Let $\Delta\subset \mathbb C$ be an open disk, $\Delta^*=\Delta\setminus\{0\}$ the punctured disk. Let $p:\mathcal{X}\to \Delta^*$ be a family of irreducible projective hypersurfaces of degree $d$. In other words, we have a diagram
$\require{AMScd}$ \begin{CD} \mathcal{X} @>i>> \mathbb P^n\times \Delta^*\\ @V{p}VV @VVV\\ \Delta^* @= \Delta^* \end{CD}
where $i$ is embedding, $p$ is proper with each fiber an irreducible hypersurface (not necessarily smooth). Moreover I assume that $p$ is an algebraic map, so you can replace $\Delta^*$ by a quasiprojective curve if you want. I'd like to know the answers to the following:
Question 1: If $F_t$ is the homogeneous polynomial defining the hypersurface $p^{-1}(t)$, with $t\in \Delta^*$. Can we argue $F_t$ depends on $t$ algebraically? In other words, can we argue that $F_t=\sum_I a_{I,t} x^I$, (where $I$ is multiindex) for $t\in \Delta^*$ and each coefficient $a_{I,t}$ is a polynomial in $t$?
Question 2: If the above is true, then we can define the limit hypersurface$$F_0:=\lim_{t\to 0}F_t.$$
Let $\bar{\mathcal{X}}$ be the closure of $\mathcal{X}$ in $\mathbb P^n\times \Delta$, I'd like to know is $\{F_0=0\}$ coincides with fiber of zero of $\bar{\mathcal{X}}$?
Note that $F_0$ can be reducible and having nonreduced components, so my goal is to understand if the closure $\bar{\mathcal{X}}$ can capture this information. The closure here is the same in both Zariski topology and analytic topology, by my assumption that $p$ is an algebraic family.
Thanks in advance for any comment or answer.
Assume $\mathbb{P}^n = \mathbb{P}(V)$. Consider the morphism of sheaves $$ a \colon L := p_*(I_{\mathcal{X}}(d)) \to p_*(\mathcal{O}_{\mathbb{P}(V) \times \Delta^*}(d)) = S^dV^\vee \otimes \mathcal{O}_{\Delta^*}. $$ By base change $L$ is a line bundle and the morphism $a$ is a fiberwise monomorphism. But $Pic(\Delta^*) = 0$, hence $L \cong \mathcal{O}_{\Delta^*}$ and the morphism $a$ is given by a collection $a_I$ of functions on $\Delta^*$. This answers the first question.
Now we can consider the morphism $a$ as a rational section of the bundle $S^dV^\vee \otimes \mathcal{O}_{\Delta}$. As such it might have a pole or zero at the origin. Multiplying by appropriate coordinate we can get rid of the pole and make sure that $a(0) \ne 0$. Then $a$ defines a hypersurface $$ \bar{\mathcal{X}} \subset \mathbb{P}^n \times \Delta $$ It remains to note it is the closure of $\mathcal{X}$. Indeed, $\bar{\mathcal{X}}$ is irreducible (because $a(0) \ne 0$) and reduced (because it is Cohen-Macaulay and generically reduced). This answers the second question.