Download Introduction to abstract algebra: definitions, examples, and exercises

Introduction to abstract algebra:
definitions, examples, and exercises
Travis Schedler
January 21, 2015
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Definitions and some exercises
Definition 1. A binary operation on a set X is a map X × X → X, (x, y) 7→ x · y, also
denoted xy.
The operation is called the multiplication or composition.
Definition 2. A magma (X, ·) is a set together with a binary operation · (a·b is the operation
for a, b ∈ X).
Definition 3. A semigroup is a magma whose operation is associative.
In other words, we could say that a semigroup is an associative magma.
Definition 4. An identity, or unit, element of a magma X is an element e such that ea =
a = ae for all a ∈ X.
Note that, if an identity e exists, it is unique, since if e and e0 are both identity elements,
then e0 = ee0 = e.
Definition 5. A monoid is a semigroup for which there exists an identity.
Definition 6. A group G is a monoid for which, for every element a ∈ G, there exists b ∈ G
such that ab = e = ba.
Such an element b is unique, since if ab0 = e = b0 a as well, then b = be = bab0 = eb0 = b0 .
Definition 7. An abelian (or commutative) group is a group for which the operation is
commutative.
Definition 8. A ring (R, +, ·), is a set R equipped with two operations, called addition +
and multiplication ·, such that (R, +) is an abelian group and (R, ·) is a monoid.
The unit for addition is denoted 0 and the unit for multiplication is denoted 1. The
additive inverse of an element a is denoted −a, so (−a) + a = 0 = a + (−a).
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Definition 9. A commutative ring R is a ring for which the multiplication · is commutative:
ab = ba for all a, b ∈ R.
Note that, since addition is always commutative, when we say the ring is commutative,
it is clear commutativity is applying to the multiplication rather than the addition.
Definition 10. A division ring (R, +, ·) is a ring such that every nonzero element a has a
multiplicative inverse.
In other words, the monoid R× , defined as the set of nonzero elements of R under
multiplication, is a group.
Definition 11. A field is a commutative division ring.
Definition 12. A homomorphism (or morphism) of magmas f : (X, cdot) → (Y, ·) is a map
f : X → Y such that f (ab) = f (a)f (b) for all a, b ∈ X.
The set of homomorphisms from X to Y is denoted Hom(X, Y ) (the same holds when
magmas are replaced by any other of the structures below for which we define homomorphisms).
Definition 13. A homomorphism (or morphism) of semigroups is a map f : X → Y which
is a homomorphism of magmas.
In other words, the condition to be a homomorphism is the same as for magmas; we only
require that the source and target are associative.
Definition 14. A homomorphism (or morphism) of monoids is a map f : X → Y which
is a homomorphism of magmas (equivalently, of semigroups) such that f (eX ) = eY , where
eX ∈ X is the unit in X and eY ∈ Y is the unit in Y .
Definition 15. A homomorphism (or morphism) of groups is a map f : G → H which is a
homomorphism of monoids.
Exercise 16. (i) Show that a homomorphism of groups also has the property that f (a−1 ) =
f (a)−1 for all a ∈ G. (ii) Show that the condition that f (eG ) = eH is redundant for a group.
More precisely, show that, if G and H are groups and f : G → H is a morphism of magmas,
then actually f (eG ) = eH and hence f is a morphism of groups (here eG ∈ G and eH ∈ H
are the identity elements).
Thus, a homomorphism of groups is the same as a map f such that f (ab) = f (a)f (b) for
all a, b ∈ G.
Exercise 17. Show that this is not true for a monoid: give an example of monoids (M, ·)
and (N, ·) with units eM and eN , and a morphism of semigroups M → N which does not
satisfy f (eM ) = eN , i.e., is not a morphism of monoids. Hint: You can try N =a monoid
with two elements e, x such that x2 = x and e is the identity.
Definition 18. An endomorphism of X is a homomorphism from X to itself. They are
denoted End(X). These are always equipped with composition: f ◦ g ∈ End(X) is given by
(f ◦ g)(x) = f (g(x)).
Exercise 19. If A is an abelian group, show that End(A) forms a ring where, for ϕ, ψ ∈
End(A), the sum is given by (ϕ + ψ)(a) = ϕ(a) + ψ(a), and the product is as usual given by
composition. Why does the formula for sum not work when A is not abelian?
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Some examples
Example 20. Some examples of fields: Q, R, C.
Example 21. Z is a commutative ring but not a division ring (hence not a field.) Z≥0 is
a commutative monoid under addition (with identity 0) and a commutative monoid under
multiplication (with identity 1). Z>0 is a commutative semigroup under addition (but not a
monoid) and a commutative monoid under multiplication (with identity 1).
Example 22. The nonzero real numbers, R× , form an abelian group under multiplication.
The same is true for Q× and C× , or more generally F × for any field F (again denoting the
nonzero elements of F under multiplication).
Example 23. The n × n matrices over R, Matn (R) (or, Matn (R) where R is any ring) form
a ring under matrix addition and multiplication. The additive identity is the zero matrix,
and the multiplicative identity is the identity matrix.
Example 24. The invertible n × n matrices over R, GLn (R) (or GLn (R) for any ring R)
form a group which is not abelian.
Example 25. The m × n matrices over any ring form an abelian group under matrix sum,
but the matrix multiplication does not make sense when m 6= n.
Example 26. A student asked for an example of a division ring which is not a field (i.e., not
commutative). Here is an example: the quaternions H := {a + bI + cJ + dK | a, b, c, d ∈ R},
with the multiplication which is R-linear satisfying I 2 = J 2 = K 2 = −1 and IJ = K, JK =
I, KI = J.
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Isomorphisms and permutations
Let X and Y have any type of structure.
Definition 27. An isomorphism is an invertible morphism, i.e., a morphism f : X → Y
such that there is an inverse morphism g : Y → X, with f ◦ g the identity on Y and g ◦ f
the identity on X.
In the case of sets, an invertible map of sets f : X → Y is the same as a bijective map,
i.e., one which is one-to-one and onto.
Definition 28. A permutation of a set X is an invertible (equivalently bijective) map p :
X → X. Let Perm(X) denote the set of permutations of X.
Exercise 29. Show that Perm(X) is a group under composition ◦.
One of the main constructions of the course is that of a group action: this is an action
G × X → X, (g, x) 7→ g · x of a group G on a set X satisfying (gh) · x = g · (h · x) and e · x = x
for all x ∈ X.
Exercise 30. Show that a group action is the same as a homomorphism a : G → Perm(X),
with a(g) the map (a(g))(x) = g · x. Hint: The main step is to see that the condition
e · x = x for all x ∈ X is equivalent to the condition that the map a(g) is bijective (hence
a permutation) for all g ∈ G. To show one implication, if e · x = x for all x ∈ X, then
a(g) ◦ a(g −1 ) = a(g ◦ g −1 ) = a(e) is the identity. To show the other implication, if a(e) is
bijective, then since also a(e) ◦ a(e) = a(e2 ) = a(e), we conclude that a(e)(a(e)(x)) = a(e)(x)
for all x ∈ X, but a(e) : X → X is surjective, so that a(e)(X) = X, and hence the identity
says that a(e) is the identity.
Exercise 31. Show that, similarly to maps of sets, a homomorphism f : X → Y of any of
the types of structures above (magmas, semigroups, monoids, groups, abelian groups, rings,
division rings, commutative rings, fields) is invertible if and only if it is bijective. Hint: The
main step is to show that, if f is bijective, and g : Y → X is the inverse map of sets, then g
is also a homomorphism of the structure in question.
Example 32. Here is an example of a type of morphism for which being bijective is weaker
than being invertible: differentiable functions from R to R. For example, if f : R → R
is the function f (x)
= x3 , this is bijective and differentiable. But the inverse function
√
f −1 is f −1 (x) = 3 x, which is not differentiable at zero. So, if we restrict our attention
to differentiable functions, then not all bijective differentiable functions are invertible (by
differentiable functions).
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Modules and vector spaces
The analogue of a group action on a set for rings is an action of a ring on an abelian group.
This is called a module:
Definition 33. A module over a ring R is an abelian group M equipped with an action
R × M → M, (r, m) 7→ r · m, satisfying the axioms
(rs)(m) = r(sm),
1(m) = m,
(r+s)(m) = rm+sm,
r(m+n) = rm+rn,
∀r, s ∈ R, m, n ∈ M.
Exercise 34. Analogously to Exercise 30, show that a module is the same as a ring homomorphism R → End(M ), where End(M ) is the ring of abelian group endomorphisms of M .
See Exercise 19 for why End(M ) is a ring.
Definition 35. A vector space V over a field F is the same thing as a module over F (it is
just called a vector space in the case the ring is actually a field).