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Let $k$ be a field of characteristic zero. Let $p,q \in k[x,y]$ have invertible Jacobian, namely, $p_xq_y-p_yq_x \in k^*$ (partial derivatives). It can be shown that each of $k[x,y]$ and $k[p,q]$ is dense in $k[[x,y]]$, see this related question. Then in the induced topology, $k[p,q]$ is dense in $k[x,y]$. (More precisely, when considering $p,q \in k[[x,y]]$, we should further assume that each of $\{p,q\}$ have no constant term).

But what if I do not like to have infinite sums:

Is it possible to find for $k[x,y]$ a metric $d$ or a topology $\mathcal{T}$such that $k[p,q]$ is dense in $(k[x,y],d)$ or in $(k[x,y],\mathcal{T})$?

Remarks: (1) See also this question.

(2) It is clear the $k[p,q]$ is not dense in the metric $d(u,v):=2^{-l}$, where $u=\sum_{i=0}^{n}u_iy^i,v=\sum_{i=0}^{m}v_iy^i \in k[x,y]$, $u_i,v_i \in k[x]$, and $l$ is the minimal such that $u_l \neq v_l$.

(3) If I am not wrong, I can show that a positive answer to my above question implies that the two-dimensional Jacobian Conjecture is true; notice that the opposite implication is trivial: If the two-dimensional JC is true, then $k[p,q]=k[x,y]$, and then trivially $k[p,q]$ is dense in $k[x,y]$ in any topology.

Important edit: In the third remark above, the accurate statement is: "If I am not wrong, I can show that a positive answer to my above question with a metric such that $k[x,y]$ is complete, implies that the two-dimensional Jacobian Conjecture is true". If $k[x,y]$ is not complete in that metric or we only found a topology, then I do not know if this implies a positive answer to the two-dimensional Jacobian Conjecture.

user237522
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  • You can talk about the topology on $k[x,y]$ induced from $k[![x,y]!]$ without mentioning power series: you take a basis for the topology to be ${f + (x,y)^n\mid f\in k[x,y], n\in\Bbb N_0}$. The completion of $k[x,y]$ with respect to this topology is $k[![x,y]!]$, but you don't complete to talk about it. Then you can just take the subspace topology on $k[p,q]$. My point is, I suppose, an induced topology on a subspace is still a topology in its own right. Does this not suit you? – Stahl Apr 23 '17 at 02:17
  • @Stahl, thanks!! What bothered me is that $x$ is a limit of a sequence of elements from $k[p,q]$ (and since we do not know if $k[p,q]=k[x,y]$, then $x$ is an infinite sum of monomials in $p$ and $q$). – user237522 Apr 23 '17 at 02:33
  • @Stahl, I think that I wanted to consider $k[x,y]$ as a metric space, in order to apply a theorem concerning extension of a continuous function-- did you suggest to forget about the metric or am I missing something? (The problem is that I do not know yet if a similar theorem holds for a general topological space; perhaps yes). – user237522 Apr 23 '17 at 03:01
  • I think that the problem is that $k[x,y]$ is not complete in that topology, and in order to extend a continuous function on $k[p,q]$ to $k[x,y]$, $k[x,y]$ must be complete. Please see https://math.stackexchange.com/questions/2233292/if-a-bijection-extends-continuously-a-finite-order-bijection-is-it-also-of-fini – user237522 Apr 23 '17 at 03:14
  • I did not mean to suggest that this was a metric space. Assuming $k$ has a metric, you can place a metric on $k[![x,y]!]$ (hence on $k[x,y]$), but I don't know of a "natural" metric to consider (you might also see the answer here). In the question, it was not clear that you wanted a metric topology on the rings; after all, you had only asked for a metric or topology such that $k[p,q]\subseteq k[x,y]$ is dense. – Stahl Apr 23 '17 at 03:17
  • However, I'll remark that the problem should not be completeness. If you have a function $f : S\to Y$, where $S\subseteq T\subseteq X$, and you know that $f$ extends [uniquely] to $g : X\to Y$, then you know that $f$ extends to $T$ (perhaps nonuniquely), as you can take an extension to be $\left. g\right|_T : T\to Y$. (Although if uniqueness is a concern, then completeness might play some role.) – Stahl Apr 23 '17 at 03:25
  • You are right that I did not ask for a metric only (I still hope that it will be possible to apply one of the theorems for not necessarily metric spaces, for example as I mentioned in the above link). My problem is that, in your notations, I need that the restriction of $g$ to $T$ will map $T$ to $T$... – user237522 Apr 23 '17 at 03:49
  • You can write your first comment as an answer, if you like. – user237522 Apr 23 '17 at 03:50
  • In that case, I will post my comment as an answer you can accept, and I would suggest asking the question again with the properties you need your topology/metric to satisfy being specified in the question, so you can hopefully find an appropriate answer. – Stahl Apr 23 '17 at 03:52
  • ok, thank you very much for all your efforts to help. – user237522 Apr 23 '17 at 03:54

1 Answers1

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An induced topology on a subspace is still a topology in its own right -- you can talk about the topology on $k[x,y]$ induced from $k[\![x,y]\!]$ without mentioning power series. A basis for the topology is $\{f+(x,y)^n\mid f\in k[x,y],n\in\Bbb N_0\}$. The completion of $k[x,y]$ with respect to this topology is $k[\![x,y]\!]$, but you don't need to complete to talk about it. Then you can just take the subspace topology on $k[p,q]$.

Stahl
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  • Perhaps it is possible to apply a theorem of Taimanov, Theorem 1 in the following article http://www.ams.org/journals/proc/1976-054-01/S0002-9939-1976-0390999-0/S0002-9939-1976-0390999-0.pdf; however, I am not sure if $k[x,y]$ is Hausdorff and condition (2) is satisfied? (it seems that $k[x,y]$ is not compact, so I am trying to apply the second option that it is Hausdorff). – user237522 Apr 24 '17 at 22:48