Z(H). Hence Z(H) = R.
✷
4. Let R be an integral domain.
(a) It is given that R[x] is a commutative ring with 1. Prove that R[x] is an integral domain.
(b) Put I = {x 2 f(x) | f(x) ∈ R[x]}.
(i) Prove that I ✂ R[x].
(ii) Prove that R[x]/I is not an integral domain.
Proof : (a) Suppose that R[x] has zero divisors. Then f(x)g(x) = 0 for some f(x), g(x) ∈ R[x] \ {0}. So
f(x) = a 0 + a 1 x + · · · + a m x m and g(x) = b 0 + b 1 x + · · · + b n x n where m, n ≥ 0, a 0 , . . . , a m , b 0 , . . . , b n ∈ R
and a m ≠ 0 ≠ b n . Put a i = 0 if i > m and b j = 0 if j > n. Then
( i∑
)
a k b i−k
k=0
0 = f(x)g(x) = (a 0 + a 1 x + · · · + a m x m ) (b 0 + b 1 x + · · · + b n x n ) =
Hence a m b n = 0 (the coefficient of x m+n ). Since R has no zero divisors (R is an integral domain), we get
that a m = 0 or b n = 0, a contradiction. Hence R[x] has no zero divisors. So R[x] is an integral domain.
(b) (i) Note that I = {g(x)x 2 | g(x) ∈ R} = (x 2 ). So I ✂ R.
(ii) Note that x ≠ 0 in R[x]/I (since x /∈ (x 2 )) but x · x = x · x = x 2 = 0 in R[x]/I. So x is a zero divisor
of R[x]/I.
✷
m+n
∑
i=0
x i
5. Which of the following subsets of Z[x] are ideals of (Z[x], +, ·) :
(a) The set of all polynomials whose constant term is a multiple of 3.
(b) The set of all polynomials whose coefficient of x 2 is a multiple of 3.
(c) Z[x 2 ] (so the set of polynomials with integral coefficients in which only even powers of x appear).
(d) The set of polynomials whose coefficients sum to zero.
(e) The set of polynomials f(x) with f ′ (0) = 0 where f ′ (x) is the usual derivative of f(x) with respect
to x.
Solution : (a) IDEAL : Note that the constant term of a polynomial p(x) is p(0). Pick f(x), g(x) ∈ S
and r(x) ∈ Z[x]. Then
f(0) − g(0) ≡ 0 mod 3 and r(0)f(0) ≡ 0 mod 3
So f(x) − g(x), r(x)f(x) ∈ S. Hence S is an ideal of Z[x].
2

(b) NOT AN IDEAL : 3x 2 + x, 3x 2 − x ∈ S but (3x 2 + x)(3x 2 − x) = 9x 4 − x 2 /∈ S. So S is not a
subring nor an ideal of Z[x].
(c) NOT AN IDEAL : x 2 ∈ Z[x 2 ] and x ∈ Z[x]. But x · x 2 = x 3 /∈ Z[x 2 ]. So Z[x 2 ] is not an ideal of
Z[x]. Note that Z[x 2 ] is a subring of Z[x].
(d) IDEAL : Let f(x), g(x) ∈ S and h(x) ∈ Z[x]. Note that the sum of the coefficients of a polynomial
p(x) is p(1). So f(1) = g(1) = 0. Then
f(1) − g(1) = 0 and h(1)f(1) = 0
So f(x) − g(x), h(x)f(x) ∈ S. Hence S is an ideal of Z[x].
(e) NOT AN IDEAL : x 2 + 1 ∈ S and x ∈ Z[x] but x · (x 2 + 1) = x 3 + x /∈ S since (x 3 + x) ′∣ ∣
x=0
=
(3x 2 + 1) ∣ ∣
x=0
= 1. So S is not an ideal of Z[x]. Note that S is a subring of Z[x]. ✷
6. Let R be a commutative ring with 1.
(a) Let I, J ✂ R with I + J = R. Prove that IJ = I ∩ J.
(b) Let I, J, K ✂ R. Prove that (I + J)K = IK + JK.
Proof : (a) Let x ∈ IJ. Then x = i 1 j 1 + · · · + i n j n where n > 0, i 1 , . . . , i n ∈ I and j 1 , . . . , j n ∈ J. Since
I, J ✂ R, we get that i k j k ∈ I (because i k ∈ I and j k ∈ R) and i k j k ∈ J (because i k ∈ R and j k ∈ J) for
k = 1, . . . , n. Since I ∩ J ✂ R, we get that x = i 1 j 1 + · · · + i n j n ∈ I ∩ J. So IJ ⊆ I ∩ J.
Since I + J = R, we have that 1 = x + y for some x ∈ I and some y ∈ J. Pick z ∈ I ∩ J. Then
z = z · 1 = z(x + y) = zx + zy = xz + zy. Note that x ∈ I, z ∈ J, z ∈ I and y ∈ J. So z ∈ IJ and
I ∩ J ⊆ IJ.
Hence IJ = I ∩ J.
(b) This property is true if R does not have 1.
Let x ∈ (I + J)K Then
x = (i 1 + j 1 )k 1 + · · · + (i n + j n )k n
where n > 0, i 1 , . . . , i n ∈ I, j 1 , . . . , j n ∈ J and k 1 , . . . , k n ∈ K. Then
Hence
i 1 k 1 + · · · + i n k n ∈ IK and j 1 k 1 + · · · j n k n ∈ JK
x = (i 1 + j 1 )k 1 + · · · + (i n + j n )k n = (i 1 k 1 + · · · + i n k n ) + (j 1 k 1 + · · · j n k n ) ∈ IK + JK
So (I + J)K ⊆ IK + JK.
Since I, J ⊆ I + J, we have that IK, JK ⊆ (I + J)K. Hence IK + JK ⊆ (I + J)K.
Thus IK + JK = (I + J)K.
✷
7. Let D be an integer such that D is not a perfect square and D ≡ 1 mod 4. Put R = Z
{
a + b 1 + √ }
D
: a, b ∈ Z .
2
[
1 + √ ]
D
=
2
3

(a) Prove that R is a ring with Z[ √ D] ⊂ R ⊂ Q( √ D). Note that these are proper inclusions!
(b) Let N be the usual norm on Q( √ D).
• Prove that N(α) ∈ Z for all α ∈ R.
• Find R × (the units of R) if D < 0.
Proof :
(a) Note that
a + b 1 + √ D
2
= a + b 2 + b 2√
D ∈ Q(
√
D) for all a, b ∈ Z
Moreover, if 1 3 = a + b 1+√ D
2
for some a, b ∈ Z then b = 0 and a = 1 3 , a contradiction. Hence R ⊂ Q(√ D).
Note that
a + b √ D = (a − b) + (2b) 1 + √ D
∈ R for all a, b ∈ Z
2
Moreover, if 1+√ D
2
= a + b √ D for some a, b ∈ Z then a = 1 2 = b, a contradiction. Hence Z[√ D] ⊂ R.
Let α, β ∈ R. Then α = a + b 1+√ D
2
and β = c + d 1+√ D
2
for some a, b, c, d ∈ Z. Hence
(
) (
)
αβ =
(
α − β =
a + b 1 + √ D
2
a + b 1 + √ D
2
) (
−
c + d 1 + √ D
2
c + d 1 + √ D
2
since D ≡ 1 mod 4. Hence R is a subring of Q( √ D).
(b) • Let a, b ∈ Z. Then
(
N a + b 1 + √ )
D
= N
2
since D ≡ 1 mod 4.
)
=
= (a − c) + (b − d) 1 + √ D
2
∈ R
(
ac + bd D − 1 )
(
1 + √ )
D
+ (ad + bc + bd)
∈ R
4
2
((
a + b )
+ b ) (
√
D = a + b 2 ( ) 2 b
− D = a
2 2
2) 2 + ab + b 2 1 − D ∈ Z
2
4
• Suppose that D < 0. For a, b ∈ Z, we have that
a+b 1 + √ (
D
∈ R × ⇐⇒ N a + b 1 + √ )
D
= ±1 ⇐⇒
2
2
(
a + b ) 2
−D b2
2 4 = ±1 ⇐⇒ a2 +ab+b 2 1 − D = 1
4
If b 2 = 0 then a 2 = 1 and so (a, b) ∈ {(−1, 0), (1, 0)}. So suppose that b 2 ≠ 0. Then |D|b 2 ≤ 4. Since
D ≡ 1 mod 4, this is only possible if D = −3 and b 2 = 1. If b = 1 then a 2 + a + 1 = 1 and so a ∈ {−1, 0};
if b = −1 then a 2 − a + 1 = 1 and so a ∈ {0, 1}.
⎧ {
⎪⎨
R × =
⎪⎩
1, −1, −1 + √ −3
2
, 1 + √ −3
2
, − 1 + √ −3
, 1 − √ }
−3
2 2
if D = −3
{1, −1} if D < −3
4

8. Find all ring homomorphisms ϕ : Q → Q.
Solution : Since ϕ is a ring homomorphism, we get that
So ϕ(1) is a solution of x 2 = x. Hence ϕ(1) ∈ {0, 1}.
ϕ(1) = ϕ(1 · 1) = ϕ(1) · ϕ(1)
Suppose first that ϕ(1) = 0. So {0} ̸= Ker ϕ ✂ Q. But Q has only two ideals since Q is a field. So
Ker ϕ = Q. Hence ϕ(x) = 0 for all x ∈ Q.
Suppose next that ϕ(1) = 1. Let n ∈ N 0 . Since ϕ is a ring homomorphism, we have that
ϕ(n) = ϕ(1 + 1 + · · · + 1) = ϕ(1) + ϕ(1) + · · · + ϕ(1) = n ϕ(1) = n
} {{ } } {{ }
n times
n times
Since ϕ is a group homomorphism from (Q, +) to (Q, +), we get that ϕ(−n) = −ϕ(n) = −n.
ϕ(0) = 0, we proved that ϕ(m) = m for all m ∈ Z.
Pick q ∈ Q. Then q = m n
So ϕ(q) = m n = q.
Since
for some m ∈ Z and some n ∈ N 0. Since ϕ is a ring homomorphism, we get that
(
m = ϕ(m) = ϕ n · m )
n
( m
)
= ϕ(n) ϕ = n ϕ(q)
n
We proved that there are only two possible ring homomorphisms from Q to Q :
ϕ 1 : Q → Q : x → 0 and ϕ 2 : Q → Q : x → x
One easily checks that ϕ 1 and ϕ 2 are indeed ring homomorphisms.
✷
5