Abstract Nonsense

Crushing one theorem at a time

Automorphisms of k(t)/k


Point of Post: This is just a fun problem which computes the automorphisms of the extension k(t)/k where k is some field and k(t) is the rational function field.

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Motivation

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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 k(t)/k where k(t) 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 \displaystyle t\mapsto \frac{at+b}{ct+d} (i.e. that \displaystyle f(t)\mapsto f\left(\frac{at+b}{ct+d}\right)) where ad-bc\ne 0. This gives us a mapping \text{GL}_2(k)\to \text{Aut}(k(t)/k) and we shall determine what it’s kernel is, giving us an alternate description of of the automorphism group of this extension.

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Automorphisms of k(t)/k

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

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Lemma: Let k be a field and consider the ratioanl function field k(x). Let then \displaystyle t=\frac{P(x)}{Q(x)}\in k(x) with (P,Q)=1 and Q\ne0. Then, \left[k(x):k(t)\right]=\max\{\deg P,\deg Q\}.

Proof: We first note that the polynomial P(X)-tQ(X)\in k(t)[X] is irreducible over k(t). Indeed, by Gauss’s lemma we need merely check that it is irreducible in k[t][X] but k[t][X]=k[X][t] and this is clearly irreducible in the latter since it’s linear. Moreover, it’s easy to see that x is a root of this polynomial and thus we see that P(X)-tQ(X) is a minimal polynomial for x over k(t) and thus

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\max\{\deg P,\deg Q\}=\deg_X(P(X)-tQ(X))=[k(t)(x):k(t)]=[k(x):k(t)]

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and so the result follows. \blacksquare

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Using this we can complete the problem then:

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Theorem: Let k be any field. Then, \text{Aut}(k(t)/k) is equal to the set of all linear fractional transformations of the form \displaystyle t\mapsto \frac{at+b}{ct+d} where ad-bc\ne 0.

Proof: Let \sigma\in\text{Aut}(k(t)/k) then we know that if \displaystyle t\mapsto \frac{P(t)}{Q(t)} (where we can clearly assume that neither P nor Q is constant) then

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\displaystyle [k(t):\text{im }\sigma]=\left[k(t):k\left(\frac{P(t)}{Q(t)}\right)\right]=\max\{\deg P,\deg Q\}

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so that evidently if \sigma is going to be surjective we must have that \deg P=\deg Q=1. Thus, we see automatically that our only hope for an automorphism k(t)/k is a linear fractional transformation.

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Assume now that a,b,c,d\in k are such that ad-bc\ne 0. Since k[t] is the free commutative k-algebra on one generator and k(t) is a commutative k-algebra we know that we can extend the set map \displaystyle t\mapsto \frac{at+b}{ct+d} to a k-algebra map \varphi:k[t]\to k(t). 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 k(t)\to k(t). To see that this map is injective assume that

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\displaystyle 0=\varphi\left(\sum_j a_j t^j\right)=\sum_j a_j \left(\frac{at+b}{ct+d}\right)^j

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and that a_j\ne 0 for some j. This then tells us that \displaystyle \frac{at+b}{ct+d} is algebraic over k. If c=0 this is all easy so assume not then note that \displaystyle \frac{at+b}{ct+d}=\frac{a}{c}+\frac{bc-ad}{c^2t+cd} and since the algebraic elements of k in k(t) form a field we have that we may subtract, multiply, and divide any algebraic elements to/from \displaystyle \frac{a}{c}+\frac{bc-ad}{c^2t+cd}. That said, constants are all algebraic and ad-bc\ne0 and so we may clearly perform the following operations

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\displaystyle \frac{a}{c}+\frac{bc-ad}{c^2t+cd}\implies\frac{bc-ad}{c^2t+cd}\implies \frac{c^2t+cd}{bc-ad}\implies c^2t+cd\implies c^2t\implies t

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to conclude that t is algebraic over k, but this is clearly ridiculous. Thus we see that all the a_j‘s must be zero, and thus \varphi has trivial kernel. Thus, (by previous comment) we are granted an extension \overline{\varphi}:k(t)\to k(t) which is clearly a k-algebra map and which satisfies \displaystyle t\mapsto \frac{at+b}{ct+d}. Since this is a ring (field) map on a field we know that \overline{\varphi} must be injective, and since we have (by the lemma) that [k(t):\text{im }\overline{\varphi}]=1 we must have that \overline{\varphi} is surjective, and thus we may conclude that \overline{\varphi} is bijective. Thus, we then see that \overline{\varphi}\in\text{Aut}(k(t)/k).

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Now, to see that such a map won’t work if ad-bc\ne 0 we merely note that if \varphi:k(t)\to k(t) were an element of \text{Aut}(k(t)/k) with \displaystyle t\mapsto \frac{at+b}{ct+d} then necessarily \varphi(dt-b)=0 but then we see that \displaystyle \varphi\left(\frac{1}{dt-b}\right) isn’t even well-defined.

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Combining the above results tells us that \text{Aut}(k(t)/k) are precisely the fractional linear transformations coming from invertible matrices as desired. \blacksquare

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From the above we clearly are able to create a mapping \lambda:\text{GL}_2(k)\to \text{Aut}(k(t)/k) given by \displaystyle \lambda\left(\begin{matrix}a & b\\ c & d\end{matrix}\right)(t)=\frac{at+b}{ct+d}. Moreover, from the above (an a little elbow grease) we actually know that \lambda is an epimorphism, and so the first isomorphism theorem tells us that \text{Aut}(k(t)/k)\cong \text{GL}_2(k)/\ker \lambda. That said it’s fairly easy to see that the matrix determines the corresponding automorphism up to scaling, so that \ker\lambda =\left\{aI:a\in k^\times\right\}=Z(\text{GL}_2(k)) (where we have proven before this last equality, where the Z stands for the center of a ring) and thus we have that \text{Aut}(k(t)/k)\cong \text{PGL}_2(k) where the \text{PGL} stands for the projective linear group.

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References:

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

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February 28, 2012 - Posted by | Algebra, Field Theory | , , , , , ,

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