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.
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1The 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 $f,g$. – hardmath Feb 07 '16 at 23:03
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@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. – justin Feb 07 '16 at 23:38
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1That'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 $k(t)$. – hardmath Feb 08 '16 at 00:24
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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 $k$).
