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I want to know if there can exist a map of $n$ polynomials in $m$ variables, over the Rational numbers, $\mathbb{Q}^m$, such that if each polynomial corresponded to a coordinate in n dimensional Rational space, the map would be surjective.

If we were working over the Real numbers (or any smooth manifold), then we could use Sard's mini theorem to show that this type of map can't exist (because any such map would have measure $0$).

If we can show that a continuous differentiable map from $\mathbb{R}^m$ to $\mathbb{R}^n$, that maps all of the Rational values in $\mathbb{R}^m$ to the Rational values in $\mathbb{R}^n$ either

  1. can not have measure zero,
  2. must be closed,
  3. must contain some ball of nonzero radius,

then my question will be answered.

Intuitively, I think that if such a map exists over the Rationals then it could be extended to a surjective map over the Reals, so I think that such a map can't exist, but I have been unable to prove it.

Any comments or ideas are appreciated.

Daniele Tampieri
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Joe42
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    Can I suggest that you look [here][https://terrytao.wordpress.com/2019/06/08/ruling-out-polynomial-bijections-over-the-rationals-via-bombieri-lang/] for Terry Tao's recent post on polynomial bijections which may be related and also [this][https://mathoverflow.net/questions/21003/polynomial-bijection-from-mathbb-q-times-mathbb-q-to-mathbb-q] on MathOverflow, which suggests that one of the simplest cases is rather harder than it looks. – Mark Bennet Jun 21 '19 at 11:16
  • As pointed out to me, my previous comment would be a case with $m\gt n$ - and you will see that injectivity is a live question. It is helpful if you put the whole of your question in the body of the text, including the conditions you want. – Mark Bennet Jun 21 '19 at 11:35

1 Answers1

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There is such a smooth map. Let $S$ be any countable subset in $\mathbb{R}^n$, consisting of points $s_i$. To build a surjective smooth map from $\mathbb{Q}$ to $A$ just send $[2i, 2i+1]$ to $s_i$ and interpolate smoothly (using bump functions) on $[2i+1, 2i+2]$. Of course if you want a map from $\mathbb{Q}^m$ you can just project to $\mathbb{Q}$ first.

Polynomial maps is another kettle of fish entirely.

You can see that there is no such polynomial map from $\mathbb{Q}$ to any $\mathbb{Q}^n$ ($n>1$) by arguing along the lines outlined in your answer, because single-variable polynomial maps are proper. Any proper, continuous, surjective map $\mathbb{Q} \to \mathbb{Q}^n$ has a continuous, proper extension to a map $\mathbb{R} \to \mathbb{R}^n$ which will also be surjective (approximate $p\in \mathbb{R}^n$ by points $p_i$ in $\mathbb{Q}^n$; pick a sequence of their preimages; it will lie in some compact, and so subconverge to some $s$; that $s$ is sent to $p$).

In fact there is never a polynomial map of the kind you ask about for very general reasons: Surjective morphism of affine varieties and dimension, Surjective Morphism of affine algebraic variety etc.

Max
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  • Polynomial maps is an entirely different thing. Existence of smooth interpolations (bump functions) is one of (if not THE) key differences between algebraic and differential geometries. – Max Jun 21 '19 at 11:17
  • Sorry, I meant to comment on the question, which specifically refers in the body of the text to polynomials. I appreciate that this is different from the title. – Mark Bennet Jun 21 '19 at 11:18
  • Now I will always think of fish when dealing with polynomials... – Dietrich Burde Jun 21 '19 at 11:20
  • Yes, thanks, I was not attentive enough, and the question mixes categories bit. It's good to keep these things straight. – Max Jun 21 '19 at 11:20
  • Note that the links you give are asking about a bijection, and it is polynomial "in the opposite direction". – Max Jun 21 '19 at 11:28
  • @Max thanks for that - I added the comment with links to the main question - I am not pretending it is a proper answer. I had missed $n\gt m$ because I tend to ignore the titles. – Mark Bennet Jun 21 '19 at 11:33