Showing automorphisms on $\mathbb{C}(x)$

Question

Let $\mathbb{C}(x)$ denote the field of rational functions over $\mathbb{C}$, the field of complex numbers.

Consider the six mappings $\phi : \mathbb{C}(x) → \mathbb{C}(x)$ defined by

$\phi_{1}:f(x) \rightarrow f(x)$

$\phi_{2}:f(x) \rightarrow f(1-x)$

$\phi_{3}:f(x) \rightarrow f(\frac{1}{x})$

$\phi_{4}:f(x) \rightarrow f(\frac{x-1}{x})$

$\phi_{5}:f(x) \rightarrow f(\frac{1}{1-x})$

$\phi_{6}:f(x) \rightarrow f(\frac{x}{x-1})$

Show that these mappings are automorphisms.

I decided to prove this by showing that they are bijective and homomorphic. The bijectivity is not a problem for any of them, but I don't know how to show that they are homomorphisms. Please advise. Thanks!

Answer 1

For the second function, let's show that $\phi_2$ is an homomorphism. You need to verify $3$ properties in order to prove that.

Let $f,g \in \mathbb{C}(x)$ : $$\phi_2((f+g)(x))=(f+g)(1-x)=f(1-x)+g(1-x)=\phi_2(f)+\phi_2(g)$$ $$\phi_2((fg)(x))=(fg)(1-x)=f(1-x)g(1-x)=\phi_2(f)\phi_2(g)$$ $$\phi_2(0_{\mathbb{C}(x)}(x))=0_{\mathbb{C}(x)}(x-1)=0_{\mathbb{C}(x)}$$

So $\phi_2$ is a field homomorphism, you can try to verify the same $3$ properties for the other examples.

Answer 2

To show these are homomorphisms, we use two standard results:

(a) evaluation map theorem for polynomial rings over an arbitrary ring

(b) Universal property for field of fractions.

We proceed as follows:

(a) Given a ring homomorphism $\mu:R\rightarrow S$ and an element $s\in S$. There exists a unique ring homomorphism $\bar{\mu}:R[x]\rightarrow S$ from the polynomial ring over $R$ to $S$ such that:

(1) $\bar{\mu}(r)=f(r)$ for all $r\in R$ and

(2) $\mu(x)=s$.

This $\bar{\mu}$ is given by $\bar{\mu}(r_{0}+r_{1}x+r_{2}x^{2}+\cdots+r_{n}x^{n})=\mu(r_{0})+\mu(r_{1})s+\mu(r_{2})s^{2}+\cdots+\mu(r_{n})s^{n}$.

Then in your problem, set $R=\mathbb{C},~S=\mathbb{C}(t),~\mu=inclusion~map$ sending $z\in\mathbb{C}$ to $z\in\mathbb{C}(x).$

Set $s=x,~1-x,~\frac{1}{x},~\frac{x-1}{x},~\frac{1}{1-x},~\frac{x}{x-1}$ in your $\phi_{1},~\phi_{2},~\phi_{3},~\phi_{4},~\phi_{5},~\phi_{6}$ respectively. This gives corresponding evaluation map $\mu_{i}$, $i=1,2,3,4,5,6$.

This has not completely solved your problem yet since we only get a map from the polynomial ring $\phi_{i}:\mathbb{C}[x]\rightarrow \mathbb{C}(x)$ for $i=1,2,3,4,5,6$ but not from the rational function field.

(b) The next thing we use is the universal property of field of fractions:

Let $R$ be an integral domain, $F$ be its field of fractions. $\psi:R\rightarrow S$ an injective ring homomorphism into a field $S$, then there exists a field homomorphism $\bar{\psi}:F\rightarrow S$ extending $\psi$.

It is given by $\bar{\psi}(\frac{r_{1}}{r_{2}})=\frac{\psi(r_{1})}{\psi(r_{2})}$.

We know that the field of fractions of $\mathbb{C}[x]$ is $\mathbb{C}(x)$.

To continue your problem, set $R=\mathbb{C}[x],~S=\mathbb{C}(x),~\psi=\mu_{i}$ from above. $\mu_{i}$ will be extended to $\phi_{i}$ for all $i=1,2,3,4,5,6$ as your desired.

To show these are bijective: (a)(b) tells u that these are homomorphism. To show these are automorphism, repeat (a)(b) to produce a homomorphism $\phi_{1}':\mathbb{C}(x)\rightarrow \mathbb{C}(x)$ sending $x$ to $-x-1$. Now show $\phi_{1}$ and $\phi_{1}'$ are mutual inverse of each other. Similar argument applies to other cases.