If $X$ a subset of ordinals, is $M[X]$ a generic extension? Suppose $M$ is a countable transitive model for ZFC. Suppose $X\subset M\cap \textbf{ON}$ is bounded in $M\cap \textbf{ON}$. Why does $M[X]$ (the least ctm $N$ such that $M \subset N$ and $X \in N$) exists? Furthermore, is it true that there exists a forcing notion $P \in M$ and a $P$-generic filter over $M$, $G$, such that $M[G]=M[X]$?
I have looked for this information in some books such as Kunen, Jech and Kanamori's but I have found nothing.
Edit: @Elliot Glazer's answer shows that $M[X]$ is not always a generic extension. So I am changing the question a bit.
Suppose that there exists a forcing notion $Q$ and a generic filter $H$ over $M$ such that $X \in M[H]$ is a subset of $M\cap \mathbf{ON}=M[H]\cap \mathbf{ON}$. Is it true that there exists a is  a forcing notion $P \in M$ and a $P$-generic filter over $M$, $G$, such that $M[G]=M[X]$?
 A: It is not true in general that $M[X]$ is a generic extension of $M,$ even in the case $X \subset \omega.$ For example, such an $X$ could code a well-ordering of $\omega$ of type $o(M),$ in which case $M[X]$ has more ordinals than $M$ and is thus not a generic extension.
A: Concerning your modified question, as it is commented by Andres Caicedo, the answer is yes; in fact we have the following theorem:
Theorem 1. Assume $V[G]$ is a set forcing extension of $V$ by a complete Boolean algebra $\mathbb{B}$ and suppose $N$ is a model of $ZFC$ with $V \subseteq N \subseteq V[G]$. Then for some regular complete subalgebra $\mathbb{C}$ of $\mathbb{B}$, such that $N=V[G \cap \mathbb{C}]$.
On the other hand, if we extend a bit the notion of genericity, then we have the following theorem of Mack Stanley (I don't recall the paper in which the result is stated):
Theorem 2. Suppose $V \subseteq W$ are countable transitive models of $ZFC$ with the same ordinals. Then $W$ is a class generic extension of $V$ (with respect to this extended notion of genericity).
