Adjoining new ordinals to a model -- a question about one of Cohen's articles

Question

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 $a$.

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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.

Answer 1

Note that part of the working assumption is that passing from $M$ to $M(a)$ does not add ordinals; earlier on that page P. Cohen states:

An important decision is that no new ordinals are to be created.

and also in the text you quoted:

However, we also made the rigid assumption that we were going to add no new ordinals.

So $M(a)$ cannot satisfy ZF because it does not contain the ordinal that $a$ encodes.