Abhyankar-Moh embedding theorem without algebraic closedness

After asking this question, I figured out that I am also interested in the following related question:

Is Abhyankar-Moh theorem 1.6 still valid if we remove the algebraic closedness assumption?

The answer is probably no, but I do not know how to find a counterexample over $\mathbb{R}$, for example.

Moreover, if I am not wrong, AM theorem 1.6 appears as Theorem 1 of van den Essen's paper; I do not see where algebraic closedness is needed in the proof.

Thank you very much for any comments or hints!

Edit: I have recently found this beautiful criterion (Theorem 3.3), which does not require algebraic closedness of the base field.

1 Answer

In the van den Essen's paper, a polynomial map $t\mapsto (x(t),y(t))$ is called an embedding if there is a polynomial $F$ such that $F(x(t),y(t))=t$. There is no problem with his proof: Theorem 1 is true over any field if we interpret embedding in this sense.

The reason why he considered $\mathbb{C}$ rather then an arbitrary field is that the above definition is weird if used for not algebraically closed fields such as reals. (In this case, one might call it something like "strong embedding" or "nice embedding", but just "embedding" would be quite misleading.)

• Thank you very much! Please can you explain what is weird in the definition if we replace $\mathbb{C}$ by $\mathbb{R}$? – user237522 Apr 11 '18 at 17:04
• It is at odds with the common usage of the term. In differential geometry, a map of this sort is an embedding if it is injective and has no singular points. Over complex numbers it is the same as above (even if it is not immediately obvious), but over reals this is a much weaker condition. – Alex Gavrilov Apr 11 '18 at 17:09
• This is terminology from differential geometry though. I am not sure what would be the meaning of "embedding" in algebraic geometry over a general field, if any. (This isn't my subject.) – Alex Gavrilov Apr 11 '18 at 17:15
• Thank you! I 'thought' that over $\mathbb{R}$ it is also true that $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$ if and only if $f'(t) \neq 0$ for all $t \in \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R}^2$ is injective. Actually, I was not sure, so I asked this here: math.stackexchange.com/questions/2698100/… – user237522 Apr 11 '18 at 17:39
• Unfortunately, your question was a bit vague. Anyway, $(t^2, t+t^3)$ is already a counterexample. – Alex Gavrilov Apr 11 '18 at 17:55