Adding Substructures by Forcing

Question

Consider a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $M$. Let $V$ be a model of ZFC (or ZF) the general question is that what would happen to classes, $Sub(M):=\{N~;~N~\text{is a substructure of}~M\}$ and $Sup(M):=\{N~;~M~\text{is a substructure of}~N\}$ in the generic extensions?

Question 1: If $M,N\in V$ are two $\mathcal{L}$-structures and $T$ an $\mathcal{L}$-theory with $M\subsetneq N$ and $M,N\models T$ such that there is no other $\mathcal{L}$-structure $K\in V$ with $K\models T, M\subsetneq K \subsetneq N$, what is a sufficient condition on $T$ to make it possible to find a forcing notion $\mathbb{P}\in V$ in which in $V^{\mathbb{P}}$ there is a $K\models T$ with $M\subsetneq K\subsetneq N$?

Question 2: Consider an infinite field $F$ in a model $V$, if $V$ is a model of $ZF$ it is possible that $F$ has no algebraic closure in $V$ because it needs $AC$ to construct an algebraic closure for an arbitrary field. Is it possible to extend $V$ to a generic extension $V[G]$ which $F$ remains a field and also have an algebraic closure $F'$? In other words, can we always add the algebraic closure of a field to the world?

Answer 1

Note that $M \models T$ is absolute (all quantifiers are bounded). In particular, things like algebraically closed fields cannot cease to be algebraically closed when you move to a different model.

Answer 2

The second question admits an easy answer. Almost yes. If a field is well-orderable then it has a canonical algebraic closure (it might not be the unique algebraic closure, by it exists). So by making $F$ well-orderable (we can always make it countable, anyway) we ensure that it will have an algebraic closure.

In fact, if $F$ has no algebraic closure, you can just look at the partial order of algebraic extensions of $F$, and force with that.

I'm not quite sure whether you can always force the existence of a unique algebraic closure, but that would have to be one strange universe of sets.