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I consider the tropical semi-ring $(\mathbb{R},\oplus,\odot)=(\mathbb{R},\max,+)$. I know that the tropiclisation of any (Laurent) polynomial $p\in\mathbb{R}[x_1^{\pm1},...,x_n^{\pm1}]$ given a valuation $\nu$ simply is given by

$$\mathrm{Trop}(p)(X_1,...,X_n)=\bigoplus_{(i_1,...,i_n)} \nu(c_{i_1,...,i_n})\odot X_{i_1}^{\pm 1}\odot...\odot X_{i_n}^{\pm 1}$$

where $c_{i_1,...,i_n}$ are the coefficients of $p$. Now the zeros of $p$ interpreted as a hypersurface are translated via tropicalization to the non-differentiable "regions" of $\mathrm{Trop}(p)$, which give for example in the $n=2$ case rise to tropical curves via a possible construction using Newton polygons. This can, as far as I understood and so from examples, be extended to Laurent polynomials without any bigger issues.

Now, I am wondering, what the inverse would look like. So Assume that you have a tropical Laurent polynomial $L$ given by

$$L(X_1,...,X_n)=\bigoplus_{(i_1,...,i_n)} d_{i_1,...,i_n}\odot X_{i_1}^{\pm 1}\odot...\odot X_{i_n}^{\pm 1}$$

and some valuation $\mu$. Is there an interpretation of the solutions to $L(X_1,...,X_n)=0$, where $0$ is the neutral element with respect to $\odot$? If I take as an example $\mu$ on $\mathbb{C}$ as $\mu(z)=-\log(|z|)$, then the "un-tropicalization" looks like we are investigating the poles of $q=\mathrm{Trop}^{-1}(L)$.

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If I understand your question correctly, then the solution locus of $L(X) = c$ for any constant $c$ (if $L(X)$ itself is not constant) is a subset of the "tropical solutions" of $L(X) \oplus c$, which as you write is the locus of points where a tropical polynomial is non-smooth. Geometrically, the "tropical solutions" are those points where at least two terms switch in dominance (with respect to $\max$ in your case). Equating two tropical polynomials (one of which could be a constant) focuses on the subset of points where the switch in dominance happens between groups of terms with at least one term drawn from each side of the equality.

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