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Dec 29, 2021 at 6:24 answer added pinaki timeline score: 1
Dec 29, 2021 at 1:10 vote accept user237522
Dec 28, 2021 at 21:39 answer added Jérémy Blanc timeline score: 3
Dec 23, 2021 at 12:56 comment added user237522 If $\deg(f)=\deg(g)=n$ and $\deg(h)=n-1$ (so $\deg(f_1)=\deg(g_1)=1$), then I think it is not difficult to show that $t \mapsto (f(t),g(t))$ is injective, so with new condition (4) we obtain that $\mathbb{C}[f,g]=\mathbb{C}[t]$. But for $\deg(h) \geq 2$ and $\deg(f_1)=\deg(g_1) \geq 2$, I do not know how to show that $t \mapsto (f(t),g(t))$ is injective.
Dec 22, 2021 at 23:56 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 23:39 comment added user237522 Therefore, and in view of idea (i), we will add the condition that $f',g'$ are not simultaneously zero. Then, the above counterexample is not a counterexample anymore, since $f'=15t^{14}+2t$, $g'=15t^{14}$ have a common zero at $0$.
Dec 22, 2021 at 23:21 comment added user237522 $f=t^{15}+t^2$, $g=t^{15}$ is a counterexample....
Dec 22, 2021 at 23:20 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 23:00 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 22:21 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 19:31 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 19:14 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 19:05 history edited LSpice CC BY-SA 4.0
Edited link to Q295158 per https://mathoverflow.net/questions/412323/a-variation-on-abhyankar-moh-suzuki-theorem#comment1056641_412323
Dec 22, 2021 at 19:04 comment added user237522 @LSpice, no, it was not intensional. Thank you for letting me know.
Dec 22, 2021 at 18:59 comment added LSpice Your second question link (mathoverflow.net/questions/295158/…) goes to an answer. Was that intentional?
Dec 22, 2021 at 18:59 history edited user237522 CC BY-SA 4.0
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Dec 22, 2021 at 18:59 history edited LSpice CC BY-SA 4.0
`|` -> `\mid`; inline question links
Dec 22, 2021 at 18:55 history asked user237522 CC BY-SA 4.0