1
$\begingroup$

Let $k$ be a field and consider the rational function field $k(t)$. I was just reading that if $f(t),g(t),h(t)\in k(t)$ are such that $f(h(t)) = g(h(t))$, then "$f = g$ in $k(h(t))$." What does that mean? I know that $k(h(t))$ is the smallest subfield of $k(t)$ containing both $k$ and $h(t)$, but what does it mean that $f = g$ in this subfield? I'm getting thrown off because I'd expect to see $f(X)$ or $f(t)$ or $f(something)$, not just $f$ by itself.

$\endgroup$
  • 1
    $\begingroup$ The idea is that $k(h(t))$ is a subfield of $k(t)$, one that you can "construct" as the image of the evaluation homomorphism $\phi: k(t)\to k(t)$ sending $t$ to $h(t)$. Of course $h(t)$ should be nonzero to make this work sensibly. $f,g$ are still elements of $k(t)$, but with a bit of notation abuse their "restrictions" (actually compositions with $h$) are still being called $f,g$. $\endgroup$ – hardmath Feb 7 '16 at 23:03
  • $\begingroup$ @hardmath Okay, that makes sense. So then if $h(t)$ is invertible (in the sense of composition) we can conclude that $f(t) = g(t)$, right? The example I was thinking of was $h$ being a linear fractional transformation. $\endgroup$ – justin Feb 7 '16 at 23:38
  • 1
    $\begingroup$ That's correct. $h( t) $ could be a moebius transformation or something else invertible in the sense of rational function composition, and that would guarantee $f=g$ on $k(t)$. $\endgroup$ – hardmath Feb 8 '16 at 0:24
1
$\begingroup$

There is a bit of abuse of notation involved here, since $f,g$ are being reused to identify their compositions $f\circ h$ and $g \circ h$ as if they were the same rational expressions in $k(t)$ but "restricted" to $k(h(t))$. But the notation needs to be taken with a grain of salt. It really means the compositions $f(h(t)) = g(h(t)$ are equal as rational expressions, which we can also express as equal images under an evaluation homomorphism on $k(t)$ sending $t$ to $h(t)$.

Now if $h(t)$ is a constant, then of course $f(h(t)) = g(h(t))$ is no guarantee that $f(t) = g(t)$ in $k(t)$. However in other cases the reverse implication will follow, e.g. if $h(t) = t+ \alpha$ were a linear function (or "birational" such as a Moebius transformation), then it would be true that:

$$ f=g \text{ in } k(h(t)) \implies f(t) = g(t) \text{ in } k(t) $$

I've got an inkling there must be more interesting examples of where the reverse implication fails, but it did not dawn on me after a good night's sleep.


Added: Heh, I feel better now. One of the "Related" links to the right was Lüroth's Theorem, according to which any subfield of $k(t)$ that is not $k$ itself is field isomorphic to $k(t)$. This result establishes that $f = g$ in $k(h(t))$ implies $f(t) = g(t)$ unless $h(t)$ is a constant (in $k$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.