The concept of prime ideal in a commutative ring 
An ideal
is called prime if for any
the following holds:
implies that at least one of these elements,
and
is in
E.g. the principal ideal 
A maximal ideal
is prime.
Indeed, let 
Analysing this proof, one can easily see that
If
Further important property of prime ideals is that they are radical, i.e.
Yet another interesting observation is that 
Rings of fractions
A subset
is called multiplicatively closed (or multiplicative) if
and
for any
Given a multiplicatively closed set 
a relation on 
It is not hard to show that 
To simplify the notation, write its equivalence class with representative 
is the ring of fractions of
w.r.t.
with addition and multiplication given by the rules
It is easy to check that this is well-defined. We also have
so that
is a ring homomorphism.
Note that 
The most well-known example is the case 
Examples:
is the field of fractions of
- for the ring of polynomials
over a field
the field
is the field of rational functions over
- Let
isa multiplicative set. Moreover, then
(it is not completely trivial to prove this, though). Intuitively, we make the variable
behave like the inverse of
as
in this ring.
Now let us look at the case 
Example: Let ![\mathbb{F}[T]](http://s2.wordpress.com/latex.php?latex=%5Cmathbb%7BF%7D%5BT%5D&bg=ffffff&fg=000000&s=0 "\mathbb{F}[T]")
![R_I=\mathbb{F}[T]_{(T-a)}](http://s3.wordpress.com/latex.php?latex=R_I%3D%5Cmathbb%7BF%7D%5BT%5D_%7B%28T-a%29%7D&bg=ffffff&fg=000000&s=0 "R_I=\mathbb{F}[T]_{(T-a)}")
The ring
It suffices to show that 
be a vector space over a field 
Denote by 
the set of linear functions 
![V=C[a,b],](http://s3.wordpress.com/latex.php?latex=V%3DC%5Ba%2Cb%5D%2C&bg=ffffff&fg=000000&s=0 "V=C[a,b],")
the space of continuous functions on ![[a,b].](http://s1.wordpress.com/latex.php?latex=%5Ba%2Cb%5D.&bg=ffffff&fg=000000&s=0 "[a,b].")
Then the function 
given by 
is linear on 
![V=\mathbb{R}[x]](http://s2.wordpress.com/latex.php?latex=V%3D%5Cmathbb%7BR%7D%5Bx%5D&bg=ffffff&fg=000000&s=0 "V=\mathbb{R}[x]")
be the vector space of polynomials with real coefficients.
given by 
is linear on 
is linear, one has 
for any 
Thus we have 
defined by 
so that 
To simplify notation, we will write 
instead 
As well, we can define 
for any 
and more generally, 
And there is the zero function 
for any 
Thus we have all the ingredients of a vector space, as can be easily checked. 
It is called the dual space of 
). Indeed, any closed under addition and multiplication set of functions 
would form a vector space.
on a basis 
of 
so we can compute 
for any 
once we know the 
) bijection between 
and 
This bijection, that is even an isomorphism of vector spaces, is defined by the dual basis of 
given by coordinate functions 
where 
’s are the coefficients of 
is the decomposition of 
![\mathbb{Z}[x].](http://s3.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D%5Bx%5D.&bg=ffffff&fg=000000&s=0 "\mathbb{Z}[x].")
It is a countable space: one can view it as the set of infinite 0-1 strings, with only finitely many 1’s occurring in each string. On the other hand, 
one can define a function 
as follows: 
It is linear, as 
Here we do not see any dependence on the choice of a basis in 
and we have
of linear functions on 
and so we need only to check that this mapping is bijective. Let 
its dual basis in 
if 
and 0 otherwise. Thus 
is the basis of 
which is dual to the basis 
and the mapping 
sends the vector with coordinates 
to the vector with the same coordinates in
Hence the latter is bijective. 
and write 
instead 
The set of 
such that 
is a subspace, called annihilator of 
More generally, the following holds.
be a subspace of 
and 
Then the annihilator 
of 
is a subspace of 
of dimension 
in 
is a basis of 
where 
Then we have the dual basis 
of 
and 
is the subspace with the basis 
for 
of rank 
be a linear transformation of 
and 
is a subspace. How can one look at 
By writing out 
in a basis 
for any 
in the dual basis 
we get equation 
Thus, considering 
as a matrix, we get 
where 
denotes the action of 
It follows that 
i.e. 
We have, considering that 
acts on 
acts on 
We work in the standard basis 
of 
, so that 
be the group of matrices 
It fixes, in its left action on 
. Let 
be the 1-dimensional subspace of 
Then 
and 
The group 
preserves 
in its action on 
As 
is 2-dimensional, there should be a nontrivial kernel in this action, and indeed, it consists of the elements of the form 
Then 
the symmetric group on 4 letters, as can be seen in its action on the 4 elements of 
On the other hand, it acts on the 3 nonzero elements of 
-
-
in a 
-letter alphabet, with beginnings and ends of the words glued together and made invisible (so one can visualise them as regular 
with letters from the alphabet placed at the vertices). Two necklaces are identical when one is obtained from another by rotations; two bracelets are identical when one can be obtained from another by rotation and a “flip”. One can view counting of necklaces and bracelets as counting of orbits of certain groups on the sets of “labelled” necklaces (resp. bracelets). For the necklace case, the group is 
and for the bracelet case it is the dihedral group of order 
of labelled necklaces/bracelets is just the set of words of length 
act on a finite set 
Denote by 
for 
the set of elements of 
fixed by 
Then the number 
of orbits of 
of 
i.e.
and the other being 
a pair 
for 
and 
is an edge iff 
Thus the total number of edges in this graph is 
is an edge iff 
the stabiliser of 
in 
Thus, for a 
-orbit 
one has the total number of edges leaving 
equal to 
Summing over all the 
completing the proof.
of the conjugacy classes 
as 
So one has
denoting the centraliser of 
in 
on the set of 
words of a 
-letter alphabet. 
and an order two “flip” permutation 
The group can be decomposed into the union of cosets 
of the subgroup 
are the “constant” ones, i.e. words consisting of a single letter, and so 
The coset 
equals 
(an unusual coincidence), and so it suffices to find
For a word 
to be in the latter set, it must have 
… 
so there are 
possibilites.
one obtains 
a bijection 
by 
by 
of a subgroup 
when we associated with each 
by 
represented by 
permutation matrices), etc.
An action of 
is a finite set and 
the symmetric group (of all permutations of 
An action satisfying 
is called faithful (or effective).
need not be faithful. E.g. it will not be faithful if 
on 
iff there exists 
such that 
The equivalence classes of 
are called orbits (more precisely, orbits of 
on 
and 
Then 
is called the stabiliser of 
in 
and denoted 
for some 
What can be said about 
consists of the elements of 
that map 
Thus any 
fixes 
and any 
that fixes 
belongs to 
as 
of 
the symmetric group on 
where 
as follows:
be written in the cyclic notation. Then for any 
one has
to any permutation with cycles of the same lengths 
and 
of a group 
can be viewed as “the same”? Obviously we need a bijection 
to exist, that also “respects” 
for all 
and 
Such an 
is called a 
Then 
for any 
where we denoted by 
the preimage of 
in 
such that 
where 
Indeed,
is obviously a bijection
iff 
iff 
iff 
is 
Then for any 
as claimed.
of 
as
for an 
the group, with respect to matrix multiplication, of invertible 
have a number of natural actions, that all come one way or another from the vectorspace 
Indeed, each 
provides a bijection 
by 
for any 
(which one can equivalently writen as 
). It is called contragradient action.
is not always an action, as 
and so 
, with 
on the same set 
were an 
for all 
and 
and $f(v)=u$ so that 
are linearly independent. In the basis 
, take 
Then 
We see that 
a contradiction.
and 
and see that 
for some 
again a contradiction. The argument showing non-equivalence in the case 
is similar. For 
these two actions are isomorphic, as 
is commutative.
on subspaces of 
of subspaces of dimension 
Given a 
and 
we define 
Then 
is a 
-dimensional subspace, as can be seen by choosing a basis 
of 
is a basis of 
Thus 
we can choose a basis in such a way, that 
block matrices, discussed earlier 
that is the kernel of the action of 
of 
acts trivially on 
as it preserves every 1-dimensional subspace (indeed, any 
equals 
with 
).
is 
when 
This set is called projective line and denoted 
and can be viewed as the set of pairs 
as follows: every 1-dimensional subspace 
of 
or by equation 
of 
on 
consider 
and 
just as if we multiply a vector by 
and then normalize either to 
or to 
depending upon whether or not 
and with 
points on each line (a 
, for short) has a lot of “regularity” – and as we learned from a 
for some 
This regularity allows for the use of quite a bit of linear algebra and algebraic graph theory in investigating them.
is the graph with the vertex set 
two vertices 
are adajcent iff they are collinear, i.e. if there exists 
such that 
such a graph is strongly regular, see 
is strongly regular if it is 
there are exactly 
vertices 
such that 
there are exactly 
vertices 
such that 
is as above, one has an 
One famous example of a srg is the 
is an 
as a function of 
and 
(namely, it is 
)
is the 
matrix 
that has 
for each 
and otherwise 
and the 
-th row sum is the number of vertices adjacent to 
This means that the all-1 vector is an eigenvector of 
with eigenvalue 
the adjacency matrix 
has at most 3 distinct eigenvalues, namely 
is the number of length 2 walks from 
to 
In our case we have 
the all-1 matrix. Any eigenvector 
with eigenvalue 
must be orthogonal to the 
Thus 
and so the other two eigenvalues have multiplicities 
and 
that satisfy
(by just discussed) and 
(as the trace of 
is 0.)
as 
must be nonnegative integers!
and 
one has 
With more work (one needs e.g. the Krein bound) the case 
can be ruled out.
gives rise to a coset decomposition of 
or, more precisely, to two decompositions, into left cosets i.e. 
have the same cardianlity:
and 
Then 
divides 
where 
is a group homomorphism. (Just as in the case of abelian groups, one can show that 
and 
)
one has 
for some 
We compute 
Hence 
for all 
or 
when 
)
for all 
in 
is normal.
defined by 
is normal.
and an observation that 
so 
Then 
is a group with operation 
and 
a group homomorphism 
with the kernel 
to establish that the multiplication in 
is well-defined: 
is a group with the identity element 
follows from the associativity in 
Indeed, 
Also, 
for 
and so 
On the other hand if 
then 
and so 
the group, with respect to matrix multiplication, of invertible 
matrices with entries in a field 
provide a rich playground for various examples of normal subgroups and quotient groups. To this end, we will work with block matrices, i.e. we will partition our matrices into rectangular blocks of appropriate size, so that this block decomposition is preserved under the matrix multiplication. The simplest case is the following partition into 4 blocks (i.e. 
block matrices).
we can view 
where 
is a 
matrix, 
a 
matrix (and certainly 
and 
being of size 
). 
:
we see that the shape is preserved, i.e. 
More precisely
and 
(that is, assuming the matrices we talk about are invertible). We can formalise this observation as follows.
be a subgroup of 
and 
and such that 
for all 
Then 
and
for 
dimensional vectorspace 
and derive that 
is a normal subgroup of 
(where 
or 
) cannot be obtained by simply setting the remaining off-diagonal block of 
to 0, and the remaining diagonal block to the identity matrix. For instance, let 
be generated by the matrix 
where 
Then 
is a cyclic group of order 4, and it satisfies the conditions of the above lemma with 
It is easy to see that 
is an isomorphism, and 
is a homomorphism with the kernel generated by 
and 
by the usual argument involving decomposition of 
computing 
for 
for 
prime, can be accomplished by computing 
and 
separately. In turn, 
so we can simplify the task of computing 
too. 
where 
is prime.
matrix in 
way (everything goes, except all zeros). Fixing this 1st row vector 
we can again fill in the 2nd row with entries of any vector 
which is linearly independent (to get an invertible matrix) of 
i.e. lies outside of the subspace 
spanned by 
As 
we get 
ways to choose 
Now we fix 
as well, and fill in the 3rd row with entries of any vector 
which is linearly independent of 
and 
spanned by 
As 
we get 
ways to choose 
-th row, we have 
choices, giving us the formula above.
Indeed, 
consists of 
whereas we can fill the entries of 
for 
and 
is a group homomorphism.) 
i.e. a proper ideal 
itself.
of all proper ideals of 
iff 
for 
such that for any 
either 
or 
We need to check that 
has an upper bound, i.e. 
such that 
for any 
Set 
i.e. that it is a proper ideal. As 
we only need to check that it is an ideal. It suffices, and is easy, to see that 
there exists 
such that 
and thus 
(Indeed, there exists 
such that 
and w.l.o.g 
as 
and 
here exists 
such that 
and thus 
(Indeed, there exists 
and 
as 
is an ideal.) 
, so it has maximal element, that is obviously a maximal proper ideal in 
QED.
is contained in a maximal proper ideal 
contains a maximal proper ideal 
and 
for a maximal ideal 
is contained in such an ideal, by applying the latter construction to 
