## Automorphisms of k(t)/k

**Point of Post: **This is just a fun problem which computes the automorphisms of the extension where is some field and is the rational function field.

**Motivation**

This is another one of those slightly interesting problems that comes up in my day-to-day life, and I thought I’d share. Roughly the problem shows the very interesting result that the only automorphisms of the extension where is the rational function field in one variable (i.e. the quotient field of the polynomial ring in one variable) are the fractional linear transformations (i.e. that ) where . This gives us a mapping and we shall determine what it’s kernel is, giving us an alternate description of of the automorphism group of this extension.

**Automorphisms of **

Before we begin we prove a certain lemma concerning the degrees of certain

**Lemma: ***Let be a field and consider the ratioanl function field . Let then with and . Then, .*

**Proof: **We first note that the polynomial is irreducible over . Indeed, by Gauss’s lemma we need merely check that it is irreducible in but and this is clearly irreducible in the latter since it’s linear. Moreover, it’s easy to see that is a root of this polynomial and thus we see that is a minimal polynomial for over and thus

and so the result follows.

Using this we can complete the problem then:

**Theorem: ***Let be any field. Then, is equal to the set of all linear fractional transformations of the form where .*

**Proof: **Let then we know that if (where we can clearly assume that neither nor is constant) then

so that evidently if is going to be surjective we must have that . Thus, we see automatically that our only hope for an automorphism is a linear fractional transformation.

Assume now that are such that . Since is the free commutative -algebra on one generator and is a commutative -algebra we know that we can extend the set map to a -algebra map . If we can show that this map is injective (i.e. has trivial kernel) then by the universal characterization of quotient fields we may extend this to a map . To see that this map is injective assume that

and that for some . This then tells us that is algebraic over . If this is all easy so assume not then note that and since the algebraic elements of in form a field we have that we may subtract, multiply, and divide any algebraic elements to/from . That said, constants are all algebraic and and so we may clearly perform the following operations

to conclude that is algebraic over , but this is clearly ridiculous. Thus we see that all the ‘s must be zero, and thus has trivial kernel. Thus, (by previous comment) we are granted an extension which is clearly a -algebra map and which satisfies . Since this is a ring (field) map on a field we know that must be injective, and since we have (by the lemma) that we must have that is surjective, and thus we may conclude that is bijective. Thus, we then see that .

Now, to see that such a map won’t work if we merely note that if were an element of with then necessarily but then we see that isn’t even well-defined.

Combining the above results tells us that are precisely the fractional linear transformations coming from invertible matrices as desired.

From the above we clearly are able to create a mapping given by . Moreover, from the above (an a little elbow grease) we actually know that is an epimorphism, and so the first isomorphism theorem tells us that . That said it’s fairly easy to see that the matrix determines the corresponding automorphism up to scaling, so that (where we have proven before this last equality, where the stands for the center of a ring) and thus we have that where the stands for the projective linear group.

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

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