x9 + 5x - 1
(2x + 3)/(x - 1)
are all rational functions in one variable
. In general
a rational function over a field k
(just think of
the real number
s if you like) is a quotient f(x)/g(x)
s, with g(x)
Rational functions do define k-valued functions but you have to be
careful with the domain. You have to restrict it to k without
the zeroes of g(x). Obviously you get a pole, a place where
the function is not defined, if you try to extend to the domain to
include one of these zeroes.
The collection of all rational functions in x is denoted by
k(x). You can add, multiply and divide rational functions
and in fact they form a field, a field extension of k.
Notice that all the polynomials are sitting inside
k(x) so that k[x] is a subring
if you take some rational function y then you can look
at the subfield k(y) of rational functions in y.
It's an interesting problem to decide when k(y)=k(x).
For example, k(x + 1)=k(x) but k(x2)
is obviously too small.
Theorem Let y be a rational function. Then k(y)=k(x)
iff y=ax+b/cx+d, for some a,b,c,d in k such that
ad-bc is nonzero.
Write y=f(x)/g(x) so that g(x) is nonzero and
f(x) and g(x) are coprime. Let
t=max(deg f, deg g). We will
show in a minute that t=[k(x):k(y)], the dimension of
k(x) considered as a k(y)-vector space.
Let's observe first that this fact gives the result. For k(y)=k(x)
iff this dimension is 1. Thus y will have the form
ax+b/cx+d (and we have taken the determinant to be nonzero
to make sure y is not a scalar).
So let's try to demonstrate the claim. Consider the polynomial
in k(y)[z] (that is the ring of polynomials in
an indeterminate z with coefficients from k(y))
h(z)=f(z)-yg(z). This polynomial has x as a zero.
Firstly, I claim that h(z) is nonzero. That's simply
because g(z) is nonzero (by assumption). As a consequence,
we see that x is algebraic over k(y).
Next I claim that h(z) is irreducible in k(y)[z].
This follows from Gauss's lemma. For, any factorization
in k(y)[z] is one in k[y,z]. Because h(z)
has degree one in y if we write it as a product of two factors
one of them contains y and the other does not.
Let's say f-yg=p(a+by), for a,b,p in k[z].
Equating coefficients of powers of y we quickly see that
p is a common factor of f and g, contradicting
our choice that they were coprime.
Thus, h(z) is the minimal polynomial of x over k(y).
This tells us that the dimension of k(x) over k(y)
is the degree of h(z) which is t.
Finally, we can rephrase the theorem in the following way.
Gal(k(x)/k) is isomorphic to
Proof: To describe a k-ring homomorphism
f of k(x) one simply has to specify f(x).
Such an f is automatically injective (because k(x)
is a field). To be surjective (and hence be
a k-automorphism) one needs that k(x)=k(f(x)).
Thus the theorem tells us that there is a surjective group homomorphism
defined by mapping a 2x2 invertible matrix
| a b |
| c d |
to the automorphism that maps x
Notice that this homomorphism has a kernel
consisting of nonzero
scalar multiples of the identity. But PGL2(k)
exactly the quotient of GL2(k)
by this subgroup.
So we are done by first isomorphism theorem