MATH 320 The Euclidean Algorithm and The Negative Statement Question

Name:Sample Problems to Review for Final Exam, Spring 2022, Math 320
1. (a) Use the Euclidean Algorithm to find gcd(111, 21).
(b) Find integers u, v such that 111u + 21v = gcd(111, 21).
(c) Find all solutions x such that [21]111 x = [18]111 in Z111
2. (a) The following statement is false. Negate it and prove the negation. “∀x ∈ Z, if x
is odd, then x > 10 or x ≤ 8.”
(b) The following statement is false. Negate it and prove the negation. “∀a, b ∈ Z, if
gcd(a, b) = 1, then 2|(a + b).”
3. Define h : R → [5, ∞) by h(x) = x2 + 5.
(a) Is h one-to-one/injective? Prove your claim.
(b) Is h onto/surjective? Prove your claim.
4. Find multiplicative inverses, if possible. Let p = 2×2 + 3x + 1 ∈ Z5 [x].
(a) [169]195
(b) [191]195
(c) [x + 4]p
(d) [4x + 4]p
5. Provide examples…
(a) Give an example of non-zero [a], [b] ∈ Z12 so that [a]x = [b] has no solution.
(b) Give an example of a unit in M2 (R).
(c) Give an example of a zero divisor in M2 (R).
(d) Give an example of a nonconstant homomorphism H : Z8 → Z4 .
(e) Give an example of an irreducible polynomial of degree 3 in Z3 [x].
(f) Give an example of a reducible polynomial with no roots in Z3 [x].
(g) Give an example of a field with 7 members. with 8 members. with 16 members.
with 25 members.
6. You will need to show a subset is a subring. You will need to prove a relation is a
well-defined homomorphism.
1
Definitions and Theorems for Final Exam, Math 320
1
Chapter 1
Theorem (1.1, The Division Algorithm for Integers). Let a, b be integers with b > 0.
Then there exist unique integers q and r such that
a = bq + r
and
0≤r 0}
In particular, ∃u, v ∈ Z such that gcd(a, b) = au + bv
Theorem (1.4, Useful Divisibility Lemma/Theorem). Suppose a, b, c ∈ Z. If a|(bc)
and gcd(a, b) = 1, then a|c.
Definition. An integer p is prime iff p ̸= 0, ±1 and the only divisors of p are ±1
and ±p. An integer that is not 0, 1, -1, or prime is called composite.
Theorem (1.5, Prime Characterization). Let p ∈ Z such that p ̸= 0, ±1. Then p is
prime iff ∀b, c ∈ Z, if p|bc, then p|b or p|c.
Theorem (1.10, Primality Testing).
Suppose n ∈ Z with n > 1. If n has no prime
√
divisors less than or equal to n, then n is prime.
2
Appendix B – Sets and Functions
Definition. We could say f is a function from A to B , written f : A → B, iff
1. For all x ∈ A, there exists y ∈ B such that f (x) = y. When f is an operation
later, we will sometimes need to check this condition calling it closure.
2. For all x1 , x2 ∈ A, if x1 = x2 , then f (x1 ) = f (x2 ). This is sometimes referred
to as the function being well-defined.
1
Definition. Suppose that f is a function from A to B.
1. f is onto/surjective iff for all y ∈ B, there exists x ∈ A such that f (x) = y
2. f is one-to-one/ injective iff for all x1 , x2 ∈ A, if f (x1 ) = f (x2 ), then
x1 = x2
3. f is invertible/bijective iff f is one-to-one and onto .
3
Chapter 2 and Ring Definitions
Definition. Let a, b, n ∈ Z with n > 0. Then a is congruent to b modulo n iff
n|(a − b). We write a ≡ b (mod n).
Corollary (2.5, think of congruence classes as remainders when dividing by n). Let
n > 1 be an integer.
1. If ∃a, q, r ∈ Z such that a = qn + r, then [a]n = [r]n .
2. There are exactly n distinct congruence classes mod n. Namely, [0]n , [1]n , . . . , [n−
1]n .
Definition. Fix an integer n > 1. Let Zn := {[0]n , [1]n , . . . , [n − 1]n }. Define operations on Zn as follows
[a]n ⊕ [c]n := [a + c]n
and [a]n ⊙ [c]n := [ac]n .
After section 2.2, the operations on Zn will use the more commmon addition and
multiplication notations.
Theorem (2.8, Zp versus Zn ). If p > 1 is an integer, then the following are equivalent:
1. p is prime.
2. ∀[a]p ∈ Zp , if [a]p ̸= [0]p , then ∃[x]p ∈ Zp such that [a]p [x]p = [1]p .
3. ∀[b]p , [c]p ∈ Zp , if [b]p [c]p = [0]p , then [b]p = [0]p or [c]p = [0]p .
Remark. Note: p prime is the same as Zp is a field is the same as Zp is an integral
domain;
and: n composite is the same as Zn is not a field is the same as Zn is not an integral
domain.
Theorem (2.9). Let a, n ∈ Z with n > 1.
Then ∃[x]n ∈ Zn such that [a]n [x]n = [1]n iff gcd(a, n) = 1.
2
4
Chapter 3
Definition. (Section 3.1) A ring R is a commutative ring iff
9. ∀a, b ∈ R we have ab = ba. (Commutative multiplication)
Definition. (Section 3.1) A ring R is a ring with identity iff
10. ∃1R ∈ R, ∀a ∈ R we have a1R = a = 1R a. (Multiplicative identity)
Definition. (Section 3.1) Suppose that R is a commutative ring with identity. We
say that R is an integral domain iff
11. ∀a, b ∈ R, if ab = 0R , then a = 0R or b = 0R .
We say that R is a field iff
12. ∀aR , if a ̸= 0R , then ∃x ∈ R such that ax = 1R .
Definition. Suppose that R is a ring and S ⊆ R is non-empty. We say that S is a
subring of R iff S is a ring with the operations from R.
Definition. (Section 3.2) Suppose R is a ring and a ∈ R − {0R }. We say that a is
a zero divisor iff ∃b ∈ R − {0R } such that ab = 0R . Further, if R is a ring with
identity, we say a is a unit iff ∃b ∈ R such that ab = 1R .
Theorem (3.6, Checking a subset is a subring, version 2 and this is better than
3.2). Suppose that S is a nonempty subset of a ring R. If
1. S is closed under subtraction
2. S is closed under multiplication
then S is a subring of R.
Theorem (3.8). Every field is an integral domain.
Definition. Suppose that R and S are rings. We say that f : R → S is a ring
homomorphism iff ∀a, b ∈ R, we have
f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b).
Definition. Suppose that R and S are rings. We say that f : R → S is an ring
isomorphism iff f is a bijective ring homomorphism. We say that two rings R and
S are isomorphic iff there exists a ring isomorphism f : R → S.
Theorem (3.10, basic properties of ring homomorphisms). Suppose f : R → S is a
ring homomorphism. Then
1. f (0R ) = 0S
2. ∀a ∈ R, f (−a) = −f (a)
3. ∀a, b ∈ R, f (a − b) = f (a) − f (b).
3
5
Chapter 4
Definition. Let R be a ring and p = an xn +· · ·+a1 x+a0 ∈ R[x] with an ̸= 0R . Then
the degree of p is n and the leading coefficient of p is an . We write deg(p) = n
for the degree of p is n. Also the p = 0R polynomial has no degree and the rest
of the constant polynomials p = a0 ̸= 0R have degree 0. Also the ai are all called
coefficients of p. We say that p is monic iff the leading coefficient is 1R . I tend to
suppress the “function notation” and write p rather than p(x).
Theorem (4.2, degree of a product – we use this lots). Suppose R is an integral
domain and f, g ∈ R[x] are non-zero. Then
deg(f g) = deg(f ) + deg(g).
Theorem (4.6, Division Algorithm for Polynomials). Let F be a field and suppose
f, g ∈ F [x] with g ̸= 0F . Then there exist unique q, r ∈ F [x] such that
f = gq + r
where r = 0F
or
deg(r) < deg(g). Definition. Let F be a field and a, b ∈ F [x] with b ̸= 0F . We say that b divides a iff ∃h ∈ F [x] such that a = bh. Also we say things like b is a factor/divisor of a and a is a multiple of b in such a case. Theorem (4.7, divisibility of polynomials is a bit different than integers). Let F be a field and a, b ∈ F [x] with b nonzero. 1. If b|a, then ∀γ ∈ F − 0F , we have (γb)|a. 2. Every divisor of a has degree less than or equal to a. Definition. Let F be a field and a, b ∈ F [x] not both zero. The greatest common divisor of a and b is the monic polynomial of greatest degree that divides both a and b. In other words, d = gcd(a, b) iff 1. d is monic 2. d|a and d|b 3. ∀c ∈ F [x], if c|a and c|b, then deg(c) ≤ deg(d). We say that a and b are relatively prime iff gcd(a, b) = 1F . Theorem (4.8 – Much like Theorem 1.2). Let F be a field and a, b ∈ F [x] not both zero. Then there is a unique d = gcd(a, b). Furthermore, there exist polynomials u, v ∈ F [x] such that d = au + bv. Definition. Let R be a ring with identity. An element a ∈ R is an associate of b ∈ R iff a = bu for some unit u ∈ R. Let F be a field and p ∈ F [x] where deg(p) > 0.
We say that p is irreducible over F (think prime) iff ∀q ∈ F [x], if q|p, then q is
a unit in F [x] or q is an associate of p. We say that p is reducible over F (think
composite) iff p is not irreducible over F .
4
Theorem (4.16, Zero/Factor Theorem). Let F be a field, f ∈ F [x], and α ∈ F.
Then α is a zero of f iff (x − α)|f.
Corollary (4.18). Let F be a field and f ∈ F [x] with deg(f ) ≥ 2. If f is irreducible
over F, then f has no roots in F.
Corollary (4.19). Let F be a field and f ∈ F [x] have degree 2 or 3. Then f is
irreducible over F iff f has no roots in F.
6
Chapter 5
Definition. Let F be a field and f, g, p ∈ F [x] where p ̸= 0F . Then f is congruent
to g modulo p written f ≡ g (mod p) iff p|(f − g).
Theorem (5.5, congruence classes are like remainder classes, see Theorem 2.5). Let
F be a field and p ∈ F [x] has deg(p) = n.
1. If ∃f, q, r ∈ F [x] such that f = pq + r, then [f ]p = [r]p .
2. Let S := {g | g ∈ F [x] and deg(g) < n} ∪ {0F }. Then ∀f ∈ F [x], ∃g ∈ S such that [f ]p = [g]p . Theorem (5.9, the units in polynomial quotient rings). Let F be a field and p ∈ F [x] with deg(p) > 0. Let f ∈ F [x]. We have gcd(f, p) = 1 iff [f ]p is a unit in F [x]/(p).
Theorem (5.10, modulo an irreducible versus reducible over F ). Let F be a field
and p ∈ F [x] with deg(p) > 0. TFAE:
1. p is irreducible over F.
2. F [x]/(p) is a field. I.e. ∀[f ]p ̸= [0]p , ∃[g]p such that [f ]p [g]p = [1]p .
3. F [x]/(p) is an integral domain. I.e. ∀[f ]p , [g]p ̸= [0]p , we have [f ]p [g]p ̸= [0]p .
5
Name:
Sample Problems to Review for Final Exam, Spring 2022, Math 320
1. (a) Use the Euclidean Algorithm to find 𝑔𝑐𝑑(111,21).
The Euclidean Algorithm is
111 = 5(21) + 6
21 = 3(6) + 3
6 = 2(3)
Thus 𝒈𝒄𝒅(𝟏𝟏𝟏, 𝟐𝟏) = 𝟑.
(b) Find integers 𝑢, 𝑣 such that 111𝑢 + 21𝑣 = 𝑔𝑐𝑑(111,21).
From the Euclidean Algorithm above
3 = 21 − 3(6)
3 = 21 − 3(111 − 5(21))
3 = 21 − 3(111) + 15(21)
3 = 16(𝟐𝟏) − 3(𝟏𝟏𝟏)
We have found 𝒖 = −𝟑, 𝒗 = 𝟏𝟔, they are not unique. Note that
3 = 𝟐𝟏(16 + 111𝑘) + 𝟏𝟏𝟏(−3 − 21𝑘), with 𝑘 ∈ ℤ
(c) Find all solutions 𝑥 such that [21]111 𝑥 = [18]111 in ℤ111 .
From the equation 𝟑 = 𝟏𝟔(𝟐𝟏) − 𝟑(𝟏𝟏𝟏) above, multiplying by 6 we have:
18 = 96(𝟐𝟏) − 18(𝟏𝟏𝟏)
then,
[18]111 = [96]111 [21]111 + [−18]111 [111]111
[21]111 [96]111 = [18]111
Subtracting or replacing
−{
[21]111 𝑥 = [18]111
[21]111 [96]111 = [18]111
we have
[21]111 (𝑥 − [96]111 ) = [0]111
that means 𝑥 − [96]111 = [37𝑘]111 with 𝑘 ∈ ℤ (because 21 ∙ 37𝑘 = 3 ∙ 111𝑘).
Then,
𝑥 = [37𝑘]111 + [96]111
𝑥 = [96 + 37𝑘]111 , with 𝑘 ∈ ℤ.
Now, note that 0 ≤ 𝑘 < 3, because, if 𝑘 = 3 + 𝑟 then [37(3 + 𝑟)]111 = [111 + 37𝑟]111 = [37𝑟]111 Thus, all the solutions are: 𝑥 = [𝟗𝟔]𝟏𝟏𝟏 𝑥 = [96 + 37]111 = [133]111 = [𝟐𝟐]𝟏𝟏𝟏 𝑥 = [96 + 37(2)]111 = [170]111 = [𝟓𝟗]𝟏𝟏𝟏 2. (a) The following statement is false. Negate it and prove the negation. “∀𝑥 ∈ ℤ, if 𝑥 is odd, then 𝑥 > 10 or 𝑥 ≤ 8.”
Negation: “∃𝑥 ∈ ℤ such that; 𝑥 is odd, 𝑥 ≤ 10 and 𝑥 > 8.”
Proof: The negative statement is true, 𝑥 = 9 is the integer.
(b) The following statement is false. Negate it and prove the negation. “∀𝑎, 𝑏 ∈ ℤ, if
𝑔𝑐𝑑(𝑎, 𝑏) = 1, then 2|(𝑎 + 𝑏).”
Negation: “∃𝑎, 𝑏 ∈ ℤ such that 𝑔𝑐𝑑(𝑎, 𝑏) = 1 and “2 doesn′t divide (𝑎 + 𝑏)” .
Proof: The negative statement is true, 𝑎 = 5 and 𝑏 = 8 are an example.
3. Define ℎ ∶ ℝ → [5, +∞) by ℎ(𝑥) = 𝑥 2 + 5.
(a) Is h one-to-one/injective? Prove your claim.
ℎ isn’t injective because ℎ(𝑥) = ℎ(−𝑥) for all 𝑥 ≠ 0. That is,
ℎ(−𝑥) = (−𝑥)2 + 5 = (𝑥)2 + 5 = ℎ(𝑥).
Then, ℎ(𝑥) = ℎ(−𝑥) but 𝑥 ≠ −𝑥 for all 𝑥 ≠ 0.
(b) Is h onto/surjective? Prove your claim.
ℎ is surjective. In fact, set 𝑦 ∈ [5, ∞[, then 𝑦 − 5 ≥ 0.
We take 𝑥 = √𝑦 − 5 ∈ ℝ, then
2
ℎ(𝑥) = ℎ(√𝑦 − 5) = (√𝑦 − 5) + 5 = 𝑦 − 5 + 5 = 𝑦
Thus, ℎ is surjective.
4. Find multiplicative inverses, if possible. Let 𝑝 = 2𝑥 2 + 3𝑥 + 1 ∈ ℤ5 [𝑥].
We have to find the solution of equation [𝑎]𝑥 = [1], and we know that it has a
solution in ℤ𝑛 if and only if (𝑎, 𝑛) = 1 in ℤ.
(a) [169]195
We have to solve [169]𝑥 = [1] in ℤ195 .
By Euclidean Algorithm
169 = 0(195) + 169
195 = 1(169) + 26
169 = 6(26) + 13
26 = 2(13)
Then (169,195) = 13 ≠ 1. Thus, [169]195 doesn’t have multiplicative inverse.
(b) [191]195
By Euclidean Algorithm
195 = 1(191) + 4
191 = 47(4) + 3
4 = 1(3) + 1
3 = 3(1)
we see that (191,195) = 1, then [191]195 has multiplicative inverse. From the
Euclidean Algorithm above, we get Bezout identity.
1 = 4 − 1(3) = 4 − 1(191 − 47(4)) = 48(4) − 191
= 48(195 − 1(191)) − 191 = 48(195) − 49(191)
That is 48(195) − 49(191) = 1. Then
[48(195) − 49(191)]195 = [1]195
[48]195 [195]195 + [−49]195 [191]195 = [1]195
[48]195 [0]195 + [195 − 49]195 [191]195 = [1]195
[0]195 + [146]195 [191]195 = [1]195
[146]195 [191]195 = [1]195
Thus [146]195 is the multiplicative inverse of [191]195 .
(c) [𝑥 + 4]𝑝
By Euclidean algorithm
2𝑥 2 + 3𝑥 + 1 = 2𝑥(𝑥 + 4) + 1
𝑥 + 4 = (𝑥 + 4)1
Then 𝑔𝑐𝑑(2𝑥 2 + 3𝑥 + 1, 𝑥 + 4) = 1. Furthermore,
(2𝑥 2 + 3𝑥 + 1) − 2𝑥(𝑥 + 4) = 1
taking 𝑚𝑜𝑑(𝑝)
[2𝑥 2 + 3𝑥 + 1]𝑝 + [−2𝑥]𝑝 [(𝑥 + 4)]𝑝 = [1]𝑝
[0]𝑝 + [−2𝑥]𝑝 [(𝑥 + 4)]𝑝 = [1]𝑝
[−2𝑥]𝑝 [(𝑥 + 4)]𝑝 = [1]𝑝
Thus [−2𝑥]𝑝 is the multiplicative inverse of [(𝑥 + 4)]𝑝 .
(d) [4𝑥 + 4]𝑝
By Euclidean algorithm
2𝑥 2 + 3𝑥 + 1 = (−2𝑥 − 1)(4𝑥 + 4),
That is 𝑔𝑐𝑑(2𝑥 2 + 3𝑥 + 1,4𝑥 + 4) = −2𝑥 − 1 ≠ 1.
Thus, [4𝑥 + 4]𝑝 doesn’t have multiplicative inverse.
5. Provide examples…
(a) Give an example of non-zero [𝑎], [𝑏] ∈ ℤ12 so that [𝑎]𝑥 = [𝑏] has no solution.
For our interests 𝑎 and 12 should not be coprime, because, if 𝑔𝑐𝑑(𝑎, 12) = 1
then [𝑎]𝑥 = [𝑏] has a solution.
We take 𝑎 = 4 and 𝑏 = 7, then
[4] ∙ [0] = [0] ≠ [7],
[4] ∙ [1] = [4] ≠ [7],
[4] ∙ [2] = [8] ≠ [7],
[4] ∙ [3] = [0] ≠ [7],
[4] ∙ [4] = [4] ≠ [7],
[4] ∙ [5] = [8] ≠ [7],
[4] ∙ [6] = [0] ≠ [7],
[4] ∙ [7] = [4] ≠ [7],
[4] ∙ [8] = [8] ≠ [7],
[4] ∙ [9] = [0] ≠ [7],
[4] ∙ [10] = [4] ≠ [7],
[4] ∙ [11] = [8] ≠ [7].
Thus [4]𝑥 = [7] doesn’t have solution in ℤ12 .
(b) Give an example of a unit in 𝑀2 (ℝ).
We take 𝐴 = (
1
1
2
−1 2
) ∈ 𝑀2 (ℝ). Then, 𝐵 = (
) ∈ 𝑀2 (ℝ) is such that
1
1 −1
𝐴𝐵 = (
Thus, 𝐴 = (
1 0
2
−1 2
)∙(
)=(
).
0 1
1
1 −1
1
1
1 2
) is a unit in 𝑀2 (ℝ).
1 1
(c) Give an example of a zero divisor in 𝑀2 (ℝ).
We take 𝐴 = (
1
0
0
0 0
) ∈ 𝑀2 (ℝ). We see that 𝐴 = (
) ∈ 𝑀2 (ℝ) is such that
0
0 1
𝐴𝐵 = (
1
0
0
0
)∙(
0
0
0
0
)=(
1
0
0
)
0
Since 𝐴 ≠ 0 and 𝐵 ≠ 0, then 𝐴 is a zero divisor.
(d) Give an example of a nonconstant homomorphism 𝐻 ∶ ℤ8 → ℤ4 .
A homomorphism 𝐻([𝑥]8 ) = 𝑥𝐻([1]8 ) is completely determined by 𝐻([1]8 ).
[0]4
[1]4
𝐻([1]8 ) =
[2]4
{[3]4
If 𝐻([1]8 ) = [0]4 is a constant null homomorphism.
If 𝐻([1]8 ) = [1]4 then 𝐻([𝑥]8 ) = [𝑥]4 , this is a no constant homomorphism, in
fact: 𝐻 is well-defined because, if [𝑎]8 = [𝑏]8 then
𝑎 − 𝑏 = 8𝑘 ⟹ 𝑎 − 𝑏 = 4𝑞
that is, [𝑎]4 = [𝑏]4 . Then 𝐻([𝑎]8 ) = [𝑎]4 = [𝑏]4 = 𝐻([𝑏]8 ).
Furthermore,
𝐻([𝑥]8 + [𝑦]8 ) = 𝐻([𝑥 + 𝑦]8 ) = [𝑥 + 𝑦]4 = [𝑥]4 + [𝑦]4 = 𝐻([𝑥]8 ) + 𝐻([𝑦]8 )
𝐻([𝑥]8 ∙ [𝑦]8 ) = 𝐻([𝑥 ∙ 𝑦]8 ) = [𝑥 ∙ 𝑦]4 = [𝑥]4 ∙ [𝑥]4 = 𝐻([𝑥]8 ) ∙ 𝐻([𝑦]8 ).
and 𝐻 is no constant because 𝐻([2]8 ) = [2]4 and 𝐻([5]8 ) = [5]4 .
(e) Give an example of an irreducible polynomial of degree 3 in ℤ3 [𝑥].
Set 𝑝(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 an polynomial of degree 3 in ℤ3 [𝑥]. Then we
have 54 possibilities to choose an irreducible. Furthermore 𝑝(𝑥) must fulfill that
𝑝(0) ≠ 0, 𝑝(1) ≠ 0 and 𝑝(2) ≠ 0
We choose 𝑝(𝑥) = 𝑥 3 + 𝑥 2 + 𝑥 + 2, then
𝑝(0) = 2 ≠ 0
𝑝(1) = 1 + 1 + 1 + 2 = 2 ≠ 0
𝑝(2) = 8 + 4 + 2 + 2 = 1 ≠ 0
Thus 𝑝(𝑥) = 𝑥 3 + 𝑥 2 + 𝑥 + 2 is irreducible of degree 3 in ℤ3 [𝑥].
(f) Give an example of a reducible polynomial with no roots in ℤ3 [𝑥].
We take an irreducible polynomials of degree 2 in ℤ3 [𝑥]
𝑞(𝑥) = 𝑥 2 + 𝑥 + 2
and the irreducible polynomial 𝑟(𝑥) = 𝑥 3 + 𝑥 2 + 𝑥 + 2 of the (g). Then
𝑝(𝑥) = 𝑞(𝑥)𝑟(𝑥) = (𝑥 2 + 𝑥 + 2)(𝑥 3 + 𝑥 2 + 𝑥 + 2 )
is a reducible polynomial in ℤ3 [𝑥] and
𝑝(0) = (2)(2) = 1
𝑝(1) = (4)(5) = 2
𝑝(2) = (8)(16) = 2
Thus 𝑝(𝑥) doesn’t have roots in ℤ3 [𝑥].
(g) Give an example of a field with 7 members, with 8 members, with 16 members,
with 25 members.
A field with 7 elements is ℤ7 because 7 is prime.
For the rest we apply Theorem 5.10.
A field with 8 members: 8 = 23
We consider ℤ2 [𝑥], to get a field with 8 members, must find an irreducible
polynomial of degree 3 in ℤ2 [𝑥].
Set 𝑝(𝑥) = 𝑥 3 + 𝑥 2 + 1 that is irreducible polynomial in ℤ2 [𝑥] , then
ℤ2 [𝑥]
〈𝑥 3 + 𝑥 2 + 1〉
is a field with 8 members.
A field with 16 members: 16 = 24
We do the same as in the previous case, now we need an irreducible polynomial
of degree 4 in ℤ2 [𝑥]
We consider 𝑝(𝑥) = 𝑥 4 + 𝑥 + 1, this is irreducible because doesn’t have linear
factors and the unique polynomial of irreducible of 2-degree 𝑥 2 + 𝑥 + 1 is not
factor of 𝑝(𝑥).
Thus
〈𝑥 4
ℤ2 [𝑥]
+ 𝑥 + 1〉
is a field with 26 members.
A field with 25 members: 25 = 52
For this, we need a polynomial irreducible of degree 2 in ℤ5 [𝑥]
We consider 𝑝(𝑥) = 2𝑥 2 + 1, this is irreducible because doesn’t have linear
factors. In fact,
𝑝(0) = 1, 𝑝(1) = 3, 𝑝(2) = 4, 𝑝(3) = 4, 𝑝(4) = 3
Thus,
ℤ5 [𝑥]
〈2𝑥 2 + 1〉
is a field with 25 members.
6. You will need to show a subset is a subring. You will need to prove a relation is
awell-defined homomorphism.
1) Set 𝑨 a ring and 𝑺 ⊂ 𝑨 a no empty subset. 𝑺 is a ring (subring of 𝑨) if it
meets the conditions:
a) 𝒂 − 𝒃 ∈ 𝑺 for all 𝒂, 𝒃 ∈ 𝑺,
b) 𝒂 ∙ 𝒃 ∈ 𝑺 for all 𝒂, 𝒃 ∈ 𝑺
Proof. We just need to prove the properties zero element and closure for
addition because all the other properties are inherited from 𝐴.
1) If 𝑎 ∈ 𝑆 then 0 = 𝑎 − 𝑎 ∈ 𝑆 ……… [zero element]
2) If 𝑎, 𝑏 ∈ 𝑆 then by 1)-a) we have −𝑏 = 0 − 𝑏 ∈ 𝑆.
Thus 𝑎 + 𝑏 = 𝑎 − (−𝑏) ∈ 𝑆 ……….[Closure for addition]
If A is a ring with identity 1𝐴 we have to add 1𝐴 ∈ 𝑆 to our statement.
2) Prove an equivalence relation is a well-defined homomorphism.
WHAT IS THAT. 6 IS NOT EXERCISE, AND TGHERE IS NO THE
WORD,”equivalence”
“
This is false because, consider ℤ and 𝐹 = {(𝑥, 𝑦) ∈ ℤ × ℤ| 𝑦 = 𝑥 2 }. F is a relation
such that
𝐹(2) = 4,
𝐹(3) = 9
𝐹(2 + 3) = 25
Then 𝐹(2 + 3) = 25 ≠ 13 = 𝐹(2) + 𝐹(3). Thus, F is not a homomorphism.