Algebra

Question 1

What is the subring criterion?

Answer 1

• 0, 1 ∈ S;
• a + b ∈ S for all a, b ∈ S;
• −a ∈ S for all a ∈ S;
• ab ∈ S for all a, b ∈ S.

Question 2

What are left and right cosets?

Answer 2

Let G be a group and H a subgroup. Let g be an element of G. We call the set
gH = {gh : h ∈ H}
a left coset of H in G and the set
Hg = {hg : h ∈ H}
a right coset of H in G.

Question 3

When are two cosets equal?

Answer 3

Let G be a group and H a subgroup. Let g1, g2 ∈ G. Then g₁H = g₂H if and only if g₁^-1g2 ∈ H.

Proof.
Suppose g₁H = g₂H. But g₂ = g₂ · 1_G ∈ g₂H as H is a subgroup of G. But g₂H = g₁H. Hence g₂ ∈ g₁H, so g₂ = g₁h for some h ∈ H. Hence g₁^-1 g₂ = h ∈ H as required.
Conversely, suppose g₁^-1g₂ = h ∈ H. We want to show that g₁H =
g₂H. Let k ∈ g₁H. Then k = g₁h₁ for some h₁ ∈ H. But h^-1 = g₂^-1g₁
so g₁ = g₂h^-1, so k = g₁h₁ = g₂h^-1h₁ ∈ g₂H. Therefore every k ∈ g₁H belongs to g₂H. By a similar argument (exercise) every k ∈ g₂H belongs to g₁H. Thus g₁H = g₂H.

Question 4

What are the left and right index?

Answer 4

Let G be a group and H be a subgroup. We shall define the left index of H in G, denoted by [G : H], to be the number of left cosets of H in G. We shall define the right index of H in G to be the number of right cosets of H in G.

Question 5

How are the amount of elements in the left and right coset related?

Answer 5

Let G be a group and H a finite subgroup. If g ∈ G then gH and Hg have the same number of elements as H.

Question 6

How are 2 different cosets related?

Answer 6

Let G be a group and H be a subgroup. Let g1, g2 be elements of G. Then the cosets g1H, g2H are either equal or disjoint.

Question 7

How do cosets form a partition?

Answer 7

Let G be a finite group and H a subgroup. The left cosets of H in G form a partition of G.

Question 8

What is Lagrange’s Theorem?

Answer 8

Let G be a finite group and H a subgroup. Then #G = [G : H] · #H.

Proof.
Let g1H, g2H, . . . , gmH be the distinct left cosets of H. These form a partition of H. Hence
#G = #g1H + #g2H + · · · + #gmH.
Now as the order of a coset is the order of the subgroup,
#g1H = #g2H = · · · = #gmH = #H.
Hence
#G = m · #H.
What is m? It is the number of left cosets of H in G. We defined this to be the index of H in G, so m = [G : H].

Question 9

How does the order of a subgroup relate to the order of the group?

Answer 9

Let G be a finite group and H a subgroup. Then
#H | #G.
Proof. This follows from Lagrange’s Theorem as the index [G : H] is
an integer.

Question 10

How does the order of an element relate to the order of the group?

Answer 10

Let G be a finite group. Let g ∈ G have order n.
Then n | #G.

Proof. Let H = 〈g〉 the cyclic group generated by g. We know that #H = n. As #H | #G, we have n | #G.

Question 11

How are the left and right index related?

Answer 11

The number of left cosets is equal to the number of right cosets
[G : H]_L = #G/#H = [G : H]_R.

Question 12

What is a Conjugate of a subgroup?

Answer 12

Let H be a subgroup of G. A conjugate of H has the form gHg−1 for some g ∈ G.

Question 13

What is a Normal subgroup?

Answer 13

We say that a subgroup H of G is normal if and only if gHg−1 = H for all g ∈ G. That is H is normal if and only if it is equal to all its conjugates. We write H ◁ G to denote that H is a normal subgroup of G.

Question 14

What are the equivalent statements about normal subgroups and conjugates?

Answer 14

Let H be a subgroup of G. Then the following are
equivalent.
(a) H is normal in G;
(b) gHg−1 = H for all g ∈ G;
(c) gH = Hg for all g ∈ G;
(d) gHg−1 ⊆ H for all g ∈ G;
(e) ghg−1 ∈ H for all g ∈ G, h ∈ H.

Proof.
It is easy to see that
(a) ⇐⇒ (b) ⇐⇒ (c), (d) ⇐⇒ (e)
and also that (b) =⇒ (d).
Let’s do (d) =⇒ (b). Suppose gHg−1 ⊆ H
for all g in G. Then, since g−1 ∈ G we have g−1Hg ⊆ H. Multiply by g on the left and g−1 on the right to get
H = g(g−1Hg)g−1 ⊆ gHg−1.
As gHg−1 ⊆ H and H ⊆ gHg−1 we have gHg−1 = H.

Question 15

What index of subgroup will always be normal?

Answer 15

Let G be a finite group and let H be a subgroup of G of index 2. Then H is normal in G.

Proof.
We want to show that gH = Hg for all g ∈ G. We know, gH = H if and only if g ∈ H (note that H = 1H).
Suppose first that g ∈ H. Then gH = H and Hg = H. Thus gH = Hg.
Suppose instead that g /∈ H. Then gH /= H. But H has index 2 in G so has exactly two left cosets, which must be H and gH. Thus
G = H ∪ gH and H ∩ gH = ∅, since cosets form a partition. Thus
gH = G \ H. Similary, Hg = G \ H. Hence gH = Hg.

Question 16

How is the quotient, of a group with a normal subgroup, a group and what is its order?

Answer 16

Let G be a group and N a normal subgroup. Then G/N with operation (gN)(g’N) = gg’N is a group with identity element N = 1GN
and inverses given by (gN)−1 = g−1N.

Moreover, if G is finite then
#(G/N) = [G : N] = #G/#N.
We call G/N the quotient group of G over N.

Question 17

What is a Homomorphism of Groups?

Answer 17

Definition. Let G, H be groups and let φ : G → H be a map. We say that φ is a homomorphism of groups if
φ(gh) = φ(g)φ(h)
for all g, h ∈ G.

Question 18

What is an Isomorphism of Groups?

Answer 18

Definition. Let G and H be groups. A map φ : G → H is an isomorphism if it is a bijective homomorphism. If G and H are isomorphic we write G ∼= H.

Question 19

What are the Kernel and Image of a group homomorphism?

Answer 19

Associated to any homomorphism φ : G → H are its kernel and image:
Ker(φ) = {g ∈ G : φ(g) = 1H}, Im(φ) = {φ(g) : g ∈ G}.

Question 20

How are the kernel and image related to the groups in the homomorphism?

Answer 20

Let φ : G → H be a homomorphism of groups. Then
(i) Ker(φ) is a normal subgroup of G.
(ii) Im(φ) is a subgroup of H.

Question 21

How can we tell if a homomorphism is injective from its kernel?

Answer 21

Let φ : G → H be a homomorphism of groups. Then
φ is injective if and only if Ker(φ) = {1G}.

Proof.
Suppose Ker(φ) = {1G}. Let g1, g2 ∈ G and suppose φ(g1) = φ(g2). Then φ(g1−1g2) = (g1)−1φ(g2) = 1H. Thus g1−1g2 ∈ Ker(φ) =
{1G} so g1−1g2 = 1G and hence g1 = g2. Therefore φ is injective.
Conversely, suppose φ is injective. Let g ∈ Ker(φ). Thus φ(g) = 1H = φ(1G). As φ is injective g = 1G. Hence Ker(φ) = {1G}.

Question 22

What is the First Isomorphism Theorem?

Answer 22

Let φ : G → H be a homomorphism of groups. Let φˆ : G/ Ker(φ) → Im(φ),
φˆ(g Ker(φ)) = φ(g)
Then φˆ is a well-defined group isomorphism.

Question 23

How is An related to Sn?

Answer 23

Let n ≥ 2. Then An is a normal subgroup of Sn.
Moreover,
[Sn : An] = 2, #An = #Sn/2 = n!/2.

Question 24

How are two cyclic groups of the same order related?

Answer 24

Let G and H be cyclic groups of order n. Then G and H are isomorphic.

Question 25

What group is a group of prime order related to?

Answer 25

Let p be a prime. Any group of order p is isomorphic to Cp.

Question 26

What is the Direct Product of two groups?

Answer 26

Let G and H be groups. We define the direct product of G and H to be G × H = {(g, h) : g ∈ G, h ∈ H}; i.e. G × H is the set of ordered pairs (g, h) where g ∈ G and h ∈ H. The binary operation on G × H is (g1, h1) · (g2, h2) = (g1g2, h1h2).

Question 27

How is G × H a group?

Answer 27

G×H is a group with identity element (1G, 1H), and inverse given by (g, h)−1 = (g−1, h−1). If G, H are finite then #(G × H) = (#G) · (#H).

Question 28

What are the groups of order 4?

Answer 28

The only groups of order 4 are C4 and C2 × C2.

Question 29

How can we write D_2n?

Answer 29

Let n ≥ 3. Then D2n = {id, r, r2, . . . , rn−1} ∪ {s, sr, sr2, . . . , srn−1} = R ∪ sR. In particular, #D2n = 2n and R is a normal subgroup of index 2. Note this tells us that [D2n : R] = 2. Thus, the subgroup R is normal in D2n.

Question 30

If a symmetry in D2n fixes vertices 1 and 2 what can we tell?

Answer 30

Let a ∈ D2n. Suppose a fixes vertices 1 and 2. Then a = id.

Question 31

If two symmetries both send 1 and 2 to the same vertex what can we tell?

Answer 31

Let b, c ∈ D2n. Suppose b(1) = c(1) and b(2) = c(2). Then b = c.

Question 32

With r defined as an anticlockwise rotation around the centre through angle 2π/n and s a reflection in the x-axis, what identities do we get?

Answer 32

With r, s as above, rⁿ = id, s2 = id, srs = r^-1. Proof. The first two relations are clear. We need to check the last one. Note that s^-1 = s. Thus srs = srs^-1. As R is normal in D2n and r ∈ R we have srs ∈ R. Hence srs = r^k for some k. We compute (srs)(1) = (sr)(s(1)) = (sr)(1) = s(r(1)) = s(2) = n. Hence srs = r^n-1 = r^-1.

Question 33

What is a Word in a group?

Answer 33

Let G be a group. Let g1, . . . , gn be elements of G. A word in g1, g2, . . . , gn is a finite product of g1, g1−1, g2, g2−1, . . . , gn, gn−1. For example, if g, h ∈ G, then the following are words in g, h: h^3, g−1h, h−2g−1h−1g3h, 1G.

Question 34

How is the subset of words in elements of a group also a group?

Answer 34

Let G be a group. Let g1, . . . , gn be elements of G. Write 〈g1, . . . , gn 〉for the subset of words in g1, . . . , gn. Then 〈g1, . . . , gn〉 is a subgroup of G.

Question 35

What is a Group Presentation?

Answer 35

It often convenient to specify a group by specifying generators and relations. This is called a group presentation. The usual notation has the form G = 〈S|R〉 where S is a set of symbols, and R is a set of relations between the symbols.

Question 36

What is the Fundamental Theorem of Group Presentations?

Answer 36

Let G = 〈S | R〉 be a group presentation where S = {s1, s2, . . . , sn} is a finite set of generators and R is a set of relations. Let H be a group and let h1, h2, . . . , hn be elements of H. There exists a homomorphism φ : G → H satisfying φ(si) = hi if and only if every relation r ∈ R holds with the si replaced by the hi. Moreover, in this case the homomorphism φ is unique.

Question 37

What is the Quarternion Group?

Answer 37

The quaternion group Q8 is defined by the presentation Q8 = 〈a, b | a⁴ = id, a² = b², bab^-1 = a^-1〉.

Question 38

How can we write elements of Q8?

Answer 38

Every element of Q8 can uniquely be written as aⁱb^j with 0 ≤ i ≤ 3 and 0 ≤ j ≤ 1. Thus #Q8 = 8.

Question 39

What is a presentation for D2n?

Answer 39

D2n = 〈r, s |rⁿ = s² = id, srs = r^-1〉

Question 40

When do elements pairwise commute and when are they independent?

Answer 40

Let G be a group. Let g1, g2, . . . , gn be elements of G. We shall say that g1, g2, . . . , gn pairwise commute if gigj = gjgi for all i and j. We shall say that they are independent if 〈gi〉 ∩ 〈g1, g2, . . . , gi−1〉 = {id} for i = 1, 2, . . . , n.

Question 41

What group is the subset of words of elements in a group isomorphic to?

Answer 41

Let G be a group. Let g1, g2, . . . , gn ∈ G, and suppose that they are pairwise commuting and independent. Then 〈g1, g2, . . . , gn〉 ∼= 〈g1〉 × 〈g2〉 × · · · × 〈gn〉.

Question 42

What is the exponent of a group?

Answer 42

Let n be a positive integer. We say that a group G has exponent n if n is the smallest positive integer such that gⁿ = id for all g ∈ G.

Question 43

What type of group is a group with exponent 2?

Answer 43

Let G be a group with exponent 2. Then G is abelian. Proof. Let g, h ∈ G. Then g = g−1 and h = h−1 (as g^2 = id = h^2). Also hg = h−1g−1 = (gh)−1 = gh since (gh)^2 = id. Hence G is abelian.

Question 44

What group is a group with exponent 2 isomorphic to?

Answer 44

Let G be a finite group with exponent 2. Then, for some positive integer n, G ∼= C₂ⁿ. In particular, #G = 2ⁿ.

Question 45

What is a group of order 6 isomorphic to?

Answer 45

Let G be a group of order 6. Then G ∼= C6 or G ∼= D6.

Question 46

What is a group of order 8 isomorphic to?

Answer 46

Let G be a group of order 8. Then G is isomorphic to one of C2 × C2 × C2, C4 × C2, C8, D8, Q8.

Question 47

What is a left and right action?

Answer 47

Let G be a group and X a set. A left action of G on X is a map G × X → X, (g, x) → g ∗ x which satisfies the following two properties (A1) 1G ∗ x = x for all x ∈ X; (A2) (gh) ∗ x = g ∗ (h ∗ x) for all g, h ∈ G and x ∈ X. A right action of G on X is defined similarly. It is a map X × G → X, (x, g) → x ∗ g which satisfies the following two properties (B1) x ∗ 1G = x for all x ∈ X; (B2) x ∗ (gh) = (x ∗ g) ∗ h for all g, h ∈ G and x ∈ X.

Question 48

What is the orbit of an element?

Answer 48

We define the orbit of x ∈ X under the action of G by OrbG(x) = {g ∗ x : g ∈ G}.

Question 49

What is a fixed point and what is Fix(G)?

Answer 49

We say that x ∈ X is fixed by G if g ∗ x = x for all g ∈ G (we also call x a fixed point of G). We write Fix(G) for the set of x ∈ X that are fixed by G. Observe that x ∈ Fix(G) ⇐⇒ x is fixed by G ⇐⇒ OrbG(x) = {x}.

Question 50

How does the set of orbits of an action on X relate to X?

Answer 50

Let G be a group acting on a set X. The set of orbits of the action is a partition of X.

Question 51

When does G act transitively on X?

Answer 51

We say G acts transitively on X if, for any x, y ∈ X, there is some g ∈ G such that g ∗ x = y. Note that G acts transitively on X if and only if OrbG(x) = X for all x ∈ X. So transitive action is the same as having only one orbit.

Question 52

What is the stabiliser of an element?

Answer 52

Suppose G acts on X. The stabilizer of x ∈ X is StabG(x) = {g ∈ G : g ∗ x = x} the set of elements of G that fix x

Question 53

How does the stabiliser relate to G?

Answer 53

StabG(x) is a subgroup of G. Moreover, StabG(x) = G if and only if x ∈ Fix(G).

Question 54

What is the Orbit-Stabiliser Theorem?

Answer 54

Let G be a finite group acting on a finite set X, and let x ∈ X. Then #G = # OrbG(x) × # StabG(x), or equivalently [G : StabG(x)] = # OrbG(x).

Question 55

What do we find if we have Cp acting on X with p prime?

Answer 55

Let p be a prime and let X be a finite set. Suppose Cp acts on X. (i) Every orbit has size 1 or p. (ii) # Fix(Cp) ≡ #X (mod p).

Question 56

What identity do we have for the order of a group to the power of p-1 with p prime?

Answer 56

Let G be a finite group and p a prime. Then (#G)^p-1 ≡ (#{g ∈ G : g^p = id}) (mod p)

Question 57

What is Cauchy's Theorem?

Answer 57

Let G be a finite group and let p be a prime. Suppose p | #G. Then G has an element of order p.

Question 58

When are two elements conjugates?

Answer 58

Let G be a group. Two elements h1, h2 ∈ G are conjugate if there is some g ∈ G such that gh1g−1 = h2.

Question 59

How does a group act on itself?

Answer 59

The group G acts on itself by conjugation G × G → G, g ∗ h = ghg−1.

Question 60

What is the Conjugacy Class of an element?

Answer 60

The conjugacy class of h ∈ G is simply the orbit of h under the conjugation action of G: ClG(h) = {ghg−1 : g ∈ G}.

Question 61

What is the Centraliser of an element?

Answer 61

The centralizer of h ∈ G is simply the stabilizer of h: CG(h) = {g ∈ G : ghg−1 = h} = {g ∈ G gh = hg}. The centralizer is a subgroup of G.

Question 62

What is Cycle Type?

Answer 62

Let σ ∈ Sn be a permutation. We say that σ has cycle type 1^r1 2^r2 3^r3 4^r4· · · (not powers, just notation) if its disjoint cycle decomposition has exactly r1 cycles of length 1, r2 cycles of length 1, r3 cycles of length 3, and so on.

Question 63

What can we find about conjugates of an r cycle in Sn?

Answer 63

Let σ = (x1, x2, . . . , xr) be an r-cycle in Sn. Let τ ∈ Sn. Then τ στ −1 = (τ (x1), τ (x2), . . . , τ (xr)).

Question 64

When are two permutations conjugate?

Answer 64

Two permutations in Sn are conjugate if and only if they have the same cycle type.

Question 65

What is a unit of a ring?

Answer 65

Let R be a ring. An element u is called a unit if there is some element v in R such that uv = vu = 1. In other words, an element u of R is a unit if it has a multiplicative inverse that belongs to R.

Question 66

What is the unit group of a ring?

Answer 66

Let R be a ring. We define the unit group of R to be the set R∗ = {a ∈ R : a is a unit in R}.

Question 67

What are the Gaussian Integers?

Answer 67

A Gaussian integer is a complex number whose real and imaginary parts are both integers. ie Z[i] = {a + bi | a,b ∈ Z}

Question 68

What is the Norm Map?

Answer 68

We define the norm map N : Z[i] → Z by N(a + bi) = a² + b², a, b ∈ Z.

Question 69

How does the norm map multiply?

Answer 69

Let α, β ∈ Z[i]. Then N(αβ) = N(α)N(β).

Question 70

What is the unit group of Z[i]?

Answer 70

The unit group of Z[i] is {1, −1, i, −i}. Proof. We want the units of Z[i]. Let α be a unit. Then there is some β ∈ Z[i] such that 1 αβ = 1. Applying the norm map, and recalling that it is multiplicative, we see that N(α)N(β) = N(αβ) = N(1) = 1. Now N(α) and N(β) are in Z (go back to the definition of the norm map to see this), and they multiply to give 1. So N(α) = N(β) = 1, or N(α) = N(β) = −1. Write α = a + bi where a, b are in Z. Then a² + b² = N(α) = ±1. Of course −1 is impossible, so a² + b² = 1. But a, b are integers. So (a, b) = (±1, 0) or (0, ±1). Hence α = a + bi = ±1 or ±i. Clearly ±1, ±i are units. So the unit group is Z[i]∗ = {1, −1, i, −i}.

Question 71

What is a Ring Homomorphism and Isomorphism?

Answer 71

Let R, S be rings. A homomorphism φ : R → S is a function that satisfies (a) φ(a + b) = φ(a) + φ(b) for all a, b ∈ R; (b) φ(ab) = φ(a)φ(b) for all a, b ∈ R; (c) φ(1R) = 1S. An isomorphism is a bijective homomorphism.

Question 72

What is the kernel and image of a ring homomorphism?

Answer 72

Let ψ : R → S be a homomorphism of rings. We define the kernel of ψ to be Ker(ψ) = {r ∈ R : ψ(r) = 0}. We define the image of ψ to be Im(ψ) = {ψ(r) : r ∈ R}.

Question 73

What is the First Isomorphism Theorem for rings?

Answer 73

Let ψ : R → S be a homomorphism of rings. (i) Ker(ψ) is an ideal of R. (ii) Im(ψ) is a subring of S. (iii) The induced map ψˆ : R/ Ker(ψ) → Im(ψ), ψˆ(r + Ker(ψ)) = ψ(r) is an isomorphism.

Question 74

When is an element in a ring a zero divisor?

Answer 74

Let R be a commutative ring. An element x /= 0 is called a zero divisor if there is y /= 0 in R such that xy = 0.

Question 75

What is an Integral Domain?

Answer 75

An integral domain is a non-zero commutative ring that has no zero divisors. Thus in an integral domain, if xy = 0 then x = 0 or y = 0.

Question 76

How are field related to integral domains?

Answer 76

Any field is an integral domain. A subring of a field is an integral domain.

Question 77

What is the Cancellation Law for Integral Domains?

Answer 77

Let R be an integral domain. Let a, b, c ∈ R with a /= 0. If ab = ac then b = c. Proof. Since ab = ac we have a(b − c) = 0. As R is an integral domain, we conclude that a = 0 or b−c = 0. But a /= 0 by hypothesis. Thus b − c = 0 so b = c.

Question 78

What are all finite integral domains also?

Answer 78

Every finite integral domain is a field.

Question 79

If R is an integral domain, what is R[x]?

Answer 79

Let R be an integral domain. Then R[x] is an integral domain.

Question 80

When do two elements of an integral domain divide each other?

Answer 80

Let R be an integral domain. Let a, b ∈ R. We say that a divides b and write a | b if b = ac for some c ∈ R. We say that a, b are associates (and write a ∼ b) if a | b and b | a.

Question 81

When are two elements of an integral domain associates?

Answer 81

Let R be an integral domain and let a, b ∈ R. Then a ∼ b if and only if there is a unit u ∈ R∗ such that a = ub.

Question 82

What are irreducibles and primes in integral domains?

Answer 82

Let R be an integral domain. An element a ∈ R is called irreducible if it satisfies the following (i) a /= 0; (ii) a is not a unit of R; (iii) if a = bc with b, c ∈ R, then either b is a unit or c is a unit. An element π ∈ R is called prime if it satisfies the following (i) π /= 0; (ii) π is not a unit of R; (iii) if π | bc with b, c ∈ R, then π | b or π | c.

Question 83

How are primes irreducible?

Answer 83

Let R be an integral domain. Then every prime of R is irreducible. Proof. Let π be a prime of R. Suppose π = ab. Then π | ab. As π is prime, π | a, or π | b. Without loss of generality we suppose π | a. Thus a = πc where c ∈ R. Thus π = ab = πcb. As R is an integral domain, we conclude that 1 = cb. Thus b is a unit. Hence π is irreducible.

Question 84

What is a Unique Factorisation Domain?

Answer 84

Let R be an integral domain. We say that R is a unique factorisation domain (UFD for short) if (i) every element a that is neither zero nor a unit can be written in the form a = p1p2 · · · pr where the pi are irreducible; (ii) whenever p1p2 · · · pr = q1q2 · · · qs with pi, qj irreducible, then r = s and we can reorder the qj so that pi ∼ qi for i = 1, 2, . . . , r.

Question 85

What is division with remainder, for the integers and polynomials?

Answer 85

Let m, n ∈ Z with n /= 0. Then there are unique q, r ∈ Z such that m = qn + r, 0 ≤ r < |n|. We call q the quotient and r the remainder obtained upon dividing m by n. Let F be a field. Let g, f ∈ F[X] with f /= 0. Then there are unique q, r ∈ F[X] with g = qf + r, r = 0 or deg(r) < deg(f). We call q the quotient and r the remainder obtained upon dividing g by f. Some people define the degree of the zero polynomial to be −∞. In that case they can simply write g = qf + r, deg(r) < deg(f).

Question 86

What is a Euclidean function?

Answer 86

Let R be an integral domain. A Euclidean function on R is a map ∂ : R \ {0} → N satisfying the following two conditions: (i) if a, b ∈ R \ {0} and a | b then ∂(a) ≤ ∂(b); (ii) if a, b ∈ R, and b /= 0, then there exists q, r ∈ R such that a = bq + r with either r = 0 or ∂(r) < ∂(b).

Question 87

What is a Euclidean ring?

Answer 87

A Euclidean ring is a pair (R, ∂) where R is an integral domain, and ∂ is a Euclidean function on R. Normally we just say “R is a Euclidean ring” when ∂ is understood.

Question 88

Is the integers a Euclidean ring?

Answer 88

Z is a Euclidean ring with Euclidean function |·|.

Question 89

Are the polynomials, with coefficients from a field, a Euclidean ring?

Answer 89

Let F be a field. Then F[X] is a Euclidean domain with Euclidean function deg.

Question 90

Are the Gaussian Integers a Euclidean ring?

Answer 90

Z[i] is a Euclidean domain with the Euclidean function being the norm.

Question 91

What is an Ideal?

Answer 91

Let R be a commutative ring. Recall the definition of an ideal I of R. A subset I of R is called an ideal if (i) 0 ∈ I; (ii) u + v ∈ I whenever u ∈ I and v ∈ I; (iii) uv ∈ I whenever u ∈ I and v ∈ R.

Question 92

When does an Ideal equal the Ring?

Answer 92

Let R be a commutative ring and I an ideal of R. Then I = R if and only if I contains a unit.

Question 93

What is the sum and product of ideals?

Answer 93

We define the sum of ideals I, J of R by I + J = {x + y : x ∈ I, y ∈ J}, and the product IJ as the set of all finite sums ∑_i=1ⁿ xiyi, xi ∈ I, yi ∈ J. Here we interpret the empty sum (with n = 0) as 0 (so 0 ∈ IJ).

Question 94

How are the sum, product and intersection of ideals also ideals?

Answer 94

Let R be a commutative ring. Let I, J be ideals of R. Then I ∩ J, I + J and IJ are ideals of R. Moreover, IJ ⊆ I ∩ J.

Question 95

What are the multiples of an element in a commutive ring?

Answer 95

Let R be a commutative ring and a ∈ R. We define Ra = {ra : r ∈ R} to be the set of multiples of a (i.e. the set of all elements divisible by a).

Question 96

How is Ra an ideal?

Answer 96

Let R be a commutative ring. Let a ∈ R. Then Ra is an ideal of R.

Question 97

What is a Principal Ideal?

Answer 97

We call Ra the principal ideal generated by a. An ideal I of R is called principal if it is of the form I = Ra for some a ∈ R.

Question 98

What is an ideal generated by multiple elements?

Answer 98

Let R be a commutative ring. If a1, a2, . . . , an ∈ R we write (a1, a2, . . . , an) = Ra1 + Ra2 + · · · + Ran and call this the ideal generated by a1, . . . , an. Note that the elements of (a1, . . . , an) have the form r1a1 + · · · + rnan where ri ∈ R.

Question 99

What are the equivalent statements about principal ideals?

Answer 99

Let R be a commutative ring. Let a, b ∈ R. The following are equivalent. (a) b ∈ Ra; (b) Rb ⊆ Ra; (c) a | b.

Question 100

When do principal ideals equal each other or the ring?

Answer 100

Let R be a commutative ring. Let a, b ∈ R. (a) Ra = Rb if and only if a and b are associates. (b) Ra = R if and only if a is a unit.

Question 101

What is a Principal Ideal Domain?

Answer 101

The ring R is called a principal ideal domain (PID) if it is an integral domain, and every ideal of R is principal. i.e. every ideal I of R has the form Ra for some a ∈ R.

Question 102

Are Euclidean domains also PIDs?

Answer 102

Let R be a Euclidean domain. Then R is a principal ideal domain.

Question 103

What are the known PIDs?

Answer 103

Z Z[i] F[X]

Question 104

How do primes and irreducibles relate in a PID?

Answer 104

Let R be a PID. The primes of R are the same as the irreducibles of R.

Question 105

When is a sequence of ideals increasing and stationary, and what does it mean for a ring to be Noetherian?

Answer 105

We say that a sequence of ideal {Ij}_j=1^∞ is increasing if I1 ⊆ I2 ⊆ I3 ⊆ · · · . We say that this increasing sequence is stationary if there is some N such that IN = IN+1 = IN+2 = · · · . A ring R is Noetherian if every increasing sequence of ideals is stationary.

Question 106

Are PIDs Noetherian?

Answer 106

Let R be a PID. Then R is Noetherian.

Question 107

How can we write every non zero, non unit element of a PID?

Answer 107

Let R be a PID. Then every element that is neither 0 nor a unit can be written as a finite product of irreducibles.

Question 108

How do PIDs relate to UFD?

Answer 108

Every PID is a UFD.

Question 109

How do EDs, PIDs and UFDs relate?

Answer 109

ED =⇒ PID =⇒ UFD.

Question 110

What is an R-Module?

Answer 110

Let R be a ring. An R-module is an additive abelian group (M, +, 0) equipped with an operation R × M → M, (r, m) → rm (scalar multiplication) that satisfies the following properties: (a) 1 · m = m for all m ∈ M; (b) (r · s) · m = r · (s · m) for all r, s ∈ R and m ∈ M; (c) (r + s) · m = r · m + s · m for all r, s ∈ R and m ∈ M; (d) r · (m + n) = r · m + r · n for all r ∈ R and m, n ∈ M.

Question 111

What is an R-submodule?

Answer 111

Let R be a ring and M an R-module. An R-submodule of M is a subgroup (N, +, 0) of (M, +, 0) that satisfies r · n ∈ N for all r ∈ R and n ∈ N. It is easy to see that an R-submodule is an R-module.

Question 112

What is the Quotient Module?

Answer 112

Let M be an R-module and N be an R-submodule of M. We define the quotient module M/N to be the set of cosets m + N with m ∈ M. Addition and scalar multiplication are given in the natural way (m1 + N) + (m2 + N) = (m1 + m2) + N, r · (m + N) = rm + N, r ∈ R for m1, m2, m ∈ M, It is easy to check that these operations are well-defined and that M/N is an R-module. The criterion for equality of cosets is also easy to check m1 + N = m2 + N ⇐⇒ m1 − m2 ∈ N

Question 113

What is the Direct Product of modules and how is it a module?

Answer 113

Let M, N be R-modules. Then M × N = {(m, n) : m ∈ M, n ∈ N} is an R-module where addition and scalar multiplication is defined by (m1, n1) + (m2, n2) = (m1 + m2, n1 + n2), r · (m, n) = (rm, rn).

Question 114

What is a homomorphism of modules?

Answer 114

Let M, N be R-modules. A map φ : M → N is a homomorphism if φ(m1 + m2) = φ(m1) + φ(m2), φ(rm) = rφ(m). An isomorphism is a bijective homomorphism.

Question 115

What is the Isomorphism Theorem?

Answer 115

Let φ : M → N be a homomorphism of R-modules. (i) Ker(φ) is an R-submodule of M. (ii) Im(φ) is an R-submodule of N. (iii) The induced map φˆ : M/ Ker(φ) → Im(φ), φˆ(m + Ker(φ)) = φ(m) is an isomorphism of R-modules.

Question 116

What is the span of a subset of a module?

Answer 116

Let M be an R-module. Let X = {x1, . . . , xn} be a finite subset of M. We define the R-span of X to be SpanR(X) = {r1x1 + · · · + rnxn : ri ∈ R, xi ∈ X} . This the set of all linear combinations of elements of X with coefficients in R. We say a finite subset X of M spans (or generates) M as Rmodule if M = SpanR(X). We say that M is finitely generated if it is the span of a finite subset X ⊆ M.

Question 117

How does a module being finitely generated relate to homomorphisms?

Answer 117

Let M be an R-module. Then M is finitely generated if and only if there is a surjective homomorphism φ : Rn → M for some n ≥ 1.

Question 118

When is a subset of a module R-linearly independent, when is it a basis, what is a free basis and what is rank?

Answer 118

A subset X = {x1, x2, . . . , xn} of M is R-linearly independent if whenever r1x1 + · · · + rmxm = 0 with ri ∈ R, xi ∈ X then r1 = r2 = · · · = rm = 0. If X both spans M and is independent then we say that M is an R-basis for M. An R-module M is called free if it has an R-basis. Sometimes an R-basis is called a free R-basis to emphasise its independence. If X = {x1, x2, . . . , xn} is a basis for M then we say that M is free of rank n.

Question 119

What is the Fundamental Theorem of Finitely-Generated Modules over Euclidean Rings?

Answer 119

Let R be a Euclidean ring. Let M be a finitely generated R-module. Then there is a finite number of elements b1, b2, . . . , bn ∈ R which are non-zero and non-units, satisfying b1 | b2 | · · · | bn, and a non-negative integer r such that M ∼= R/Rb1 × R/Rb2 × · · · × R/Rbn × Rr. Moreover, if there are finitely many elements c1, c2, . . . , cm ∈ R satisfying c1 | c2 | · · · | cm which are non-zero and non-units, and an non-negative integer s such that M ∼= R/Rc1 × R × R/Rc2 × · · · × R/Rcm × Rs then s = r, m = n and ci ∼ bi for i = 1, . . . , n.

Question 120

What is the Fundamental Theorem of Finitely Generated Abelian Groups?

Answer 120

Let G be a finitely generated abelian group. Then there is a uniquely determined non-negative integer r, and a finite number of uniquely determined integers b1 | b2 | · · · | bn, all > 1, such that G ∼= Z/b1Z × Z/b2Z × · · · × Z/bnZ × Zr. Could also have, G ∼= Cb1 × Cb2 × · · · × Cbn × Cr∞.

Question 121

What will a finite subgroup of F* be if F is a field?

Answer 121

Let F be a field. Let G be a finite subgroup of F∗. Then G is cyclic.

Question 122

How is a submodule of Rn generated?

Answer 122

Let R be a PID. Let n ≥ 1. Let N be an R-submodule of Rn. Then there are v1, v2, . . . , vn ∈ N such that N = SpanR(v1, v2, . . . , vn). In particular, N is finitely generated. Note that we’re not claiming v1, . . . , vn is linearly independent, or even distinct, or even non-zero. So it is possible that N is spanned by fewer than n elements.

Question 123

How can we generate a submodule from the ring in matrix form?

Answer 123

Let R be a PID. Let n ≥ 1. Let N be an R-submodule of Rn. Then there is a matrix A ∈ Mn(R) such that N = ARn = {Ar : r ∈ M}. Recall that Mn(R) denotes the ring of n × n matrices with entries in R.

Question 124

How can a matrix with entries from R be written in SNF?

Answer 124

Let R be a Euclidean ring. Let A ∈ Mn(R). Then there are matrices U, V ∈ GLn(R) UAV = (Picture SNF with bi along the diagonal and 0 everywhere else, could be 0 bottom right also). where b1 | b2 | · · · | bm, where the bi are non-zero elements of R.

Question 125

How can we find the SNF of a matrix?

Answer 125

(Always using row and column operations) 1. Take the smallest (in absolute value) non-zero entry and move it to the top left corner. 2. Our next step is to make the entries in the same row and column as this value smaller in absolute value. 3. Repeat from 1 if there is a value smaller than the initial one taken.

Question 126

What elementary matrices do we use to find the SNF of a matrix?

Answer 126

We consider three types of elementary matrices: * Ei,j (with i ̸= j) denotes the identity matrix with the i-th and j-th rows swapped. * Fi,j,q (with i ̸= j) is the identity matrix with an additional q ∈ R in the (i, j)-th position. * Gi,u is the identity matrix with the (i, i) entry replaced byu ∈ R∗ .