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What properties of tropical geometry (Starting from a valued Field) can be proven to be true using their analogue in algebraic geometry? For example, using the valuation on the Puiseux series $\mathbb{C}((t))$ what can be said about $\Gamma ^n$ where $\Gamma$ is the valuation group.

That is what pairs of properties $(X,X')$ such that $X$ is valid on an algebraic variety then $X'$ is valid on the tropicalization, and $X$ is a first order property in the language $L=\{ 0,1,+,*,U,|\}$ where $U$ stands for the valuation ring, and $x|y \leftrightarrow \exists z U(z) _\wedge xz=y$

A silly example would be the following: One can prove that when the support (the set of exponents corresponding to non zero coefficients) of a polynomial $f$ in two variables is equal to {$(i,j)\in \mathbb{N}^2 | i+j \leq d$} for some $d\in \mathbb{Z}^+$ and the support of another polynomial $g$ is {$(i,j)\in \mathbb{N}^2 | i+j \leq c$} for another $c\in \mathbb{Z}^+$ then the tropical curves generated by those polynomials intersects in exactly $cd$ points or in infinitely many(Bézout).

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    $\begingroup$ It is difficult to understand what would be a satisfactory answer, given the presentation of your question. Do you want a property $P$ that may be satisfied by an algebraic variety, and a corresponding property $P'$ that may be satisfied by a tropical variety, such that if $(X,X')$ is a pair given by an algebraic variety and a tropicalization, then $X$ satisfies $P$ if and only if $X'$ satisfies $P'$? Also, your notation for Puiseaux series looks remarkably similar to the usual notation for Laurent series. $\endgroup$
    – S. Carnahan ♦
    Commented Mar 21, 2011 at 7:53
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    $\begingroup$ Also of interest is the converse. What can one learn in ordinary algebraic geometry from its tropical analogue? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Mar 22, 2011 at 15:06
  • $\begingroup$ there are similar questions mathoverflow.net/questions/53306/… mathoverflow.net/questions/35553/… and by tag tropical geometry as well =) $\endgroup$
    – Nikita Kalinin
    Commented Nov 16, 2011 at 20:37

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