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The current question is motivated by this question.

Call a polynomial map $f: \mathbb{C} \to \mathbb{C}^2$, $t \mapsto f(t):=(f_1(t),f_2(t))$ an embedding of $\mathbb{Z}$ in $\mathbb{Z}^2$ if $\mathbb{Z}$ is isomorphic to its image under $f$ (see A. van den Essen, page 2, for fields). By a polynomial map we mean that $f_1(t),f_2(t) \in \mathbb{Z}[t]$.

Is the following claim true: $f$ is an embedding if and only if $f'(t) \neq 0$ for all $t \in \mathbb{Z}$ and the map $f: \mathbb{Z} \to \mathbb{Z}^2$ is injective.

Notice that if we replace $\mathbb{Z}$ by $\mathbb{C}$, then the claim (is true and) appears in A. van den Essen's paper, while if we replace $\mathbb{Z}$ by $\mathbb{R}$, then the claim (is true and) appears in the comments to my above mentioned question.

Remarks: (1) I guess that I should carefully check the arguments in the comments to my previous question and see if they are still valid if we work with $\mathbb{Z}$.

(2) I really apologize if it happens that my question is not reasonable. (Perhaps only working over fields make sense? So what if we take $\mathbb{Q}$? Or certain properties of $\mathbb{R}$ are necessary?).

user237522
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  • What do you mean by isomorphic? Isomorphic as what? – MooS Mar 20 '18 at 19:06
  • Good question. I am not sure I know the answer. Isomorphic as rings will not be enough? – user237522 Mar 20 '18 at 19:10
  • How is the image of $f$ supposed to be a ring? I'm afraid your remark (2) applies and this question is not reasonable. – MooS Mar 20 '18 at 19:15
  • Consider $f:\mathbb{Z} \rightarrow \mathbb{Z}\times\mathbb{Z}$ with $f(t)=(t^2,t^3)$. Then $f$ is an embedding, but its derivative is not nonzero everywhere.

    (here I consider the embedding as a topological embedding)

     – Mihail Mar 20 '18 at 19:16
  • Thanks for both of you. What about $\mathbb{Q}$? – user237522 Mar 20 '18 at 19:21
  • @Mihail, what I actually wanted is to obtain that $A[t]=A[f_1(t),f_2(t)]$, where $A$ is some ring $A \in {\mathbb{Z}, \mathbb{Q}}$. In the definition $f$ having an inverse on its image, perhaps (and perhaps not) working with sets is enough? – user237522 Mar 20 '18 at 19:28
  • In other words, is Abhyankar-Moh theorem valid over any commutative integral domain? But I also liked the criterion with the derivatives. – user237522 Mar 20 '18 at 19:32

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