be a field and let 
be a polynomial in 
. We denote by 
the formal derivative of . The formal derivative has the properties
1998/99 GCW
Let be a field and let be a polynomial in . We denote by the formal derivative of . The formal derivative has the properties
if the degree of is zero
when 
.
Recall that given two polynomials 
we can use the Euclidean Algorithm to find the highest common factor 
of and 
, and that there are polynomials 
such that 
.
Proposition: Let 
be a field and let 
be a polynomial of positive degree. Let be the highest common factor of and . Then the following are equivalent:
is nonzero,
the polynomial has a repeated root,
the polynomial is divisible by the square of a polynomial of nonzero degree.
Proof: (ii) 
(iii) is trivial. (i) 
(ii) because if the degree of 
is nonzero then 
has a root, say 
, in some extension of 
, and since 
divides both and then
is a common root of both. But if 
and so 
, whence 
in some extension of 
. (ii) 
(i) by a reverse argument. For if has a repeated root, and have a common root, which must therefore be shared by 
.
We call a polynomial separable if the highest common factor of the polynomial and its formal derivative has degree zero. It follows from the arguments above that a polynomial is separable if it has no repeated roots in any extension of the field in which its coefficients lie.
Proposition: Let 
be a field of characteristic zero. Let be an irreducible polynomial in . Then is separable.
Proof: The highest common factor of and has to divide , which is irreducible. So it is either a constant, in which case is separable, or itself. But this is not possible unless 
. But if 
has characteristic zero and the degree of is nonzero this does not happen.
It can happen in fields of nonzero characteristic. Suppose that 
has characteristic 
and that 
where 
is not a 
-th power in 
. Then 
, so is not a separable polynomial. There is an extension of 
in which 
where 
.
Proposition. Suppose that 
is a field of characteristic 
, and that is such that 
. Then 
for a unique 
.
Proof. Let 
. Then 
implies that 
unless 
divides 
. Take 
.
Proposition. Let 
be a field of characteristic 
and let be an irreducible polynomial. Then is separable if and only if 
.
Proof. Since is irreducible, the highest common factor of and must be either 1 or . In the first case is separable. In the second case, since the degree of is less than the degree of and divides , we must have 
.
If 
is a field and is an element algebraic over 
, we say that is separable over 
if its minimal polynomial 
over 
(which is necessarily irreducible in ) is separable. If is algebraic of degree 
, so that 
is of degree 
, and separable, then in 
, 
will have 
distinct roots, one of them being . We call these roots the conjugates of over 
.
An algebraic extension 
is normal if every polynomial over 
that is the minimal polynomial of an element of 
splits into linear factors over (one out, all out). We can say that for every element of all its conjugates over also belong to .
A standard example of an algebraic extension that is not normal is provided by
where 
denotes the real cube root of 2. The point is that all the elements of 
are real, whereas the conjugates of 
are not all real, being
The minimal polynomial over 
of 
is 
, which factorizes as
over 
.
Proposition: Any extension of degree 2 is normal.
Proof: Let 
be the minimal polynomial of an element in the extension. Then the conjugates of are 
and so belong to the extension.
For any algebraic extension we can find a minimal extension 
so that 
is normal. We simply adjoin all the conjugates over of elements of . We call 
the normal closure of . The normal closure of 
can be written as 
.
Theorem: Let be a field with an infinite number of elements. Then any extension which is finite and separable is simple. That is to say, there is an element 
such that 
.
In fact this theorem is also true for finite fields, but a different proof is required.
Proof: First we show that if 
and 
are separable algebraic over then 
for some 
. Let 
be the minimal polynomial of 
over and let 
be the minimal polynomial of 
over 
. Choose an element 
of 
different from any of the numbers
where 
are conjugates of 
over and 
are conjugates of 
over . The assumption of separability tells us that the numerator and denominators of these fractions are nonzero, and because there are only finitely many such expressions the assumption that be infinite tells us that it is possible to choose such a 
. In consequence the numbers 
are all distinct as 
ranges over the conjugates of 
and 
ranges independently over the conjugates of 
. Let 
. The polynomial 
has 
as a root, and so the highest common factor 
of 
and 
is not of degree zero, since 
. But all the roots of 
have to be conjugates of 
since they are also roots of 
. If 
were a root of 
we would have 
. But this implies that 
for some conjugate of 
. But our choice of 
makes this impossible if 
is different from 
. Hence 
.
But as 
is the highest common factor of two polynomials in 
we deduce that 
. Hence 
. So each of 
and 
is contained in the other.
As is finite, we can choose a basis 
of over . Using induction and the result just proved we deduce that 
for some 
in .
This is useful for simplifying arguments. We see that if 
is a finite separable extension, and if 
is the minimal polynomial of over , then the extension is normal precisely when every root of 
, i.e. conjugate 
of over , can be expressed in the form 
for some polynomial 
. These are called the root polynomials of 
over .