### Puiseux Series and Tropical Varieties

Puiseux series- This field is denoted by $\Bbb{C}[[t]]$. Note that we have a double brace “[[ ]]” instead of “[]”. This implies that we have infinite series instead of finite ones (which would be polynomials). The Puiseux laurent series is denoted as $\Bbb{C}((t))$. This means that $t$ is also allowed to have negative powers. Now $\mathcal{K}=\bigcup\limits_{n\geq 1} \Bbb{C}((t^{1/n}))$, which just means that $\mathcal{K}$ contains all rational powers of $t$ now, and not just integral powers, as in the Laurent series. We seem to be generalizing in every successive step.

Now we define a valuation: $v:\mathcal{K}^{\times}\to\Bbb{Q}: v(\sum a_it^{i/N})=\min(i/N)$, for $a_i\neq 0$. So what we’ve essentially done is that we’ve written all rational powers (including integral ones) as fractions with denominator $N$. Clearly, amongst all the denominators in the rational powers of $t$, $N$ has to be the largest denominator.

If the Puiseux series converges, then we have $v(f)=\lim\limits_{t\to 0^+}\frac{\log (f(t))}{\log t}$

Why is that? It seems to me that $\lim\limits_{t\to 0^+}\frac{\log (f(t))}{\log t}$ would give us the sum of all rational powers of $t$, which could possibly be infinite. Then why do we just get the lowest one?

Now for $X\in (\mathcal{K}^\times)^n$, set Trop$(X)$ to be $v(f)\subset\Bbb{Q}^n$. Take $X=x+y+1=0$ for example. One should think about this variety as $x(t)+y(t)+1=0$ instead, where $x$ and $y$ are power series in $t$ (with rational powers). Then there are three possibilities:

i) $v(x)> 0$ and $v(t)=0$.

ii) $v(x)=0$ and $v(y)> 0$

iii) $v(x)=v(y)\leq 0$.

These cases can be easily deduced to contain all possibilities. For instance, if $v(x)<0$, then $v(y)<0$ too. This is because $v(x)<0$ implies that $x$ has negative powers of $t$. This implies that $y(t)$ too has to have negative powers of $t$, as $x(t)+y(t)=-1$. When one of them contains strictly positive powers, the other has to contain $-1$ and no negative powers of $t$, which implies that if $x(t)>0$, then $y(t)=0$.