# Abstract Algebra

#### - a brief survey by Bengt Månsson

The elements of abstract algebra may look like a heap of definitions, and when it starts to become really interesting it also gets quite difficult. This page is just a start. Nevertheless I hope you will understand how earlier known results get a common ground. This page will also provide some necessary background for coming surveys on Galois Theory and Algebraic Number Theory. My intention is to make at least some essential part of these important and interesting areas of mathematics understandable to interested students at senior high school level or slightly above.

In examples we will use the standard notations Z, Q, R, C for the set of integers, rational numbers, real numbers, and complex numbers, respectively.

### Binary Operations

or composition rules are e g addition and multiplication. We will make this precise and more general.

#### Definition

 A binary operation on a non-empty set M is a function from M×M to M.

So, a binary operation combines two elements of M, taken in a certain order, to get one element of M. Instead of using a function letter it is denoted something like * and written between the arguments,  a*b  instead of  *(a,b) . You recognize, of course, that this is exactly how the usual operations like addition and multiplication are written. If the order of the arguments does not matter, i e  a*b = b*a  for all a and b in M, the operation is said to be commutative. If  (a*b)*c = a*(b*c)  for all a and b in M, the operation is said to be associative. Examples of these are usual addition and multiplication but not subtraction and division.

A binary operation can be defined by means of a table if the set is finite. An example where the set is {a,b,c,d}:

*   a   b   c   d
a  a  c  d  b
b  c  d  a  b
c  d  a  b  c
d  b  b  c  a
Any combination of values in the green cells is possible, although not always interesting. Notice that the operation in the example is commutative.

A non-empty set together with one or several binary operations is called an algebraic structure. An algebraic structure with set M and operation * is denoted  (M,*)  or just M. Instead of * the usual multiplication dot is common (although maybe not with its usual meaning). As usual the dot can often be left out. A commutative operation is often denoted + and if there are two operations defined, one of them commutative, + and · are common. Then the structure may be denoted  (R,+,·) .  We will now have a look at three basic structures, groups, rings, and fields.

### Groups

#### Definition

 A group is an algebraic structure (G,*) such that  (i)   * is associative,  (ii)   there is a neutral element e in G such that a*e = e*a = a for all a in G,  (iii)  for each a in G there is an inverse a' in G such that a*a' = a'*a = e.

Note: If * is commutative the group is said to be commutative or abelian. after the Norwegian mathematician Abel. In an abelian group + is often used for * and 0 for e. If · is used for * then 1 is often used for e. It can be proved that there is exactly one neutral element, and that there is exactly one inverse for each element. The inverse of a is often written a-1 in general and -a if + is used for *. The equation  ax = b  has the unique solution  x = a-1b ,  and  xa = b  has the unique solution  x = ba-1 . We prove this:

From  ax = bx = ex = (a-1a)x = a-1(ax) = a-1b ,  and similarly for the second equation.

Example 1:  (Z,+) ,  (Q,+) ,  (R,+) ,  (C,+) ,  (Q - {0},·) ,  (R - {0},·) ,  and  (C - {0},·)  are abelian groups.

Example 2:  No infinite subset of Z is a group under multiplication, but  ({1,-1},·)  is a group.

Example 3:  Let Z3 denote the algebraic structure with elements 0,1,2 and the operation + defined by the group table

+   0   1   2
0  0  1  2
1  1  2  0
2  2  0  1

(Z3,+)  is an abelian group. Since there are just three elements this easily checked, going through all possibilities. Also  (Z3 - {0},·)  is a group if · is defined by  a·1 = 1·a = a ,  2·2 = 1 .  As you can see + and · can be described as usual addition and multiplication where there is added such an integer multiple of 3 to the sum or product that the result is one of the numbers 0, 1, 2.

In a similar way a group  (Zn,+)  can be defined for any positive integer n. However, generalizing  (Z3 - {0},·)  to  (Zn - {0},·)  results in a group if and only if n is a prime number.

In the last example we could go through all possible products ab, (ab)c, and a(bc) to check that the algebraic structure is a group. Conversely, starting with a finite set we can try all possible group tables to find which groups can be constructed from the given set. As an example let us choose the set {e,a,b,c}. We start with  ee = eae = ea = abe = eb = b ,  and  ce = ec = c  since there must be a neutral element. This gives one row and one column of the table. Further, the existence of an inverse for each element means that each row and each column must be a permutation of the elements. Apart from the ordering of the elements this gives two distinct tables if  aa = e:

  ·   e   a   b   c    ·   e   a e a b c e a b c a e c b a e c b b c e a b c a e c b a e c b e a
There remain the possibilities  aa = b  or  aa = c . However the corresponding group tabled will be identical with the second one above if only a, b are renamed b, a, and b, c are renamed c, b, respectively. If two groups tables become identical on a suitable renaming the groups are said to be isomorphic, and are in many respects interchangeable. They are then "essentially identical". In the first table, however, each element "times" itself equals one and the same element (e), a fact which cannot be changed by any renaming.

We have found that there are exactly two non-isomorphic groups of order 4, both abelian. The first one is called Klein's Four-group while the second one is isomorphic to (Z4,+).

Up to now all groups have been abelian. The smallest non-abelian group is of order 6. If the elements are {e,p,q,r,s,t} its group table can be written:

*   e   p   q   r   s   t
e  e  p  q  r  s  t
p  p  q  e  t  r  s
q  q  e  p  s  t  r
r  r  s  t  e  p  q
s  s  t  r  q  e  p
t  t  r  s  p  q  e

({e,p,q},*)  is a also a group, a so called subgroup of  ({e,p,q,r,s,t},*) .  This group is isomorphic to the symmetric group S3, consisting of all permutations of three objects. Such a permutation is a function defined on a set of three elements and the group operation is composition of functions. If, as usual, functions are acting to the right then, e g, pr corresponds (in S3) to the permutation obtained by first acting on the three objects with r, then with p. Thinking like this one can easily construct the group table. The subgroup mentioned is abelian and is isomorphic to the so called alternating group A3, consisting of all even permutations of three objects. A3 is, of course, a subgroup of S3. That * is associative is obvious, since this is always so for compositions of functions. (An even permutation can be written as a composition of an even number of transpositions, i e permutations of two objects.)

Permutation groups played a central role in the development of group theory and are essential in Galois theory. The symmetric group Sn is defined as the permutation group of n elements, and has n! elements. The even permutations make up a subgroup, the alternating group An, with n!/2 elements. Also more general groups of functions will appear. Notice that the order in which functions are applied is in general essential so such groups are normally non-abelian.

A subgroup is defined as a group, whose elements form a subset of a group, with the same group operation. There are always at least two subgroups, one being the entire group, the other consisting of just the neutral element. These are called trivial subgroups. By a common abuse of notation, the latter is often denoted 0 or 1. All other subgroups are called proper subgroups. A non-empty subset H of a group G is a subgroup of G if and only if  ab-1  is in H whenever a and b are in H. Here b-1 denotes the inverse (in G) of b. We prove this:

First, suppose  ab-1  is in H whenever a and b are in H. Then, if a is in Haa-1 = e  is in H, so e and a are in H. This implies that  ea-1 = a-1  is in H. So, the neutral element is in H, and any a in H has its inverse in H. We must also show that  ab  is in H whenever a and b are in H, so that the group operation is really an operation on H. This follows since any element in H has its inverse in H, and  ab = a(b-1)-1 .

Conversely, suppose H is a subgroup of G. Then there is a neutral element e in H, and any b in H has an inverse b-1 in H. Since H is a subgroup, it is closed under the group operation of G, so ab-1 is in H if a and b are in H, since b-1 is then in H. Notice, that e and b-1 are neutral element and inverse, respectively, in G also, since the group operation is the same.

For finite groups Lagrange's Theorem says that the order (number of elements) of any subgroup divides the order of the entire group. We will not prove this. As a simple application, this implies that a group of prime order has no proper subgroups.

Now, let a be any element of a group G. Then all powers an (n a non-negative integer) will make up an abelian subgroup of G. Such a group is said to be generated by a, and is called a cyclic group. If G has prime order and a is not the neutral element then the generated subgroup must be the entire group. So, a group of prime order is cyclic and abelian.

Finally, let us consider the group of symmetry operations of the square. A symmetry operation can be described as a displacement of the square such that it occupies exactly the same place after as before being displaced. The operations are definied by the final position relative to the original only, not how the motion is performed. There are three different clockwise rotations of 90, 180, and 270 degrees, respectively. Notice that a counter-clockwise rotation equals some clockwise rotation giving the same resulting position. Further, there are reflections in the diagonals and in the horisontal and vertical lines bisecting the square. No displacement at all is also a symmetry operation, making 8 in all. A symmetry operation can be regarded as a function, taking one position of the square as its input and the resulting relative position as its value. It is then obvious that the symmetry operations constitute a group, the symmetry group of the square, denoted D8. The group operation is composition of functions.

If the corners are numbered as in the above picture, each symmetry operation corresponds to a permutation of 1,2,3,4. However, not all permutation occur since the ordering of the corners cannot be changed. Thus, the symmetry group must be (isomorphic to) a proper subgroup of S4. Since S4 has  4! = 24  elements this is in agreement with Lagrange's theorem.

Similarly, there are symmetry groups for other figures. These groups play a role in Galois theory. For example, the so called Galois Group of the polynomial  t4 - 2  over Q is isomorphic to D8.

We will return to permutation groups and subgroups later, in particular so called normal subgroups.

### Rings and Fields

Rings and fields are algebraic structures with two binary operations.

#### Definition

 A ring is an algebraic structure (R,+,·) such that  (i)   (R,+) is an abelian group,  (ii)   · is associative,  (iii)  a·(b + c) = a·b + a·c  and  (a + b)·c = a·c + b·c  for all a, b, c in R. A field is a ring (K,+,·) such that  (iv)  (K - {0},·)  is an abelian group.

Example 1:  (Z,+,·)  is a ring but no field.  (Q,+,·) ,  (R,+,·)  and  (C,+,·)  are fields.

Example 2:  (Zn,+,·)  is a ring for any positive integer n. It is a field if and only if n is a prime number.

Example 3:  In the same way as for groups, the operations + and · can be defined by tables. The following tables define the operations for (Z4,+,·), which is a ring but not a field:

 +   0   1   2   3    ·   0   1 0 1 2 3 0 0 0 0 1 2 3 0 0 1 2 3 2 3 0 1 0 2 0 2 3 0 1 2 0 3 2 1

Non-zero elements with product 0 are called zero-divisors. We see that 2 is a zero-divisor in Z4. In a field ther can, on the other hand, not be any zero-divisors.

There are several types of rings. We will assume that rings are commutative, ie that multiplication · is commutative, that there is a neutral element 1 for multiplication, and that 0 and 1 are different. If also there are no zero-divisors the ring is called a domain. A field is always a domain but the converse does not hold. However, a finite domain is a field.

Example 1:  (Z,+,·)  is a domain but not a field.

Example 2:  (Zn,+,·)  has zero-divisors unless n is a prime number.

Polynomials play an important role in Galois theory and in algebraic number theory. A polynomial in one indeterminate t with coefficients in a ring R is an expression of the form

an tn + an-1 tn-1 + ... + a1 t + a0

where an, an-1, ..., a1, a0 are elements in R. Provided an is non-zero n is called the degree of the polynomial. As a general notation f, g, ... or f(t), g(t) are used. With addition and multiplication defined formally as for real polynomials these polynomials form a ring  R[t] ,  called a polynomial ring over R. If R is a field K, as will often be the case, K[t] is a domain.

Part of the theory of factorization is taken over from real polynomials and part is new. For example any polynomial in K[t] can, apart from a constant factor, be uniquely factorized into irreducible polynomials. An irreducible polynomial is one which cannot be written as a product of non-constant polynomials of lower degree.

Example:  In  Z3[t] ,  2t2 + 2t + 2 = (t + 2)(2t + 1) = 2(t + 2)(t + 2) ,  since  2·2 = 1  in Z3.

A subring is defined as a ring, whose elements form a subset of a ring, with the same binary operations. There are always at least two subrings, the entire ring and {0}. These are called trivial subrings. By a common abuse of notation, the latter is often denoted 0. All other subrings are called proper subrings. A non-empty subset S of a ring R is a subring of R if and only if  a - b  and ab are in S whenever a and b are in S and 1 (of R) is in S. This is proved similarly to the corresponding theorem for subgroups.

### Homomorphism and Isomorphism

Definition:

 A group homomorphism f is a function from a group G to a group G' such that  f(ab) = f(a)f(b)  for all a, b in G. A ring homomorphism f is a function from a ring R to a ring R' such that  (i)   f(a + b) = f(a) + f(b)  for all a, b in R,  (ii)   f(a·b) = f(a)·f(b)  for all a, b in R.

An injective (invertible) homomorphism is called a monomorphism. A homomorphism that is injective and surjective (onto) is called an isomorphism. A homomorphism from a group or ring to itself is called an endomorphism. An isomorphism from a group or ring to itself is called an automorphism. Two groups or rings are said to be isomorphic if there is an isomorphism between them.

For a group homomorphism f(e) is the neutral element of G' and  f(a-1) = (f(a))-1 .  We prove this:

Let the neutral element of G' be e'. Then  f(e)f(e) = f(ee) = f(e) = f(e)e' ,  from which  f(e) = e' ,
and  f(a)f(a-1) = f(aa-1) = f(e) = e' = f(a)(f(a))-1 ,  from which  f(a-1) = (f(a))-1 .

Similarly, for a ring homomorphism f(0) is the additive neutral element of R' and  f(-a) = -f(a) .  However, f(1) need not equal the multiplicative neutral element of R'. Instead this is sometimes required as an additional condition for a ring homomorphism, which we will do.

If two finite groups or rings are isomorphic the tables defining their operations are the same except for possible renaming and rearrangement of the elements. In this sense they are essentially the same, and often their difference does not matter. The same goes for infinite groups and rings although the tables cannot actually be completely written out.

Example 1:  The complex numbers with imaginary part 0 are usually thought of as R. More precisely, they form a subfield of C, isomorphic to R. This distinction does not matter, e g when solving polynomial equations. On the contrary it is easiest to identify the subfield with R.

Example 2:  For any complex number  z = a + bi  let  f(z) = a - bi . Then f is an automorphism of C. Since  f(z) = z  when z is real, but not else, f is called an R-automorphism of C.

Example 3:  Let f be as in example 2, and let g be the identity function on C. Then g is, of course, also an R-automorphism of C. f and g are, in fact, the only R-automorphisms of C. The constitute a group with composition of functions as group operation, the so called Galois Group of the field extension C:R or of the polynomial  t2 + 1 over R. This group is isomorphic to (Z2,+) and to S2.

Example 4:  In connection with the symmetry operations of a square, we noted that these operations can be put into a 1-1 correspondence with eight of the permutations of 1,2,3,4. Now, let  f(x) for the symmetry operation x equal the corresponding permutation of 1,2,3,4. Then f is a homomorphism from D8 to S4. It is injective but not surjective, so it is a monomorphism but no isomorphism. Thus D8 is isomorphic to a proper subgroup of S4.

For future reference we define:

(i)  If f is a group homomorphism from G to G' then the kernel  ker(f)  is the set of values in G which are mapped to the neutral element of G'. It is non-empty since  f(e) = e' .

(ii)  If f is a ring homomorphism from R to R' then the kernel  ker(f)  is the set of values in R which are mapped to the additive neutral element 0 of R'. It is non-empty since  f(0) = 0 .

(iii) If f is a group or ring homomorphism the image  im(f)  is the set of values taken on by f.

### Normal Subgroups and Quotient Groups

Suppose H is a subgroup of a group G. Then, for any a in G, the set aH is defined as the set {ah:h in H}. A quotient group is a group whose elements are sets such as aH. The group operation is defined by  aH·bH = abH ,  which may look reasonable enough. However, a1H may equal a2H with a1 and a2 different, but the product  aH·bH  must depend on the sets aH and bH only, not the way they are written. So, a1bH must equal a2bH. This is so if G is abelian but otherwise maybe not. To make the group operation well-defined we will now introduce the concept of a normal subgroup.

#### Definition:

 H is a normal subgroup of the group G if  ghg-1  is in H  for all g in G and h in H.

Obviously any subgroup of an abelian group is normal.

Example 1:  The trivial subgroups are always normal.

Example 2:  ({e,p,q},*)  is a normal subgroup of  ({e,p,q,r,s,t},*) .

Example 3:  If f is a group homomorphism from G to G' then  ker(f)  is a normal subgroup of G. We prove this:

Suppose h is in ker(f) and g in G. Then  f(h) = e'  and  f(ghg-1) = f(g)f(h)f(g-1) = f(g)e'(f(g))-1 = f(g)(f(g))-1 = e', which means that ghg-1 is in ker(f).

#### Theorem:

 If H is a normal subgroup of the group G, then the set {gH: g in G}  with a binary operation · defined by  aH·bH = abH  for all aH, bH in {gH: g in G}, forms a group. This group is called a quotient group, and is denoted G/H.

#### Proof:

We must first show that the operation · is well-defined, i e independent of how aH and bH are written.

Since H is normal  aH = Ha, and  bH = Hb. From this  a1b1H = a1Hb1 = a2Hb1 = a2b1H = a2b2H  if
a1H = a2H, and  b1H = b2H.

eH = H  is obviously neutral element, and  a-1H  inverse of  aH, as usual unique.

Finally  (aHbH)cH = (abH)cH = (ab)cH = a(bc)H = aH(bcH), so · is associative.

Example 1:  Let  G = ({e,p,q,r,s,t},*)  and  H = ({e,p,q},*), which is a normal subgroup of G. From the group table for G we have  eH = pH = qH = H, and  rH = sH = tH. So,  G/H = ({H,rH},·)  with this group table:

·   H rH
H  H rH
rH rH  H
G/H is isomorphic to (Z2,+) and to S2. It is abelian although G is not.

Example 2:  Let f be a group homomorphism from G to G'. Then there is always a quotient group G/ker(f) since the kernel ker(f) is a normal subgroup of G.

In example 2 we noticed the existence of the group G/ker(f). If, in particular, f is injective, i e a monomorphism, then G is isomorphic to im(f), and  ker(f) = ({e},*). We prove the latter statement:

e is always in ker(f), since  f(e) = e'. Suppose h is in ker(f). Then  f(h) = e' = f(e). Thus  h = e, since f is injective, and ker(f) contains just the element e.

When f is injective ker(f) has just one element. This implies that there is a 1-1 correspondence between the elements g of G and the elements gker(f) of G/ker(f). Therefore G/ker(f) is isomorphic to G, so also to im(f). However, this is an important result, holding for all group homomorphisms, not just monomorphisms. We state this as a theorem.

#### Theorem:

 If f is a group homomorphism from G to G', then  G/ker(f) is isomorphic to im(f).

#### Proof:

We define a function F from G/ker(f) to im(f) by  F(xker(f)) = f(x)  for all x in G. We will prove that F is an isomorphism.

F is obviously surjective, since f(x) takes all values of im(f) when x goes through all values of G.

Next, let  F(xker(f)) = F(xker(f)), i e  f(x) = f(y). Then, since f is a homomorphism,  f(xy-1) = f(x)f(y-1) = f(x)(f(y))-1 = f(x)(f(x))-1 = e', so xy-1 is in ker(f). But then  xker(f) = yker(f), i e the same element in G/ker(f), so F is injective.

Finally, F is also a homomorphism because  F(xker(f)·yker(f)) = F(xyker(f)) = f(xy) = f(x)f(y) = F(xker(f))F(yker(f)) .

Example 1:  Let the function f from C to R be defined by  f(z) = Im z  for all z in C. Then f is a group homomorphism from (C,+) onto (R,+) with  ker(f) = {z in C:Im z = 0} = R. Thus, (C,+)/(R,+) is isomorphic to (R,+).

Example 2:  Let G be a finite group with a normal subgroup H. Then H and G/H are also finite. We use | | to denote the number of elements in a finite set. From Lagrange's theorem, |G|/|H| is an integer. Now, all cosets aH have the same number of elements, equal to |H|. Thus, the number of cosets must be |G|/|H|, but it is also |G/H|, so |G|/|H| = |G/H|. Further, let f be a homomorphism defined on G. Then im(f) and G/ker(f) have the same number of elements, since they are isomorphic. From this we deduce  |G| =  |ker(f)|·|im(f)|.

### Ideals and Quotient Rings

In the previous section section we defined G/H where H is a normal subgroup of G. To be able to do something similar for rings we will now introduce the concept of an ideal.

The ideals have their roots in the 19th century attempt to restore unique prime factorization in so rings of algebraic integer.

#### Definition:

 A non-empty subset I of a ring R is an ideal in the ring R if  (i)   x - y  is in I for all x, y in I,  (ii)  rx is in I for all r in R and x in I.

Note: We are assuming that R is a commutative, and has different elements 0 and 1. Apart from the 1, an ideal is a special kind of subring. It is closed under multiplication with any element in the entire ring. Any ideal contains the element 0, because of (i). Then  -y = 0 - y  is in I, and so are also  x + y = x - (-y). An ideal different from R, which is not strictly enclosed by any ideal, except R, is called a maximal ideal. There need not be any maximal ideal. An ideal, strictly enclosed in R, is called a proper ideal.

Example 1:  In any ring R there are always at least the ideals {0} and R.

Example 2:  For any natural number n, let  nZ = {nx:x in Z}. The nZ are ideals of Z, in fact the only ones. For positive n, nZ can be described as the set of integers divisible by n. In particular 2Z is the set of even numbers.

Example 3:  If f is a ring homomorphism from R to R' then  ker(f)  is an ideal of R. We prove this:

First, suppose x, y are in ker(f). Then  f(x) = f(y) = 0  and  f(x - y) = f(x) - f(y) = 0, which means that x - y is in ker(f). Next, suppose x is in ker(f) and r in R. Then  f(x) = 0  and  f(rx) = f(r)f(x) = f(r)·0 = 0, so rx is also in ker(f).

#### Theorem:

 If I is an ideal in the ring R, then the set {r + I: r in R} with binary operations + and · defined by  (i)   (r + I) + (s + I) = (r + s) + I  (ii)  (r + I)·(s + I) = (rs) + I forms a ring. This ring is called a quotient ring, and is denoted R/I.

#### Proof:

We must first show that the operations are well-defined, i e independent of how  r + R  and  s + are written.

For + this is in principle as for quotient groups since (R,+) is an abelian group, and thus (I,+) a normal subgroup.
For · suppose  r1 + I = r2 + I  and  s1 + I = s2 + I, i e  r1 - r2  and  s1 - s2  are in I, or  r1 = r2 + i1 and  s1 = s2 + i2 where i1, i2 are in I. We must show that  (r1s1) + I = (r2s2) + I. This is true if and only if  r1s1 - r2s2  is in I, which holds since  r1s1 - r2s2 = (r2 + i1)(s2 + i2) - r2s2 = r2i2 + s2i1 + i1i2, where every term has a factor in I.

0 + I = I  is obviously neutral element for +, and  (-r) + I  inverse of  r + I, as usual unique.

Since  ((r + I) + (s + I)) + (t + I) = ((r + s) + I) + (t + I) = ((r + s) + t) + I = (r + (s + t)) + I = (r + I) + ((s + t) + I) = (r + I) + ((s + I) + (t + I)), so + is associative. It is also commutative since + in R is. Thus, (R/I,+) is an abelian group.

Similarly  ((r + I)·(s + I))·(t + I) = (r + I)·((s + I)·(t + I)), so · is associative.

Finally  (r + I)·((s + I) + (t + I)) = (r + I)·((s + t) + I) = r(s + t) + I = (rs + rt) + I = ((rs) + I) + ((rt) + I) = (r + I)·(s + I) + (r + I)·(t + I), which means that the distributive laws hold.

Notice that R/I is commutative as well as R, and that it has a neutral element  1 + I  for multiplication (·). 1 + I  is different from  0 + I  unless  I = R.

Example 1:  Z/4Z is a ring, where 4Z is the ideal in Z consisting of all integers divisible by 4. It is isomorphic to (Z4,+,·) as can be seen from these tables:

 +   4Z 1+4Z 2+4Z 3+4Z   ·   4Z   1+4Z 4Z 1+4Z 2+4Z 3+4Z 4Z 4Z 4Z 4Z 1+4Z 2+4Z 3+4Z 4Z 4Z 1+4Z 2+4Z 3+4Z 2+4Z 3+4Z 4Z 1+4Z 4Z 2+4Z 4Z 2+4Z 3+4Z 4Z 1+4Z 2+4Z 4Z 3+4Z 2+4Z 1+4Z

Example 2: Let I, J be ideals in the ring R, and J contained in I. Define  I/J = {x + J:x in I}. Then I/J is an ideal in the quotient ring R/J.

Example 3:  Let f be a ring homomorphism from R to R'. Then there is always a quotient ring R/ker(f) since the kernel ker(f) is an ideal in R.

#### Theorem:

 If f is a ring homomorphism from R to R', then  R/ker(f) is isomorphic to im(f).

#### Proof:

We define a function F from R/ker(f) to im(f) by  F(x + ker(f)) = f(x)  for all x in R. We will show that F is an isomorphism.

F is obviously surjective, since f(x) takes all values of im(f) when x goes through all values of R.

Next, let  F(x + ker(f)) = F(x + ker(f)), i e  f(x) = f(y). Then, since f is a homomorphism,  f(x - y) = f(x) - f(y) = f(x) - f(x) = 0, so  x - y  is in ker(f). But then  x + ker(f) = y + ker(f), i e the same element in R/ker(f), so F is injective.

Finally, F is also a homomorphism because  F((x + ker(f)) + (y + ker(f))) = F((x + y) + ker(f)) = f(x + y) = f(x) + f(y) = F(x + ker(f)) + F(y + ker(f)),   F((x + ker(f))·(y + ker(f))) = F((xy) + ker(f)) = f(xy) = f(x)f(y) = F(x + ker(f))·F(y + ker(f)), and  F(1 + ker(f)) = f(1) = 1'.

Example 1:  For any positive integer n > 1, Z/nZ is a ring, where nZ is the ideal in Z consisting of all integers divisible by n. Define  f(x) = x mod n  for any integer x. Then, f is a homomorphism from Z onto the ring Znf(x) = 0  if and only if  x mod n = 0  or, equivalently, x is in nZ. Thus,  ker(f) = nZ, so Z/nZ is isomorphic to Zn.

Example 2: Let I, J be ideals in the ring R, and J contained in I. Then there is a homomorphism f from R/J onto R/I, defined by  f(x + J) = x + If(x + J) = I, the zero element of R/I, if and only if x is in I, so  ker(f) = I/J. Thus, (R/J)/(I/J) is isomorphic to R/I.

### Field Extensions

This section is mainly intended as a presentation of the very easiest aspects of field extensions, including one or two essential theorems. The next section introduces the use of so called vector spaces as a further tool in the theory of field extensions. In all, these two section serve as prerequisites for the Galois theory.

In most cases a field extension can be considered as an ordered pair of two fields K and L, where K is a subfield of L. The extension is then denoted L:K.The precise definition says that a field extension is a monomorphism from a field K to a field L. However, usually isomorphic fields can be identified, so that K is a subfield of L. We also say that L is an extension of K. K and L are called the small field and the large field, respectively.

Example 1:  R:Q and C:Q are field extensions.

Example 2:  Let Q(sqrt(2)) denote the smallest field containg Q and sqrt(2). Being a field, Q(sqrt(2)) must contain elements such as  1/(1 + sqrt(2)) = sqrt(2) - 1. It is not difficult to prove that all elements of Q(sqrt(2)) can be written  a + bsqrt(2)  with rational numbers a, b.

Example 3:  Let  <t2 + 1> = (t2 + 1)R[t]. Then <t2 + 1> is an ideal in the ring R[t], and R[t]/<t2 + 1> is a ring. This ring is isomorphic to C, so it is a field. We prove this:

Define the function F from R[t] to C by  F(f(t)) = f(i), where i is the imaginary unit. F is obviously a homomorphism. If f(t) can be written as a product with one factor  t2 + 1, then  f(i) = 0. Conversely, suppose  f(i) = 0. Then f(t) has the factor  t - i, as a polynomial over C. But to be in R[t] it must then have the factor  t + i  as well, so it has the factor  t2 + 1. Thus,  ker(F) = <t2 + 1>.

We must also show that  im(F) = C. To this end let z be any complex number and  f(t) = Re z + tIm z. Then f(t) is a polynomial in R[t] with f(i) = z. So, for any z in C there is a f(t) in R[t] such that  F(f(t)) = f(i) = z. This means that F is surjective, i e  im(F) = C. Since  ker(F) = <t2 + 1>, it follows that R[t]/<t2 + 1> is isomorphic to C.

Now, consider the element  j = t + <t2 + 1>  in the field R[t]/<t2 + 1> of the last example. From the definition of multiplication in a quotient ring we get  j2 = (t + <t2 + 1>)(t + <t2 + 1>) = t2 + <t2 + 1>. The element  1 + <t2 + 1>  is multiplicative neutral element, and the definition of addition gives   j2 + (1 + <t2 + 1>) = (t2 + <t2 + 1>) + (1 + <t2 + 1>) = (t2 + 1) + <t2 + 1> = <t2 + 1>. This is the zero element. From this we can see that if the field R[t]/<t2 + 1> is identified with C, then j assumes the role of the imaginary unit.

We will use K(a) to denote the smallest field including the field K and the element a. Then K(a) is an extension of K, and we say that it has been obtained from K by adjoining a. An important case is when a is a zero of a polynomial over K, especially when a is not an element of K. Then K is strictly included in K(a).

Definition:

 A simple algebraic extension is an extension K(a):K such that a is a zero  of a non-zero polynomial over K. a is then said to be algebraic over K. A simple transcendental extension is an extension K(a):K where a is not  a zero of any non-zero polynomial over K. a is then said to be  transcendental over K.

Example 1:  Q(sqrt(2)) and R(i) are simple algebraic extensions of Q and R, respectively. R(i) is identical to C. sqrt(2) is a zero of the polynomial t2 - 2 over Q, and i is a zero of the polynomial t2 + 1 over R.

Example 2:  R[t]/<t2 + 1> is a simple algebraic extension of R. In this case R is not a subfield, but R is isomorphic to a subfield, namely {a + <t2 + 1>:a in R}.

Complex numbers, which are algebraic over Q, are simply called algebraic numbers. They constitute a field, and are studied in algebraic number theory. Complex numbers, which are not algebraic, are called transcendental. Some of the more famous transcendental numbers are e, pi, and 2sqrt(2).

For the next theorem we notice that a polynomial over a field, with highest degree coefficient 1, is said to be monic.

#### Theorem:

 If K(a):K is a simple algebraic extension, then there is a unique,  monic polynomial m(t) over K of lowest degree, such that m(a) = 0.  The polynomial m(t) is irreducible, and divides every polynomial  over K, of which a is a zero.

Note: The polynomial m(t) is called the minimum polynomial of a over K. A polynomial over K is said to be irreducible over K if it cannot be written as a product of two non-constant polynomials over K, of lower degree. That m(t) divides a polynomial f(t) over K means that f(t) can be written as a product of polynomials over K, where one factor is m(t).

#### Proof:

It is obvious that there exists at least one monic polynomial of lowest degree with zero a. Suppose f(t) and g(t) are such polynomials. Then  f(a) - g(a) = 0, i e a is a zero of  the polynomial  f(t) - g(t). But this polynomial is of lower degree, since f(t) and g(t) have the same highest degree term. This implies a contradiction unless  f(t) - g(t) = 0, i e  f(t) = g(t).

Next, suppose  m(t) = f(t)g(t). Then  f(a)g(a) = 0, so  f(a) = 0  or  g(a) = 0, since f(a) and g(a) are elements of the field K. This implies that a is a zero of a polynomial of lower degree than m(t), which means a contradiction. Thus, m(t) is irreducible.

Finally, suppose a is a zero of a polynomial f(t) with higher degree than m(t). Then, according to the division algoritm,  f(t) = q(t)m(t) + r(t), where r(t) has lower degree than m(t). From this  r(a) = 0, so r(t) is the zero polynomial, and m(t) is a factor of f(t).

Example:  t2 + 1  is the minimum polynomial of i over R, since it is monic, has zero i, and linear polynomials over R have real zeros only.

In an earlier example we proved that R[t]/<t2 + 1> is isomorphic to C. However, we do not need the complex numbers to prove that R[t]/<t2 + 1> is a field and to find the element  j = t + <t2 + 1>. In fact, all properties of complex numbers can be translated into properties of elements of R[t]/<t2 + 1>. Since the construction of this field does not actually use C, we have here a method of defining the complex numbers algebraically. Since the subset {a + <t2 + 1>:a in R} is isomorphic to R, we have then defined C as an extension of R, and we have actually constructed this extension.

What we have just described is a particular case of a general procedure for construction of an extension of a field K. In the general case an irreducible polynomial over K will have a zero in the extension field. This is described in the following theorem by Kronecker.

#### Theorem:

 If m(t) is a monic, irreducible polynomial over the field K, then there is  an extension K(a):K, such that a has minimum polynomial m(t) over K.

In fact, K(a) can be chosen as K[t]/<m(t)>, where <m(t)> denotes the ideal in K[t], consisting of all multiples of m(t).

Before proving the theorem, we state the following. Let f and g be coprime polynomials over a field. This means that they have no common non-constant factor. Then there exist polynomials a(t) and b(t) over the same field, such that  a(t)f(t) + b(t)g(t) = 1. This statement is analogous to a theorem about integers, and we do not prove it.

#### Proof:

First we prove that K[t]/<m(t)> is a field. Let f(t) + <m(t)> be any non-zero element in K[t]/<m(t)>. Then f(t) is not a multiple of m(t), i e it is not in <m(t)>. Since m(t) is irreducible f(t) and m(t) are then coprime, so there exist polynomials a(t) and b(t), such that  a(t)f(t) + b(t)m(t) = 1. In K[t]/<m(t)> this implies that

(a(t) + <m(t)>)(f(t) + <m(t)>) + (b(t) + <m(t)>)(m(t) + <m(t)>) = 1 + <m(t)>.

Now,  m(t) + <m(t)>  is the zero element of K[t]/<m(t)>, i e  (a(t) + <m(t)>)(f(t) + <m(t)>) = 1 + <m(t)>. Since  1 + <m(t)>  is the identity element (1) of K[t]/<m(t)>, this means that every non-zero element  f(t) + <m(t)>  has an inverse. Thus, K[t]/<m(t)> is a field.

Secondly, we show that K[t]/<m(t)> is an extension of K. Define a function F from K to K[t]/<m(t)> by  F(x) = x + <m(t)>  for all x in K. Then, if  F(x) = F(y),  x + <m(t)> = y + <m(t)>,  i e  x - y  is in <m(t)>. Since x and y are elements of K this can only happen if  x - y, as element of K[t]/<m(t)>, equals the zero polynomial. Then  x = y, so F is injective. The requirements for a homomorphism are obviously fulfilled, so F is a monomorphism, and K[t]/<m(t)> is an extension of K.

Thirdly, let  a = t + <m(t)>. Then  an = tn + <m(t)>, (k + <m(t)>)an = ktn + <m(t)>, etc. From this,  m(a) = m(t) + <m(t)> = <m(t)> = 0  in K[t]/<m(t)>, so a is a zero of m(t). For a polynomial n(t) of lower degree we get  n(a) = n(t) + <m(t)>  which is non-zero, since n(t) cannot be a multiple of m(t).

Finally, we show that  K(a) = K[t]/<m(t)>. K(a) consists of all rational expressions in  t + <m(t)>  with coefficients in K. All such expressions can be written as polynomials in  t + <m(t)>  of degree less than the degree of m(t). We do not show this in detail, but the essential point is that  f(t) + <m(t)>  has an inverse  g(t) + <m(t)>  unless f(t) is a multiple of m(t). This is shown as in part one of the proof.

If a polynomial over a field has a zero in the field, then the polynomial can be factorized with one linear factor. This, together with the above theorem, applied a finite number of times, implies that for any polynomial over a field there exists an extension field where the polynomial splits into linear factors, a so called splitting field. This also means that any algebraic equation (polynomial equation), if it is considered in a large enough field, has a number of solutions equal to its degree, if multiplicities are counted. If multiplicities are not counted the number of solutions is at least one and at most equal to the degree. Anyway, an algebraic equation has at least one solution in a sufficiently large field.

Example 1:  The polynomial  t2 + 1  over R splits in R[t]/<t2 + 1>.

Example 2:  Any polynomial over C splits in C. This statement is sometimes called the fundamental theorem of algebra. This name is also used for a couple of related statements, such as "any polynomial with complex coefficients has at least one complex zero".

Example 3:  Over the field Z3 the polynomial  t2 + 1  is irreducible since none of the three elements in Z3 is a zero. According to the divison algorithm any polynomial f(t) over Z3 can be written  f(t) = q(t)(t2 + 1) + at + b, where a and b are elements of Z3. Thus, any element in Z3[t]/<t2 + 1> can be written  at + b + <t2 + 1>. Since a and b take the values 0, 1, 2 independently of each other there are 9 combinations, so the field Z3[t]/<t2 + 1> has exactly 9 elements. It is isomorphic to the so called direct product  Z3×Z3.

### Vector Spaces

As a measure of the size of a simple algebraic extension K(a):K, the degree of the minimum polynomial of a over K can be used. That this is reasonable becomes clearer when the concept of a vector space is introduced.

#### Definition:

 V is a vector space over a field F if (V,+) is an abelian group, and there is  defined a multiplication of elements in V with elements in F, such that  (i)   a(u + v) = au + av  for all a in F and u, v in V,  (ii)  (a + b)u = au + bu  for all a, b in F and u in V,  (iii)  a(bu) = (ab)u  for all a, b in F and u in V,  (iv)  1u = u  for all u in V,  (v)   0u = 0  for all u in V.

Note: The multiplication of elements in V with elements in F can be defined as a function from F×V to V, so it is a generalized binary operation. The products are elements of V, also called vectors.
Example:  Let  R2 = {(x,y):x, y in R} with component-wise addition and multiplication with a real number. Then (R2,+) is an abelian group, and (i)-(v) in the definition hold with  V = R2  and  F = R. Thus, R2 is a vector space over R.

A set of vectors are said to be linearly independent if no linear combination equals 0 unless all coefficients (in F) are 0. If any vector in a vector space V can be written as a linear combination of a set of linearly independent vectors, then this set is said to constitute a basis for V. It can be proved that all bases of a vector space V have the same number of elements. This number is called the dimension of V.

If L:K is a field extension, then L is a vector space over K. The dimension of this vector space is called the degree of the extension, and is denoted [L:K].

#### Theorem:

 If K(a):K is a simple algebraic extension, then [K(a):K] equals  the degree of the minimum polynomial of a over K.

We will just scetch the proof. Let the the degree of the minimal polynomial m(t) be n. By means of the relation  a(t)f(t) + b(t)g(t) = 1  mentioned earlier, where f(t) and g(t) are coprime polynomials over K, it is shown that every element of K(a) can be written as a polynomial in a. The degree of this polynomial can be taken as  n - 1  or less, because powers of a with exponent larger than  n - 1  can be reduced by means of the equation  m(a) = 0.  But  1, a, a2, ..., an-1  are linearly independent, since otherwise a would have been a zero of a polynomial of degree less than n. Thus, they form a basis for K(a) as a vector space over K. Since they are n in number the dimension is n.

This theorem shows why the degree n of the minimal polynomial is a kind of size measure for the extension field K(a), compared with K. A field K(a) with a larger n has greater size than a K(a) with a smaller n, in the same sense that an Rn with a larger n has greater size than an Rn with a smaller n.

Example 1:  C = R(i)  has a basis {1,i} as a vector space over R, so  [C:R] = 2. Thus, the minimum polynomial of i over R has degree 2. This agrees, of course, with  t2 + 1  being the minimum polynomial.

Example 2:  [Q(21/3):Q] = 3  and  [Q(sqrt(2)):Q] = 2, since the minimal polynomials are  t3 - 2  and  t2 - 2, respectively. Thus Q(21/3) is, so to speak, as much larger than Q(sqrt(2)), that R3 is larger than R2, or a solid body is larger than a surface.

The great advantage of the connection between field extensions and vector spaces is that it allows the theory of vector spaces, linear algebra, to be used in the study of field extensions and their applications.

### Conclusion

We are almost at the end of this little survey of basic abstract algebra. Since one of my intensions was to provide a ground for a survey of Galois theory, I would like to mention one more concept, the Galois group.

If L:K is a field extension, then a K-automorphism A of L is an automorphism of L, such that  A(k) = k  for all k in K.
The K-automorphisms of L form a group under composition of functions, the so called Galois group of L:K, denoted G(L:K). Under rather general conditions the number of elements in G(L:K) equals [L:K]. However, the structure of the Galois group gives more detailed information on the extension than the vector space structure.

A central theme in Galois theory is algebraic equations. Consider an extension K(a):K. While the dimension of K(a) as a vector space says, so to speak, how far a is from being in K, the Galois group says something on how a can be expressed by operations on the elements of K. Let S be a splitting field for f(t) over K. Then G(S:K) is isomorphic to a subgroup of the symmetric group Sn, where n is the degree of f(t). Further, the zeros of f(t) can be expressed by means of rational operations and root extractions on the coefficients of f(t) if and only if G(S:K) is a so called soluble group. Thus, the question of which algebraic equations can be solved by radicals is transformed into a problem of group theory.

The most famous example is certainly the fifth degree equation. The corresponding Galois group is a subgroup of S5, sometimes the entire group. Now, it can be proved that S5 is not soluble, so fifth degree equations cannot, at least not always, be solved by radicals. However, Sn is soluble for all  n < 5, so equations of degree up to 4 can be solved by radicals, as was of course known before the emergence of Galois theory.

If, in particular, KQ we see that algebraic numbers are more complicated than the results of root extractions. Here we have a connection to algebraic number theory, too.

For those who want to go deeper into the subject, the book Galois Theory by Ian Stewart is recommended. Information on the more elementary parts of subject can be found in any textbook on abstract algebra and in mathematical encyclopedias.