# Abstract Nonsense

## 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.