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I'm stuck at the very beginning of an exercise I have to do for my algebra class.

We're looking at the field of $\mathbb{C}(X)$ and it's automorphisms.

Let $a \in \mathbb{C}^*$, $ b \in \mathbb{C} $ and $ \sigma : \mathbb{C}(X) \to \mathbb{C}(X) $ given by $ X \mapsto aX + b$.

a) Explain why $\sigma$ is a field-automorphism.

I wrote that $\sigma$ is a coordinate transform in $\mathbb{C}(X)$ and therefore must be a field-automorphism.

Is this explanation correct?

b) Let $ H = \langle \sigma \rangle \subseteq Aut(\mathbb{C}(X)) $. Find the field fixed by $H$, i.e. $\mathbb{C}(X)^H$.

I know that $ \mathbb{C}(X)^H = \{ f \in \mathbb{C}(X) \; | \; \tau(f) = f \;\; \forall \tau \in H \} $ but I did not really get around how to decode this automorphism into something I can use to find the field fixed.

Any hints how to find it would be greatly appreciated.

Edit:

Since I am still stuck I wanted to add some thoughts of me. I'm working on b).

I tried some cases for $a$, $a=1$ is obvious.

$a = -1$:

When we only take a look at monoms $f(X) = X^i$ we have to conclude that $\sigma^i(f) = f$ for $i \in \mathbb{N}$ holds only for even exponents of $X^i$ and $\mathbb{C}(X)^H = \mathbb{C}(X^2)$. (Where we write $\sigma^i$ for $\sigma(\sigma(...))$.)

$a = i$:

With the same argumentation we see that $\mathbb{C}(X)^H = \mathbb{C}(X^4)$.

I now assumed that: $|a| = 1$ (hence $a$ is a root of unity) or else the field fixed would only consist of $\mathbb{C}$. I came up with the explanation, that when $|a| \not = 1$ you may take a look at any monom $X^j$. When you evaluate $\sigma$ $i$-times on $X^j$ you get: $\sigma^i(X^j) = a^{i+j} X$.

If $|a| > 1 $ or $|a| < 1$ you are either making the coefficient very large or close to zero by evaluation $\sigma$ many times. It can't hold that $\sigma^i(f) = f$ therefore.

What I couldn't find out if there are other fixed fields (besides $a = -i$).

As Jyrki Lahtonen pointed out, there are solutions where $ b \not = 0$. For example $ X \mapsto -X - 1$ and $ -X(X+1) $ is invariant under $\sigma$. I generalized that a bit and came up with $$ \sigma : X \mapsto -X - b \\ \sigma(-X(b+X)) = -(-X-b)(b - X + b) = -X(b+X) \\ \text{And clearly: } ord \; \sigma = 2 $$ For $b=1$ The field fixed for this is $ \mathbb{C}(X^2+X) $.

What I'm now trying to do is proof that: $$ \mathbb{C} \subsetneq \mathbb{C}(X)^H \implies ord \; \sigma \lt \infty $$ The argument for that lies within the answer of Michael Zieve though I couldn't figure it out until now: Intersection of two subfields of the Rational Function Field in characteristic $0$

Community
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kummerer94
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    If, for example $\sigma(X)=-1-X$, then $\sigma$ is of order two, and the product $-X(1+X)$ is invariant under $\sigma$. It follows that the fixed field is $\Bbb{C}(X^2+X))$. Because all the intermediate fields have $\Bbb{C}(X)$ as a finite extension, you should concentrate on finding automorphisms $\sigma$ that have a finite order. Clearly $a$ has to be a root of unity, but as the above example shows, there are surprise options for $b$. Too late here for me to spend more time with this. – Jyrki Lahtonen Jan 13 '15 at 23:06
  • Thank you so much! This at least leaves me back with a start. :) – kummerer94 Jan 13 '15 at 23:09
  • The argument for showing that if the fixed field is a proper extension of $\Bbb{C}$, then $\sigma$ must have finite order, is from Michael Zieve's answer to the linked question. – Jyrki Lahtonen Jan 13 '15 at 23:14
  • I will take a look at it tomorrow! – kummerer94 Jan 13 '15 at 23:18
  • @JyrkiLahtonen Sorry to bother you again, but I couldn't determine where exactly Michael Zieve made that argument in his answer. Is it within the first paragraph? – kummerer94 Jan 14 '15 at 12:17

1 Answers1

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About a), you should probably say better that $\sigma$ is an invertible coordinate transformation, but why not just checking the axiomatic definition of field automorphism?

About b), let $f\in\Bbb C(X)$ such that $f(aX+b)=f(X)$. Recall that $f$ is a quotient of polynomials. What can we say about the set of zeroes and poles of $f$?

Andrea Mori
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  • About a), I'm not quite sure how to proof it according to the axiomatic definition.

    I know that I have to show that $\sigma(f(X) \cdot g(X)) = \sigma(f(X)) \cdot \sigma(g(X)) $ (the same for $+$) and that $\sigma$ is bijective.

    Am I allowed to say that: $f(\sigma(X)) = \sigma(f(X))$?

     – kummerer94 Jan 11 '15 at 16:08
  • The definition is $\sigma(f(X))=f(aX+b)$ and you need to check that $\sigma(f\cdot g)=\sigma(f)\cdot\sigma(g)$. On the other hand $\sigma((f\cdot g)(X))=f\cdot g(aX+b)$ by definition so that .... (same with $+$) – Andrea Mori Jan 11 '15 at 16:33
  • Thank you! Now that is understood. I just tried showing in according to the definition and quickly ran into the problem of:

    $$ \sigma(f \cdot g) = \sigma ( \frac k l \cdot \frac m n ) $$

    I don't really know how to cope with the fact that $f$ is a quotient of two polynomials and what the automorphism does with it.

     – kummerer94 Jan 11 '15 at 16:39
  • Ok. I understand a) since I used the evaluation homomorphism and got rid of my problems.

    Now to b): I thought about your hint and came to the conclusion that if there is any $f\in \mathbb{C}(X)$ which is not constant and $f \in \mathbb{C}(X)^H $ then $b = 0$.

    This must hold but I couldn't come up with a proof so far.

    Since we are in algebraic closure we can say that $f$ degenerates into linear factors. $\sigma$ therefore has to permute the roots of $f$ if we want $f(X) = f(aX + b)$ to be true.

    But how can I use this fact to show that $b = 0$?

     – kummerer94 Jan 12 '15 at 15:59