I am struggling with the following problem:
Let $F$ be the field of rational functions of $x$ and $y$ with coefficients in $\mathbb{Q}$. Let $A$ be the subring of $F$ generated by $x, y, \frac{y+1}{x}, \frac{x+1}{y}, \frac{x+y+1}{xy}$. Show that the map $f: A \longrightarrow A$ defined by $x \longmapsto \frac{y+1}{x}$ and $y \longmapsto y$ gives a ring automorphism of $A$.
I first thought to check that the properties $f(a + b) = f(a) + f(b)$ and $f(ab) = f(a) \cdot f(b)$ are satisfied for all $a,b \in A$. Is it enough to check that these properties are satisfied on the set of generators for $A$ ? If so, how can I check, for example, that $f(x+y) = f(x) + f(y)$ ? I'm only given information about $f(x)$ and $f(y)$ -- but not about $f(x + y)$. Similarly, how can I check that $f(xy) = f(x) \cdot f(y)$, if I don't know what $f(xy)$ is? Furthermore, how can I check what $f(\frac{x}{x})$ is? I need to know that the identity in $A$ is sent to the identity in $A$.
Thanks!
The $f$ you are given should have the phrase "extended by linearity" following the specification of its action on $x$ and $y$. Those words tell you that the two specified rewrites happen to every $x$ and to every $y$ in an element of $A$. For instance $f(x+y) = f(x)+f(y)$ and $f(x \cdot y) = f(x)f(y)$. Unfortunately, sometimes those words are omitted. And strictly, you are correct: without those words, $f$ is only defined at two points of $F$.
However, extension by linearity is required and may (when context demands) be assumed. Otherwise, this map has no hope of being a homomorphism, much less an automorphism.