*1/x*

*x*

^{9}+ 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 numbers if you like) is a quotient

*f(x)/g(x)*of polynomials, with

*g(x)*nonzero.

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
of *k(x)*.

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(x ^{2})*
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.

**Proof:**
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.

**Corollary**
Gal(*k(x)/k*) is isomorphic to
*PGL _{2}(k)*.

**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
*h:GL _{2}(k)-->*Gal(

*k(x)/k*) defined by mapping a 2x2 invertible matrix

- - | a b | | c d | - -to the automorphism that maps

*x*to

*ax+b/cx+d*. Notice that this homomorphism has a kernel consisting of nonzero scalar multiples of the identity. But

*PGL*is exactly the quotient of

_{2}(k)*GL*by this subgroup. So we are done by first isomorphism theorem.

_{2}(k)