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4 votes
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Does the tropical semiring admit a universal property?

Each of the rings $\mathbf{Q}$, $\mathbf{Z}_p$, $\mathbf{Q}_p$, $\mathbf{R}$ admits a universal construction: the rationals are the field of fractions of the integers: $\mathbf{Q}:=\mathrm{Frac}(\...
Emily's user avatar
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2 votes
0 answers
164 views

Algorithms for "Ideals" in polynomial algebras over the max-plus semi-ring

I'm a beginner in tropical geometry, and I'm running into the following question: In the usual polynomial ring over a field, one has algorithms (i.e. using a Groebner basis) for determining whether ...
Drew's user avatar
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1 vote
0 answers
162 views

Integral points on "complex exponential surface" in $\mathbb{C}^3$

I encountered the following object in $\mathbb{C}^3$ defined for $m\in\mathbb{N}$ by $$A_m:=\lbrace (z_1,z_2,z_3)\in\mathbb{C}^3|(2^{2z_3}m-1)2^{2z_1+z_2+1}+3^{z_2-1}(2^{2z_1}-2^2-3^{z_3+1}m)=0\rbrace$...
Jens Fischer's user avatar
1 vote
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Proving equivalence of tropical polynomials

I am new to the world of tropical mathematics. I am wondering if there is an algorithm to prove the equivalence of two tropical polynomials (in the plus-min semiring let's say), say over multivaribles?...
Atugo's user avatar
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1 vote
0 answers
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Generators for Ideals in ring of multivariate Laurent Polynomials

Consider the following problem: Find an ideal $I \subset \mathbb{Q}[x^{\pm}_1,x^{\pm}_2,x^{\pm}_3]$ such that $I_{aff} \subset \mathbb{Q}[x_1, x_2, x_3] = I \cap k[x_1, x_2, x_3]$ requires more ...
b_dobres's user avatar
1 vote
0 answers
102 views

Notion of transversality over the field of Puiseux series.

To a given a Laurent polynomial $f$ over the field of Puiseux seris with parameter $t$, $f \in \mathbb{C} \lbrace\lbrace t \rbrace\rbrace[z_1^{\pm1},...,z_n^{\pm 1}]$, one can associate the ...
Farhad's user avatar
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1 vote
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138 views

Bases of Ideals With no Monomials

Let $K$ be an algebraically closed field and $K[\underline{x}]$ its ring of polynomials in $n$ variables $x_1,\cdots, x_n$. Let $J\leq K[\underline{x}]$ be an ideal such that there are no monomials in ...
Santiago's user avatar
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