Algebra 1

Question 1

What is a binary operation?

Answer 1

Let S be a set. A binary operation on S is a rule by which any two elements of S can be combined to give another element of S.
We are going to use the symbol * for binary operations. It’s use will mostly be
reserved for when we are talking about a general, non-specified, binary operation.
So given s1, s2 ∈ S we have a further element s1 * s2 ∈ S.

Question 2

What does it mean for a binary operation to be associative and commutative?

Answer 2

Let S be a set and * a binary operation.
We say that the binary operation * is commutative on S if a * b = b * a for all
a, b ∈ S.
We say that the binary operation * is associative on S if (a * b) * c = a * (b * c)
for all a, b, c ∈ S.

Question 3

What is a group?

Answer 3

A group is a pair (G, *) where G is a set and * is a binary operation on G, such that the following four properties hold:
(i) (closure) for all a, b ∈ G, a * b ∈ G;
(ii) (associativity) for all a, b, c ∈ G, a * (b * c) = (a * b) * c;
(iii) (existence of the identity element) there is an element e ∈ G such that for all a ∈ G, a * e = e * a = a;
(iv) (existence of inverses) for every a ∈ G, there is an element b ∈ G (called the inverse of a) such that a * b = b * a = e.

Question 4

What is an Abelian group?

Answer 4

We say that a group (G, *) is abelian if , in addition to properties (i)->(iv) it also satisfes
(v) (commutativity) for all a, b ∈ G, a * b = b * a.

Question 5

What is GL2(R)?

Answer 5

The group of 2x2 invertible matrices with real entries.
It is a group under matrix multiplication.

Question 6

Is the identity element of a group Unique?

Answer 6

Let (G, *) be a group. Then (G, *) has a unique identity element.

Proof. Suppose that e and e’ are identity elements. Thus, for all a ∈ G we have
a * e = e * a = a (1)
and
a * e’ = e’ * a = a. (2)
Now let us try evaluating e * e’ . If we let a = e and use (2) we find
e * e’ = e.
But if we let a = e’ and use (1) we find
e * e’ = e’
Thus e = e’. In other words, the identity element is unique.

Question 7

Is the inverse in a group Unique?

Answer 7

Let (G, *) be a group and let a be an element of G. Then a has a unique inverse.

Proof. Our proof follows the same pattern as the proof that the identity is unique, and you’ll see this pattern again and again during your undergraduate career. Almost all uniqueness proofs follow the same pattern: suppose that there are two of the thing that we want to prove unique; show that these two must be equal; therefore it is unique.
For our proof we suppose that b and c are both inverses of a. We want to show that b = c. By definition of inverse (property (iv) in the definition of a group) we know that
a * b = b * a = e, a * c = c * a = e,
where e is of course the identity element of the group. Thus
b = b * e by (iii) in the definition of a group
= b * (a * c) from the above a * c = e
= (b * a) * c by (ii) in the definition of a group
= e * c from the above b * a = e
= c by (iii) again.
Thus b = c. Since any two inverses of a must be equal, we see that the inverse of a is unique.

Question 8

What is the inverse of an inverse element of a group?

Answer 8

Let G be a group and a ∈ G. Then (a^−1)^−1 = a.
Proof. We’re being asked to prove that a is the inverse of a^−1. Thinking carefully about what this would mean, we want to show that a^−1a = 1 = aa−1
But this is clearly true because a^−1 is the inverse of a.

Question 9

What is the inverse of (ab)?

Answer 9

Let G be a group and a, b ∈ G. Then(ab)^−1 = b^−1a^−1.

Proof. We’re being asked to prove that b−1a−1is the inverse of ab. So we want to show that (b^−1^a−1)(ab) = 1 = (ab)(b^−1a^−1).
Now (b^−1a^−1)(ab) = b^−1(a^−1a)b by associativity
= b^−11b
= 1,
and similarly (ab)(b^−1a^−1) = 1.

Question 10

What is the theorem for a^n (ie composition of a group element n times with itself)?

Answer 10

Let G be a group, and let a ∈ G. Then
1. a^n ∈ G for all n ∈ Z.
2. If n ∈ Z then (a^−1^)n = (a^n)^−1 = a^−n.
3. Moreover, if m, n are integers then (a^m)^n = a^mn, a^ma^n = a^(m+n).
4. Further, if the group G is abelian, a, b ∈ G and n an integer then (ab)^n = a^nb^n.

Question 11

What is the order of a group element?

Answer 11

The order of an element a in a group G is the smallest positive integer n such that a^n = 1. If there is no such positive integer n, we say a has infinite order.
In a finite group (a group with a finite number of elements) every element must have finite order.

Question 12

Does every element in a finite group have finite order?

Answer 12

Let G be a finite group and g be an element of G. The g has finite order.
Proof. Suppose g ∈ G has infinite order. Then g0, g1, g2, g3, . . . are elements of
G which are distinct from one another, since if gm = gn for some natural numbers m,n with m < n then e = g0 = g^(m−m) = g^(n−m) and g has finite order.
This cannot happen in a finite group and so g must have finite order

Question 13

What are the rules for orders of an element?

Answer 13

Let G be a group and g be an element of G.
(i) g has order 1 if and only if g is the identity element.
(ii) Let m be a non-zero integer. Then g^m = 1 if and only if g has finite order d with d | m.

Proof. Let G be a group. Suppose g has order 1. By definition of order, g^1 = 1.
Thus g = 1 which is the identity element of G. Conversely, the identity element
clearly has order 1. This proves (i).
Part (ii) is an `if and only if’ statement. Suppose that g has order d and d | m.
Then g^d = 1 and m = qd where q is an integer. So g^m = (g^d)^q = 1. Let us prove
the converse. Suppose g^m = 1 where m is a non-zero integer. Then g^|m| = 1, and
|m| is a positive integer. Thus g has finite order, which we denote by d. By the
division algorithm which you met in Foundations we may write
m = qd + r, q, r ∈ Z and 0 ≤ r < d.
Now g^d = 1 by definition of order, so 1 = g^m = (g^d)^q · g^r = g^r. But 0 ≤ r < d.
As d is the order, it is the least positive integer such that g^d = 1. So g^r = 1 is
possible with 0 ≤ r < d if and only if r = 0. This happens if and only if m = qd
which is the same as d | m.

Question 14

What is the order of a Group?

Answer 14

Let G be a group. The order of G is the number of elements that G has. We denote the order of G by |G| or #G.

Question 15

When does an element have order 2?

Answer 15

Let G be a group and 1 /= g ∈ G. Then g has order 2 if and only if g = g^−1.

By definition, if g has order 2 then g^2 = 1. Multiplying both sides of this on the left by g^−1 gives g = g^−1g^2 = g^−1.
For the converse suppose that g = g^−1. Since g 6= 1, the order of g is not 1.
Multiplying both sides of g = g^−1 on the left by g gives g^2 = gg−1 = 1. Therefore
the order of g is 2.

Question 16

What is a subgroup?

Answer 16

Let (G, *) be a group. Let H be a subset of G and suppose that (H, *) is also a group. Then we say that H is a subgroup of G (or more formally (H, *) is a subgroup of (G, *)).

For H to be a subgroup of G, we want H to a group with respect to the same binary operation that makes G a group.

Question 17

What are the subgroup criterion?

Answer 17

Let G be a group. A subset H of G is a subgroup if and only if it satisfies the following three conditions:
(a) 1 ∈ H,
(b) if a, b ∈ H then ab ∈ H,
(c) if a ∈ H then a^−1 ∈ H.

Question 18

What is Un?

Answer 18

Un = {1, ζ, ζ^2, . . . , ζ^(n−1)}.
That is, Un is the set of n-th roots of unity.

Question 19

What are proper non-trivial subgroups?

Answer 19

{1} is the trivial subgroup
For a group G, any subgroup not equal to G is a proper subgroup.

Question 20

What is a Cyclic Subgroup?

Answer 20

Theorem Let G be a group, and let g be an element of G. Write <g> for the set</g>

<g> = {g^n | n ∈ Z} = {. . . , g−2, g−1, 1, g, g2, g3, . . . }.
Then <g> is a subgroup of G.

For the proof just go through the subgroup criterion.

We call <g> the cyclic subgroup of G generated by g.
</g></g></g>

Question 21

Are cyclic groups abelian?

Answer 21

Theorem Cyclic groups are abelian.
Proof. Let G be a cyclic group generated by g. Let a, b be elements of G. We want to show that ab = ba. Now, a = g^m and b = g^n for some integers m and n.
So, ab = g^mg^n = g^m+n and ba = g^ng^m = g^n+m.
But m + n = n + m (addition of integers is commutative). So ab = ba.

Question 22

How does the order of a cyclic subgroup relate to the order of its generator?

Answer 22

Let G be a group and let g be an element of finite order n. Then

<g> = {1, g, g2, . . . , gn−1}.
In particular, the order of the subgroup <g> is equal to the order of g.

For the proof show that <g> and {1, g, g2, . . . , gn−1} are equal by showing both are subsets of one another.
=> is trivial
<= use division with remainder
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Question 23

Are congruence classes groups?

Answer 23

Let n be an integer satisfying n ≥ 1. Then (Z/nZ, +) is an abelian group.

Question 24

What is Euclidean Distance?

Answer 24

Let a, b ∈ R^2 so that a = (a1, a2) and b = (b1, b2) where a1, a2, b1, b2 ∈ R. Then the (Euclidean) distance between a and b is given by
|a − b| = sqrt((a1 − b1)^2 + (a2 − b2)^2).

Question 25

What is an Isometry?

Answer 25

Let f : R^2 → R^2 be a function such that for any a, b ∈ R^2,
|f(a) − f(b)| = |a − b|. Then f is called an isometry of R^2.
This just means that, in an isometry, the distance between any two given points
and the distance between their image points is the same. You may hear such maps
referred to as being distance preserving.

Question 26

How do isometries commute?

Answer 26

Let f : R^2 → R^2 and g : R^2 → R^2 be isometries of R^2. Then
g ◦ f : R^2 → R^2 is an isometry of R^2

For proof just use definition of isometry with g(f(a)).

Question 27

What is the special isometry case?

Answer 27

Let f : R^2 → R^2 be a function such that
1. f((0, 0)) = (0, 0) and f((1, 0)) = (1, 0)
2. f is an isometry of R^2
Then f is either the identity map (i.e. f((x, y)) = (x, y) for all x, y ∈ R or it is a reflection in the x-axis (i.e f((x, y)) = (x, −y) for all x, y ∈ R).

Question 28

When is an isometry a reflection or rotation?

Answer 28

Let f : R^2 → R^2 be a function such that
1. f((0, 0)) = (0, 0)
2. f is an isometry of R^2
Then f is either a rotation about the origin or it is a reflection in the x-axis
followed by a rotation about the origin.

Question 29

What is the orthogonal group?

Answer 29

The set of isometries of R^2, here identified with C, which fix the origin is a group under composition of functions. We call this group O2(R), the orthogonal group on R^2

Question 30

What is the special orthogonal group?

Answer 30

The set of isometries, f : R^2 → R^2 of the form f(z) = e^iθz, where R^2 has been identified with C, is a subgroup of O2(R). This is the subgroup of all the rotations about the origin. It is called SO2(R), the special orthogonal group on R^2
.

Question 31

What is the MAP of a set A?

Answer 31

The set of all functions from A to itself.

Question 32

What is the SYM of a set A?

Answer 32

The set of all bijections from A to itself.

Question 33

Is SYM(A) a group with respect to composition?

Answer 33

Let A be a set. Then (Sym(A), ◦) is a group with idA (the bijection that takes every element to itself) as the identity element.

For proof just go through usual group checks.

Question 34

What is Sn?

Answer 34

We define Sn to be the group Sym({1, 2, . . . , n}). We call Sn the n-th symmetric group.
The elements of Sn are called permutations.

Question 35

What is the order of Sn?

Answer 35

Sn has order n!.

Proof is simple.

Question 36

What is Cycle Notation?

Answer 36

Let a1, a2, . . . , am be distinct elements of the set {1, 2, . . . , n}. By the notation
(a1, a2, . . . , am) (4)
we mean the element of Sn that takes a1 to a2, a2 to a3, . . . , am−1 to am and am
back to a1, and fixes all other elements of {1, 2, . . . , n}. The permutation (4) is
called a cycle of length m.

Question 37

What is a Transposition?

Answer 37

A cycle of length 2 is called a transposition.

Question 38

How can Permutations be written in terms of Cycles?

Answer 38

Every permutation can be written as a product of disjoint cycles.

Question 39

How do cycles commute?

Answer 39

Disjoint cycles commute.

Proof:
Let σ and τ be disjoint cycle in Sn and write σ = (a1, a2, . . . , ak), τ = (b1, b2, . . . , bl).
Since σ and τ are disjoint ai /= bj for i = 1, . . . , k and j = 1, . . . , l.
We want to show that στ = τ σ. This means that στ x = τ σx for all x ∈ {1, 2, . . . , n}. We subdivide into three cases:
Case 1: x does not equal any of the ai or bj. Then τ x = x and σx = x. Therefore
στ x = σx = x = τ x = τ σx.
Case 2: x = ai for some i = 1, . . . , k. Thus x does not equal any of the bj, and so
τ x = x. Hence στ x = σx = σai = ai+1; here ak+1 is interpreted as being a1. Let’s
compute τ σx. This is τ σai = τ ai+1 = ai+1 since ai+1 does not equal any of the bj.
Hence στ x = τ σx.
Case 3: x = bj for some j = 1, . . . , l. This is similar to Case 2.
We conclude that στ = τ σ as required.

Question 40

What does it mean for cycles to be disjoint?

Answer 40

Two cycles (a1, a2, . . . , an) and (b1, b2, . . . , bm) are said to be disjoint if ai /=/ bj
for all integers i, j with 1 ≤ i ≤ n and 1 ≤ j ≤ m.
i.e. No element appears in both cycles.

Question 41

How can permutations be written in terms of transpositions?

Answer 41

Every permutation can be written as a product of transpositions.
Proof. We know that every permutation can be written a product of cycles. So
it is enough to show that a cycle can be written as a product of transpositions.
Check for yourself that (a1, a2, . . . , am) = (a1, am)· · ·(a1, a3)(a1, a2).

Question 42

How can permutations be written as even/odd transpositions?

Answer 42

Every permutation in Sn can be written as a product of either an even number of
transpositions, or an odd number of transpositions but not both.

Question 43

When is a permutation even/odd?

Answer 43

We shall call a permutation even if we can write it as a product of an even number of transpositions, and we shall call it odd if we can write it as a product of an odd number of transpositions.

Question 44

What is An?

Answer 44

Let n ≥ 2. We define the n-th alternating group to be
An = {σ ∈ Sn | σ is even}.

Question 45

How does An relate to Sn?

Answer 45

An is a subgroup of Sn.

We’ve already seen that the identity element id is even, so id ∈ An. If σ, ρ ∈ An then we can write each as an even number of transpositions. Therefore the product σρ can be written as an even number of transpositions (even+even=even). Hence σρ ∈ An.
Finally we must show that the inverse of an even permutation is even. Suppose
σ is even. We can write σ = τ1τ2 . . . τm
where the τi are transpositions, and m is even. Now
σ^−1 = (τ1τ2 · · · τm)^−1= τm^-1 τm−1^-1· · · τ1^-1= τmτm−1 · · · τ1.
Here you should convince yourself that τ−1 = τ for any transposition τ . Since m
is even, we find that σ^−1 is even and so σ^−1 ∈ An.
Hence An is a subgroup of Sn.

Question 46

What is an Isomorphism?

Answer 46

Let (G, *) and (H, #) be groups. We say that the function
φ : G → H is an isomorphism if it is a bijection and it satisfies
φ(g1 * g2) = φ(g1) # φ(g2)
for all g1, g2 in G. In this case we say that (G, *) and (H, #) are isomorphic.

Ie. isomorphic groups may look different but the way the elements interact is the same. An isomorphism is a way of relabelling the elements of one group to obtain another group.

Question 47

What is a binary operation on a product of groups?

Answer 47

Given groups (G, *), (H, #) define a binary operation . on G × H = {(g, h) | g ∈ G, h ∈ H} as follows.
If (g1, h1),(g2, h2) ∈ G × H then
(g1, h1) . (g2, h2) = (g1 * g2, h1 # h2).
Then (G × H, .) is a group called the the direct product of G and H.

Question 48

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

Answer 48

Let G be a finite group. Let g ∈ G. Then the order of g divides the order of G.

Question 49

What is a Homomorphism?

Answer 49

Let (G, *) and (H, #) be groups. We say that the function
φ : G → H is a homomorphism if it satisfies
φ(g1 * g2) = φ(g1) # φ(g2) for all g1, g2 in G.

Question 50

What is the Kernel and Image of a homomorphism?

Answer 50

Let G and H be groups and let φ : G → H be a homomorphism. Then
1. Ker φ = {g ∈ G | φ(g) = 1H} is called the kernel of φ.
2. Im φ = {h ∈ H | there exists g ∈ G with φ(g) = h} = {φ(g) | g ∈ G} is called the image of φ

Question 51

How does the identity in both groups map to each other in a homomorphism?

Answer 51

Let G and H be groups and let φ : G → H be a homomorphism. Then φ(1G) = 1H (here we are using 1G and 1H for the respective identity elements rather than the usual 1 for emphasis).

Proof. We have 1G1G = 1G. Therefore, since φ is a homomorphism, we have
φ(1G) = φ(1G1G) = φ(1G)φ(1G) ().
But φ(1G) is just an element of H and as such has an inverse in H, φ(1G)^−1.
Multiplying both sides of () on the left by φ(1G)^−1 gives
1H = φ(1G)^−1φ(1G) = φ(1G)^−1φ(1G)φ(1G) = 1Hφ(1G) = φ(1G)

Question 52

How do the Kernel and Image of a homomorphism relate to its initial groups?

Answer 52

Let G and H be groups and let φ : G → H be a homomorphism. Then
1. Ker φ = {g ∈ G | φ(g) = 1} is a subgroup of G
2. Im φ = {θ(g) | g ∈ G is a subgroup of H.

Proof
Just run through the subgroup criterion.

Question 53

What is a ring?

Answer 53

A ring is a triple (R, +, ·), where R is a set and +, · are binary operations on R such that the following properties hold:
(i) (closure) for all a, b ∈ R, a + b ∈ R and a · b ∈ R;
(ii) (associativity of addition) for all a, b, c ∈ R
(a + b) + c = a + (b + c);
(iii) (existence of an additive identity element) there is an element 0 ∈ R such that for all a ∈ R,
a + 0 = 0 + a = a.
(iv) (existence of additive inverses) for all a ∈ R, there an element, denoted by
−a, such that
a + (−a) = (−a) + a = 0;
(v) (commutativity of addition) for all a, b ∈ R,
a + b = b + a;
(vi) (associativity of multiplication) for all a, b, c ∈ R,
a · (b · c) = (a · b) · c;
(vii) (distributivity) for all a, b, c ∈ R,
a · (b + c) = a · b + a · c; (b + c) · a = b · a + c · a;
(viii) (existence of a multiplicative identity) there is an element 1 ∈ R so that for
all a ∈ R,
1 · a = a · 1 = a.

Question 54

When is a ring said to be commutative?

Answer 54

A ring (R, +, ·) is said to be commutative, if it satisfies the following additional property:
(ix) (commutativity of multiplication) for all a, b ∈ R,
a · b = b · a.

Question 55

How are equivalence classes rings?

Answer 55

Let m be an integer satisfying m ≥ 2. Then Z/mZ is a ring

Question 56

What is a Subring?

Answer 56

Let (R, +, ·) be a ring. Let S be a subset of R and suppose that (S, +, ·) is also a ring with the same multiplicative identity. Then we say that S is a subring of R (or more formally (S, +, ·) is a subring of (R, +, ·))

The easiest way to show that a set is a ring is to show that it is a subring of a known ring.

Question 57

What are the Subring Criterion?

Answer 57

Let R be a ring. A subset S of R is a subring if and only if it satisfies the following conditions:
(a) 0, 1 ∈ S (that is S contains the additive and multiplicative identity elements
of R);
(b) if a, b ∈ S then a + b ∈ S;
(c) if a ∈ S then −a ∈ S;
(d) if a, b ∈ S then ab ∈ S.

Question 58

What is an Ideal?

Answer 58

Let R be a ring. Let I be a subset of R. Then I is said to be an (two-sided) ideal of R if
1. (I, +) is a subgroup of (R, +)
2. For every x ∈ I and r ∈ R, both xr ∈ I and rx ∈ I

Question 59

When does an Ideal equal its Ring?

Answer 59

Let R be a non-zero ring and let I be an ideal of R. Then I = R if and only if 1 ∈ I.

Also, let R be a non-zero ring and let I be an ideal of R. Then I = R if and only if I contains a unit in R.

Question 60

What is a Unit?

Answer 60

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 61

What are the Ideals of Z?

Answer 61

Let I be an ideal of Z. Then I = nZ = {nm |m ∈ Z} for some n ∈ Z.

Question 62

What is a Field?

Answer 62

A field (F, +, ·) is a commutative ring which is not the zero ring such that every non-zero element is a unit.

Question 63

What is the degree of a polynomial?

Answer 63

Let F be a field. Let f(x) ∈ F[x]. Suppose
f(x) = anx^n + an−1x^n−1 + · · · + a1x + a0
where an /= 0. Then we define the degree of f(x) to be n. We write deg(f(x)) = n. If f(x) = 0, the zero polynomial, we define deg(f(x)) = −∞.

Question 64

What are the known propositions about polynomials?

Answer 64

Let F be a field. Then
1. if f(x), g(x) ∈ F[x] then deg(f(x)g(x)) = deg(f(x)) + deg(g(x)) (here we apply the conventions that −∞ + n = −∞ = n + (−∞) for any n ∈ N and that −∞ + (−∞) = −∞).
2. f(x) ∈ F[x] is a unit in F[x] if and only if deg(f(x)) = 0.
3. if f(x), g(x) ∈ F[x] are such that f(x)g(x) = 0 then either f(x) = 0 or g(x) = 0.
4. if f(x), g(x) ∈ F[x] with 0 ≤ deg(g(x)) ≤ deg(f(x)) then there exist q(x), r(x) ∈ F[x] with deg(r(x)) < deg(g(x)) and f(x) = g(x)q(x) + r(x).

Question 65

What is the Factor Theorem?

Answer 65

Let F be a field. Let 0 /= f(x) ∈ F[x] and α ∈ F. Then the evaluation of f(x) at x = α is zero (f(α) = 0) if and only if there exists a polynomial q(x) ∈ F[x] such that f(x) = (x − α)q(x).

Question 66

When is a polynomial said to be irreducible?

Answer 66

Let 0 /= f(x) ∈ F[x]. Then f(x) is said to be irreducible over F if whenever f(x) = g(x)h(x) with g(x), h(x) ∈ F[x] then either g(x) or h(x) is a constant polynomial (but not both).

Note that the above definition says that we can’t have both g(x) and h(x) being constant polynomials which means that n irreducible polynomial f(x) cannot be a constant polynomial and, since f(x) /= 0 either, it has to have degree greater than or equal to 1.

Probably the easiest way of all to think about this is that 0 /= f(x) ∈ F[x] is irreducible if it has degree at least one and you cannot write it as the product of two polynomials in F[x] both with smaller degree.

Question 67

When does a polynomial divide another?

Answer 67

Let F be a field. Let f(x), g(x) ∈ F[x]. We say that f(x) divides g(x) and write f(x)|g(x) if there exists h(x) ∈ F[x] such that f(x)h(x) = g(x).

Question 68

When are two polynomials relatively prime?

Answer 68

Let F be a field. Let f(x), g(x) ∈ F[x] be non-zero polynomials. We say that f(x) and g(x) are relatively prime if whenever 0 /= h(x) ∈ F[x] with h(x)|f(x) and h(x)|g(x) then deg(h(x)) = 0, i.e. h(x) is a unit.

Question 69

What is the equivalent of Bezout’s Lemma for polynomials?

Answer 69

Let F be a field. Let f(x), g(x) ∈ F[x] be non-zero polynomials with f(x) and g(x) relatively prime. Then there exist h∗(x), k∗(x) ∈ F[x] such that
f(x)h∗(x) + g(x)k∗(x) = 1.

Question 70

How does a polynomial divide a product of others?

Answer 70

Let F be a field. Let f(x) ∈ F[x] be an irreducible over F. Suppose that g(x), h(x) ∈ F[x] and f(x)|g(x)h(x). Then f(x)|g(x) or f(x)|h(x).

Question 71

How can a polynomial be written as irreducible polynomials?

Answer 71

Let F be a field. Let f(x) ∈ F[x] with deg(f(x)) ≥ 1. Then f(x) is expressible as a product of polynomials which are irreducible over F. This expression is unique in the following sense. Suppose f(x) = g1(x)g2(x). . . gm(x) = h1(x)h2(x). . . hn(x)
where gi(x), hj (x) are all irreducible over F then m = n and, after possibly renumbering the hj (x) polynomials, g1(x) = a1h1(x), g2(x) = a2h2(x),. . . gn(x) = anhn(x), where 0 6= ai ∈ F, i.e. each ai is a unit in F[x].

Question 72

What is the Fundamental Theorem of Algebra?

Answer 72

Let f(x) ∈ C[x] be a polynomial with deg(f(x)) ≥ 1. Then there exists α ∈ C such that f(α) = 0. In other words, the equation f(x) = 0 has a root in the complex numbers.

Question 73

What is Eisenstein’s Criterion for Irreducibility?

Answer 73

Let f(x) = anx^n + an−1x^n−1 + · · · + a1x + a0 ∈ Z[x].
Suppose there exists a prime number p such that
1. p|ai for each i with 0 ≤ i < n
2. p does not divide an
3. p^2 does not divide a0.
Then f(x) is irreducible over Q.

Question 74

What is a Coset of an Ideal in a Ring?

Answer 74

Let R be a ring and I an ideal of R. Let r be an element of R. We call the set r + I = {r + x | x ∈ I} a coset of I in R.

Question 75

When are two cosets equal?

Answer 75

Let R be a ring and let I be an ideal. Let x, y ∈ R so that x + I and y + I are cosets. Then x + I = y + I if and only if x − y ∈ I.

Question 76

What is a Quotient Ring?

Answer 76

Let R be a ring and I an ideal of R. We define the quotient ring R/I to be the set of cosets
R/I = {x + I | x ∈ R}
with addition and multiplication defined by
(x + I) + (y + I) = (x + y) + I.
(x + I)(y + I) = (xy) + I.

Question 77

Is a Quotient Ring a Ring?

Answer 77

Let (R, +, .) be ring and I an ideal of R. Then (R/I, +, .) is a ring.

For proof just go through ring criterion.

Question 78

What is a Ring Homomorphism?

Answer 78

Let (R1, +R1, ×R1) and (R2, +R2, ×R2) be rings. We say that the function φ : R1 → R2 is a (ring) homomorphism if
φ(1R1) = 1R2 and it satisfies
φ(r +R1 s) = φ(r) +R2 φ(s) and φ(r ×R1 s) = φ(r) ×R2 φ(s)
for all r, s in R1.
If, in addition, φ is a bijection. Then it is said to be an (ring) isomorphism.

Question 79

What is the Kernel and Image of a Ring Homomorphism?

Answer 79

Let R1 and R1 be rings and let φ : R1 → R2 be a ring homomorphism. Then
1. Ker φ = {r ∈ R1 | φ(r) = 0R2 } is called the kernel of φ.
2. Im φ = {s ∈ R2 | there exists r ∈ R1 with φ(r) = s} = {φ(r) | r ∈ R1} is called the image of φ

Question 80

How are the kernel and image of a ring homomorphism related to ideals and subrings?

Answer 80

Let R1 and R2 be rings and let φ : R1 → R2 be a homomorphism. Then
1. Ker φ = {g ∈ G | φ(g) = 1} is an ideal of R1
2. Im φ = {θ(g) | g ∈ G} is a subring of R2.