The following is an excerpt from Paul Cohen's "The Discovery of Forcing", pp 1091, in which he explains why we do not want to add new ordinals to a countable transitive model $M$ when extending it using forcing:
Suppose $M$ were a countable model. Up until now we have not discussed the role countability might play. This means that all the sets of $M$ are countable, although the enumeration of some sets of $M$ does not exist in $M$. The simplest example would be the uncountable ordinals in $M$. These of course are actually countable ordinals, and hence there is an ordinal $I$, not in $M$, which is countable, and which is larger than all the ordinals of $M$. Since $I$ is countable, it can be expressed as a relation on the integers and hence coded as a set $a$ of integers. Now if by misfortune we try to adjoin this $a$ to $M$, the result cannot possibly be a model for ZF. For if it were, the ordinal $I$ as coded by $I$1 would have to appear in $M(a)$. However, we also made the rigid assumption that we were going to add no new ordinals. This is a contradiction, so that $M(a)$ cannot be a model. From this example, we learn of the extreme danger in allowing new sets to exist. Yet $a$ itself is a new set. How then can we satisfy these two conﬂicting demands?
1 I think this is a typo and he meant to write $a$.
What I understand: If $M$ is countable then it cannot contain all countable ordinals (since the set of all countable ordinals is itself uncountable) hence there is at least one countable ordinal not in $M$. Since it is countable it is a subset $a$ of $\omega$. If we adjoin $a$ to $M$ then in particular $a \in M(a)$ so that we have added an ordinal that was previously not in $M$.
What I don't understand:
Why can't $M(a)$ possibly satisfy ZF? If $M$ is a countable model of ZF and we add an ordinal $a$ not in $M$, why is it impossible for $M(a)$ to still satisfy ZF?
Thank you for your help.