# 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 hypersurface $$V_{f} =\overline{\lbrace(t_1,...,t_n ) \in (\mathbb{C} \lbrace\lbrace t \rbrace\rbrace)^n : f(t_1,...t_n)=0 \rbrace }$$.

Then, what it would mean if two of such hypersurfaces are transversal(in terms of generic $t$ and varieties over $\mathbb{C}$)?

If for an ideal $I=\; < g_t,h_t>\; \subset \mathbb{C} \lbrace\lbrace t \rbrace\rbrace[z_1^{\pm1},...,z_n^{\pm 1}]$,

$$V(I)= \overline{\lbrace(t_1,...,t_n ) \in (\mathbb{C} \lbrace\lbrace t \rbrace\rbrace)^n : f(t_1,...t_n)=0 , \forall f \in I \rbrace }$$ is irreducible, does imply the transversality of $V_f$ and $V_g$ over $\mathbb{C}$ for generic parameter $t \in \mathbb{R}$?

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I think if they are not transversal at a point $a(t)$ then for any $t_0$ they are not transversal at the point $a(t_0)$ –  Nikita Kalinin Apr 19 '13 at 9:20
@Nikita: If one puts t=0 then the question becomes transversality over C and what you are saying implies that it will be stable under change of coefficients by polynomials of t... –  Farhad Apr 21 '13 at 9:10