Algebra Flashcards

1
Q

What is the subring criterion?

A
  • 0, 1 ∈ S;
  • a + b ∈ S for all a, b ∈ S;
  • −a ∈ S for all a ∈ S;
  • ab ∈ S for all a, b ∈ S.
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2
Q

What are left and right cosets?

A

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.

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3
Q

When are two cosets equal?

A

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 · 1G ∈ 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.

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4
Q

What are the left and right index?

A

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.

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5
Q

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

A

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.

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6
Q

How are 2 different cosets related?

A

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.

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7
Q

How do cosets form a partition?

A

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

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8
Q

What is Lagrange’s Theorem?

A

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].

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9
Q

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

A

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.

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10
Q

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

A

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.

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11
Q

How are the left and right index related?

A

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

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12
Q

What is a Conjugate of a subgroup?

A

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

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13
Q

What is a Normal subgroup?

A

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.

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14
Q

What are the equivalent statements about normal subgroups and conjugates?

A

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.

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15
Q

What index of subgroup will always be normal?

A

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.

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16
Q

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

A

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.

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17
Q

What is a Homomorphism of Groups?

A

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.

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18
Q

What is an Isomorphism of Groups?

A

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.

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19
Q

What are the Kernel and Image of a group homomorphism?

A

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}.

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20
Q

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

A

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.

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21
Q

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

A

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}.

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22
Q

What is the First Isomorphism Theorem?

A

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.

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23
Q

How is An related to Sn?

A

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

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24
Q

How are two cyclic groups of the same order related?

A

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

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25
Q

What group is a group of prime order related to?

A

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

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26
Q

What is the Direct Product of two groups?

A

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).

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27
Q

How is G × H a group?

A

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).

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28
Q

What are the groups of order 4?

A

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

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29
Q

How can we write D2n?

A

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.

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30
Q

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

A

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

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31
Q

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

A

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

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32
Q

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?

A

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 = rk for some k. We compute
(srs)(1) = (sr)(s(1)) = (sr)(1) = s(r(1)) = s(2) = n.
Hence srs = rn-1 = r-1.

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33
Q

What is a Word in a group?

A

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.

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34
Q

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

A

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.

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35
Q

What is a Group Presentation?

A

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.

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36
Q

What is the Fundamental Theorem of Group Presentations?

A

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.

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37
Q

What is the Quarternion Group?

A

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

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38
Q

How can we write elements of Q8?

A

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

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39
Q

What is a presentation for D2n?

A

D2n = 〈r, s |rn = s2 = id, srs = r-1

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40
Q

When do elements pairwise commute and when are they independent?

A

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.

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41
Q

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

A

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〉.

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42
Q

What is the exponent of a group?

A

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.

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43
Q

What type of group is a group with exponent 2?

A

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.

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44
Q

What group is a group with exponent 2 isomorphic to?

A

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

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45
Q

What is a group of order 6 isomorphic to?

A

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

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46
Q

What is a group of order 8 isomorphic to?

A

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

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47
Q

What is a left and right action?

A

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.

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48
Q

What is the orbit of an element?

A

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

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49
Q

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

A

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}.

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50
Q

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

A

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

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51
Q

When does G act transitively on X?

A

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.

52
Q

What is the stabiliser of an element?

A

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

53
Q

How does the stabiliser relate to G?

A

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

54
Q

What is the Orbit-Stabiliser Theorem?

A

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).

55
Q

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

A

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).

56
Q

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

A

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

57
Q

What is Cauchy’s Theorem?

A

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

58
Q

When are two elements conjugates?

A

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

59
Q

How does a group act on itself?

A

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

60
Q

What is the Conjugacy Class of an element?

A

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

61
Q

What is the Centraliser of an element?

A

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.

62
Q

What is Cycle Type?

A

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.

63
Q

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

A

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

64
Q

When are two permutations conjugate?

A

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

65
Q

What is a unit of a ring?

A

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.

66
Q

What is the unit group of a ring?

A

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}.

67
Q

What are the Gaussian Integers?

A

A Gaussian integer is a complex number whose real and imaginary parts are both integers.

ie Z[i] = {a + bi | a,b ∈ Z}

68
Q

What is the Norm Map?

A

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

69
Q

How does the norm map multiply?

A

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

70
Q

What is the unit group of Z[i]?

A

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}.

71
Q

What is a Ring Homomorphism and Isomorphism?

A

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.

72
Q

What is the kernel and image of a ring homomorphism?

A

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}.

73
Q

What is the First Isomorphism Theorem for rings?

A

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.

74
Q

When is an element in a ring a zero divisor?

A

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.

75
Q

What is an Integral Domain?

A

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.

76
Q

How are field related to integral domains?

A

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

77
Q

What is the Cancellation Law for Integral Domains?

A

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.

78
Q

What are all finite integral domains also?

A

Every finite integral domain is a field.

79
Q

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

A

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

80
Q

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

A

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.

81
Q

When are two elements of an integral domain associates?

A

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.

82
Q

What are irreducibles and primes in integral domains?

A

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.

83
Q

How are primes irreducible?

A

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.

84
Q

What is a Unique Factorisation Domain?

A

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.

85
Q

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

A

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).

86
Q

What is a Euclidean function?

A

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).

87
Q

What is a Euclidean ring?

A

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.

88
Q

Is the integers a Euclidean ring?

A

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

89
Q

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

A

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

90
Q

Are the Gaussian Integers a Euclidean ring?

A

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

91
Q

What is an Ideal?

A

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.

92
Q

When does an Ideal equal the Ring?

A

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

93
Q

What is the sum and product of ideals?

A

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).

94
Q

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

A

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.

95
Q

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

A

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).

96
Q

How is Ra an ideal?

A

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

97
Q

What is a Principal Ideal?

A

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.

98
Q

What is an ideal generated by multiple elements?

A

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.

99
Q

What are the equivalent statements about principal ideals?

A

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

100
Q

When do principal ideals equal each other or the ring?

A

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.

101
Q

What is a Principal Ideal Domain?

A

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.

102
Q

Are Euclidean domains also PIDs?

A

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

103
Q

What are the known PIDs?

A

Z
Z[i]
F[X]

104
Q

How do primes and irreducibles relate in a PID?

A

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

105
Q

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

A

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.

106
Q

Are PIDs Noetherian?

A

Let R be a PID. Then R is Noetherian.

107
Q

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

A

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

108
Q

How do PIDs relate to UFD?

A

Every PID is a UFD.

109
Q

How do EDs, PIDs and UFDs relate?

A

ED =⇒ PID =⇒ UFD.

110
Q

What is an R-Module?

A

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.

111
Q

What is an R-submodule?

A

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.

112
Q

What is the Quotient Module?

A

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

113
Q

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

A

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).

114
Q

What is a homomorphism of modules?

A

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.

115
Q

What is the Isomorphism Theorem?

A

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.

116
Q

What is the span of a subset of a module?

A

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.

117
Q

How does a module being finitely generated relate to homomorphisms?

A

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.

118
Q

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

A

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.

119
Q

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

A

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.

120
Q

What is the Fundamental Theorem of Finitely Generated Abelian Groups?

A

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∞.

121
Q

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

A

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

122
Q

How is a submodule of Rn generated?

A

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.

123
Q

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

A

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.

124
Q

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

A

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.

125
Q

How can we find the SNF of a matrix?

A

(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.

126
Q

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

A

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∗
.