Project in Math
216
Modern Algebra II
Antabo, Irene Mae Y.
Caay, Mark L.
Pamplona, Jill Maegan B.
Rodesendo, John Daryl L.
Saromines, Chrisley Jade C.
Submitted to:
Jocelyn P. Vilela, Ph.D
Please provide a clear and concise proof in each of the following:
First exam:
1. Find the units, zero divisors and ideals of Z6 . Define units, zero divisors and ideals.
Definitions:
Given a ring R with 1R 6= 0 an element a ∈ R, with a 6= 0 is said to be a left (resp. right) invertible if
there exists b ∈ R (resp. c ∈ R ) such that ba = 1R (resp. ac = 1R ).
Elements b and c are called left(resp. right) inverse of a. An element a is said to be invertible or a unit
if is both left and right invertible.
A nonzero element a of a ring R is said to be left(resp. right) zero divisors if there exists nonzero b ∈ R
such that ab = 0 (resp. ba = 0).
An element a in a ring is said to be a zero divisor if it is both left and right zero divisor.
A subring M of a ring R is said to be
(i) left ideal if ∀r ∈ R and ∀a ∈ M , ra ∈ M .
(ii) right ideal if ∀r ∈ R and ∀a ∈ M , ar ∈ M .
(iii) ideal if it is both left and right ideal of R.
Solution:
Note that Z6 = 0, 1, 2, 3, 4, 5. The units of Z6 are 1 and 5. Since
1 · 1 = 1 ⇒ the multiplicative inverse of
1
5 · 5 = 1 ⇒ the multiplicative inverse of
2 is not a unit since
0·2=0
1·2=2
2·2=4
3·2=0
4·2=2
5·2=4
1
is
1
5
is
and
5
3 is not a unit since
0·3=0
1·3=3
2·3=0
3·3=3
4·3=0
5·3=3
4 is not a unit since
0·4=0
1·4=4
2·4=2
3·4=0
4·4=2
5·4=4
The zero divisors of Z6 are 2, 3 and 4 since
2 · 3 = 0 ⇒ 2 6= 0, 3 6= 0
3 · 4 = 0 ⇒ 3 6= 0, 4 6= 0
The elements of Z6 such as 1 and 5 are not zero divisors since there exists no nonzero elements that can
be multiplied to these no. such that the product will be zero or the product will be equal to 6, that is,
1·0=0
1·1=1
1·2=2
1·3=3
1·4=4
1·5=5
5·0=0
5·1=5
5·2=4
5·3=3
5·4=2
5·5=1
2
Note that,
0·2=0=2·0
1·2=2=2·1
2·2=4=2·2
3·2=0=2·3
4·2=2=2·4
5·2=4=2·5
0·3=0=3·0
1·3=3=3·1
2·3=0=3·2
3·3=3=3·3
4·3=0=3·4
5·3=3=3·5
0·4=0=4·0
1·4=4=4·1
2·4=2=4·2
3·4=0=4·3
4·4=4=4·4
5·4=2=4·5
Consider the sets I1 = {0} , I2 = {0, 3}, I3 = {0, 2, 4} and Z6 .
I1 = {0} is a subring since
0 + 0 = 0 ∈ {0} and0(0) = 0 ∈ {0}
I2 = {0, 3} is a subring since
0 + 0 = 0 ∈ {0, 3}
0 + 3 = 3 ∈ {0, 3}
3 + 3 = 6 = 0 ∈ {0, 3}
0(0) = 0 ∈ {0, 3}
0(3) = 0 ∈ {0, 3}
3(3) = 9 = 3 ∈ {0, 3}
I3 = {0, 2, 4} is a subring since
0 + 0 = 0 ∈ {0, 2, 4}
0 + 2 = 2 ∈ {0, 2, 4}
0 + 4 = 4 ∈ {0, 2, 4}
2 + 2 = 4 ∈ {0, 2, 4}
2 + 4 = 6 = 0 ∈ {0, 2, 4}
4 + 4 = 8 = 2 ∈ {0, 2, 4}
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0(0) = 0 ∈ {0, 2, 4}
0(2) = 0 ∈ {0, 2, 4}
0(4) = 0 ∈ {0, 2, 4}
2(2) = 4 ∈ {0, 2, 4}
2(4) = 8 = 2 ∈ {0, 2, 4}
4(4) = 16 = 4 ∈ {0, 2, 4}
Note that Z6 is a subring of itself. Hence, I1 , I2 , I3 and Z6 are ideals of Z6 .
2. Prove that Zn is an integral domain if and only if n is prime.
Proof:
Let Zn = {0, 1, ..., n − 1} be a ring.
(⇒) Suppose that Zn is an integral domain. Then Zn has no zero divisor.
To show: n is prime.
Assume that n is not prime. Then n = ab, ∃a, b ∈ Zn with a 6= 0 and b 6= 0. Now, ab = n implies ab = 0
since n ≡ 0(modn). Hence a and b are zero divisors. This is a contradiction to the assumption that Zn is an
integral domain. Hence, n is prime.
(⇐) Suppose n is prime and Zn has a zero divisor. Let 0 6= m ∈ Zn . Then the sequence 1m, 2m, ..., (n −
1)m contains atleast one term such that im ≡ 0(modn), ∃i ∈ Zn , 1 ≤ i, m < n.
Now, im ≡ 0(modn) ⇒ n|im. Since, n is prime then n|i or n|m. This is not possible since 1 ≤ i, m < n.
Hence, Zn has no zero divisor. Note that, Zn with 1R 6= 0 is commutative. Since ∀a, b ∈ Zn , ab = ba and
1R a = a1R = a. Therefore, Zn is an integral domain.
3. Let R be a commutative ring with identity and K be an ideal of R.
(a) Discuss and prove how R/K becomes a commutative ring.
Proof:
Let R be a commutative ring with identity and K be an ideal of R. Since K is an ideal of R, then K is
a subring of R which also implies that K is a subgroup of R. Since R is abelian group then K is normal in
R by Math 225. Hence, R/K is a well-defined abelian group.
Now, we defined R/K = {a + K : a ∈ R} and defined a binary operation + and · as
(a + K) + (b + K) = ((a + b) + K)
and
(a + K)(b + K) = (ab + K)
respectively ∀a, b ∈ R
Claim 1: R/K is closed under multiplication.
Let (a + K), (b + K) ∈ R/K. Now, (a + K)(b + K) = (ab + K). Since R is a ring and by Closure property,
ab ∈ R which implies ab + K ∈ R/K. Therefore R/K is closed under multiplication.
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Claim 2: R/K is commutative.
Note that,
(a + K)(b + K) = (ab + K)
= (ba + K), since R is commutative
= (b + K)(a + K)
Therefore, R/K is commutative.
Claim 3: R/K is associative.
Let a = a + K, b = b + K and c = c + K. Now,
(ab)c = [(a + K)(b + K)] · (c + K)
= [(ab) + K] · (c + K)
= (ab)c + K
= a(bc) + K, since associativity holds in R
= (a + K) [(bc) + K]
= (a + K) [(b + K)(c + K)]
= a(bc)
Therefore, R/K is associative under multiplication.
Claim 4: LDL and RDL holds.
By Claim 3,
RDL :
(a + b)c = [(a + K) + (b + K)] (c + K)
= [(a + b) + K] (c + K)
= [(a + b)c + K]
= [(ac + bc) + K] , since RDL holds in R
= [(ac + K) + (bc + K)]
= [(a + K)(c + K)] + [(b + K)(c + K)]
= ac + bc
LDL :
a(b + c) = (a + K) [(b + K) + (c + K)]
= (a + K) [(b + K) + K]
= [a(b + c) + K]
= [(ab + ac) + K] , since LDL holds in R
= [ab + K] + [ac + K]
= [(a + K)(b + K)] + [(a + K)(c + K)]
= ab + ac
By the definition of a ring, R/K is a commutative ring.
(b) Consider the ring epimorphism φ : R → R/K defined by φ(r) = r + K and let I be a principal ideal
of R containing K. Prove that φ(I) is a principal ideal of R/K
Proof:
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Let a, b ∈ φ(I) and x ∈ R/K. Then a = φ(a), ∃
a ∈ I and b = φ(b), ∃ b ∈ I and x = r + K, r ∈ R.
Since I is ideal, a − b ∈ I. So, φ(a − b) ∈ φ(I). But since φ is epi, φ(a − b) = φ(a) − φ(b) ∈ φ(I). So,
a − b ∈ φ(I). Also, r · a ∈ I. Since φ is epi,
φ(r · a) = φ(r)φ(a)
= (r + K)a
= xa ∈ φ(I)
and a · r ∈ I, since I is ideal. Thus,
φ(ar) = φ(a)φ(r)
= a(r + K) = ax ∈ φ(I).
Thus, φ(I) is an ideal of R/K.
Now, since I is principal ideal in R containing K, ∃a ∈ R such that I = (a).
Claim: φ(I) = (φ(a)).
Now, if a ∈ I, then φ(a) ∈ φ(I) since φ is epi. By closure property, (φ(a)) ⊆ φ(I). Let x ∈ φ(I). Then
x ∈ φ(I) ⇒ x = φ(y), y ∈ I
⇒ x = φ(y), y = aK
⇒ x = φ(aK ) = φ(a)K
⇒ x ∈ (φ(a))
Therefore, φ(I) = (φ(a)).
4. Definition: A nonzero ring is simple if its only ideals are {0} and R.
(a) Prove that a field is simple.
Proof:
Let I 6= {0} be an ideal of F . Then I ⊆ F .
Claim: F ⊆ I
Since I 6= {0} for some 0 6= a ∈ I. Since F is a field, a−1 ∈ F . Thus,
1F = a−1 a ∈ I
since I is ideal.
Let u ∈ F . Then u = u1F ∈ I. Since I is an ideal.
Hence, F ⊆ I. Thus F = I.
Therefore the ideals of a field F are {0} and F .
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(b) Let R be a commutative ring with identity and I be an ideal of R. Prove that R/I is simple if and
only if I is a maximal ideal of R.
Let R be a commutative ring with identity and I be an ideal of R.
Proof:
Claim: R/I is a field iff I is maximal ideal of R.
(⇒) Suppose R/I is a field and B is any ideal of R with I ⊂ B. Let 0 6= b ∈ B such that b 6= I. Then
b + I ∈ R/I. Since b 6= 0, b + I 6= I. Also, R/I is a field implies that there exist c + I ∈ R/I such that
(b + I)(c + I) = (1 + I). Now,
(1 + I) = (b + I)(c + I) ⇒ (1 + I) = (b · c) + I
⇒ 1 − bc ∈ I ⊂ B
Note that bc ∈ B, since B is ideal and b ∈ B. Thus,
1 = 1 − bc + bc
= (1 − bc) + bc ∈ B.
Hence, 1 ∈ B implies, R = (1) ⊆ B ⊆ R. Therefore B = R and so I is a maximal ideal.
(⇐) Suppose I is a maximal ideal of R. Assume that R/I is not a field. Then ∃x ∈ R such that
x + I 6= 0 + I. This would imply that x ∈
/ I and x is nonunit. Thus
I ( x + I ( R.
This contradicts the maximality of I. Therefore, R/I is a field. This proves the claim.
By (a), since R/I is a field, then R/I is simple. Therefore, R/I is simple if and only if I is maximal.
5. (a) Define a prime ideal.
An ideal P of a ring R is prime, if P 6= R and for any ideals A, B ∈ R, AB ⊆ P implies A ⊆ P or B ⊆ P .
(b) Let f : R → S be an epimorphism of rings with kernel K. Prove that if Q is a prime ideal in S,
then f −1 (Q) is a prime ideal in R that contains K. Show everything here.
Proof:
Let f : R → S be an epimorphism of rings with kernel K.
Claim 1: f −1 (Q) is an ideal of R.
Suppose that Q is a prime ideal in S. Since Q is a prime ideal in S, 0s ∈ Q and since f is homo(since f
is epi), f (0R ) = 0s ∈ Q implies 0R ∈ f −1 (Q). Thus, f −1 (Q) 6= φ. Also, f −1 (Q) ⊆ R.
Let a, b ∈ f −1 (Q) and r ∈ R. Then , f (a) ∈ Q and f (b) ∈ Q and f (r) ∈ S. Since Q is an ideal,
f (a) − f (b) ∈ Q. Since f is homo, f (a) − f (b) = f (a − b) ∈ Q. Thus, a − b ∈ f −1 (Q). Also, since Q is
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an ideal of S, f (r)f (a) ∈ Q. Since f is homo, f (ra) ∈ Q. Thus, ra ∈ f −1 (Q). Also, f (a)f (r) ∈ Q implies
f (ar) ∈ Q, since f is homo. Thus, ar ∈ f −1 (Q). By the ideal criterion, f −1 (Q) is ideal of R.
Claim 2: f (f −1 (Q)) = Q.
If x ∈ f −1 (Q). Then ∃y ∈ Q such that x = f (y) ∈ f −1 (Q). So, f (f −1 (Q)) ⊆ Q.
If x ∈ Q then x = f (y), y ∈ f −1 (Q). This implies that f (y) ∈ f (f −1 (Q)). So, Q ⊆ f (f −1 (Q)). Hence,
f (f −1 (Q)) = Q.
Claim 3: f (Q) is prime ideal of R.
Let A, B be ideals in R. If AB ⊆ f −1 (Q), then f (AB) ⊆ f (f −1 (Q)). Since f is onto and by claim 2,
f (A)f (B) ⊆ Q, where f (A) f (B) ∈ Q. So, f (A)f (B) implies that f (A) ⊆ Q or f (B) ⊆ Q. Thus,
A ⊆ f −1 (Q)andB ⊆ f −1 (Q).
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