22. A quick primality test
Prime numbers are one of the most basic objects in mathematics
and one of the most basic questions is to decide which numbers are
prime (a clearly related problem is to nd the prime factorisation of a
number). Given a number n o
21. Polynomial rings
Let us now turn out attention to determining the prime elements
of a polynomial ring, where the coecient ring is a eld. We already
know that such a polynomial ring is a UFD. Therefore to determine
the prime elements, it suces to deter
23. Group actions and automorphisms
Recall the denition of an action:
Denition 23.1. Let G be a group and let S be a set.
An action of G on S is a function
G S S
(g, s) g s,
(gh) s = g (h s)
In fact, an action of G on a set S
17. Field of fractions
The rational numbers Q are constructed from the integers Z by
adding inverses. In fact a rational number is of the form a/b, where a
and b are integers. Note that a rational number does not have a unique
representative in this way.
4. Cyclic groups
Lemma 4.1. Let G be a group and let Hi , i I be a collection of
subgroups of G.
Then the intersection
is a subgroup of G
Proof. First note that H is non-empty, as the identity belongs to every
Hi . We have to check that H is cl
8. Homomorphisms and kernels
An isomorphism is a bijection which respects the group structure,
that is, it does not matter whether we rst multiply and take the
image or take the image and then multiply. This latter property is so
important it is actually
20. Euclidean Domains
Let R be an integral domain. We want to nd natural conditions on
R such that R is a PID. Looking at the case of the integers, it is clear
that the key property is the division algorithm.
Denition 20.1. Let R be an integral domain. We
15. Basic Properties of Rings
We rst prove some standard results about rings.
Lemma 15.1. Let R be a ring and let a and b be elements of R.
(1) a0 = 0a = 0.
(2) a(b) = (a)b = (ab).
Proof. Let x = a0. We have
x = a0
= a(0 + 0)
= a0 + a0
= x + x.
5. Permutation groups
Denition 5.1. Let S be a set. A permutation of S is simply a
bijection f : S S.
Lemma 5.2. Let S be a set.
(1) Let f and g be two permutations of S. Then the composition of
f and g is a permutation of S.
(2) Let f be a permutation of
Look at the groups D3 and S3 . They are clearly the same group.
Given a symmetry of a triangle, the natural thing to do is to look at
the corresponding permutation of its vertices. On the other hand, it
is not hard to show that every permu
12. Presentations and Groups of small order
Denition-Lemma 12.1. Let A be a set. A word in A is a string of
elements of A and their inverses. We say that the word wi is obtained
from w by a reduction, if we can get from w to wi by repeatedly
10. The isomorphism theorems
We have already seen that given any group G and a normal subgroup
H, there is a natural homomorphism : G G/H, whose kernel is
H. In fact we will see that this map is not only natural, it is in some
sense the only such map.
13. Sylow Theorems and applications
In general the problem of classifying groups of every order is com
pletely intractable. Lets imagine though that this is not the case.
Given any group G, the rst thing to do to understand G is to look for
subgroups H. I
Now in total there are at most six dierent permutations of the
letters A, B and C. We have already given six dierent symmetries,
so we must in fact have exhausted the list of symmetries.
Note that given any two symmetries, we can always consider what
6. Conjugation in Sn
One thing that is very easy to understand in terms of Sn is conjuga
Denition 6.1. Let g and h be two elements of a group G.
The element ghg 1 is called the conjugate of h by g.
One reason why conjugation is so important, is beca
Consider the chain of inclusions of groups
Z Q R C.
where the law of multiplication is ordinary addition.
Then each subset is a group, and the group laws are obviously com
patible. That is to say that if you want to add two integers together,
18. Prime and Maximal Ideals
Let R be a ring and let I be an ideal of R, where I = R. Consider
the quotient ring R/I. Two very natural questions arise:
(1) When is R/I a domain?
(2) When is R/I a eld?
Denition-Lemma 18.1. Let R be a ring and let I be an i
16. Ring Homomorphisms and Ideals
Denition 16.1. Let : R S be a function between two rings. We
say that is a ring homomorphism if for every a and b R,
(a + b) = (a) + (b)
(a b) = (a) (b),
and in addition (1) = 1.
Note that this gives us a category, the ca
11. The Alternating Groups
Consider the group S3 . Then this group contains a normal subgroup,
generated by a 3-cycle.
Now the elements of S3 come in three types. The identity, the prod
uct of zero transpositions, the transpositions, the product of one tr
9. Quotient Groups
Given a group G and a subgroup H, under what circumstances can
we nd a homomorphism : G G' , such that H is the kernel of ?
Clearly a necessary condition is that H is normal in G. Somewhat
surprisingly this trivially necessary condition
19. Special Domains
Let R be an integral domain. Recall that an element a = 0, of R is
said to be prime, if the corresponding principal ideal (p) is prime and
a is not a unit.
Denition 19.1. Let a and b be two elements of an integral domain.
We say that a
We introduce the main object of study for the second half of the
Denition 14.1. A ring is a set R, together with two binary opera
tions addition and multiplication, denoted + and respectively, which
satisfy the following axioms. Firstly
Consider the group of integers Z under addition. Let H be the
subgroup of even integers. Notice that if you take the elements of H
and add one, then you get all the odd elements of Z. In fact if you take
the elements of H and add any odd integer