Isomorphism class of the zero module is not a set

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I'm having a bit of trouble with this one question (3A.2) in Bruce A. Magurn's An Algebraic Introduction to K-Theory. It's about set theory, which has always managed to confuse me more than it should.

Specifically, we are to prove that if $R$ is a ring and 0 is the zero module of $R$, then the isomorphism class cl(0) cannot be a set. The hint that Magurn provides is that if it were a set, then:

(1.) We should be able to construct an argument from the Axiom of Replacement to say that there exists a set $T$ of all sets having only one member.

(2.) From this it would follow that the union of all elements in $T$ is also a set, but this would be a contradiction, since $T$ then would be the set of all sets.

I think that I understand the second part. If $V$ is any given set, then $\{ V \}$ is a set with only one member, and so, since $\{ V \} \in T$, then $V \in \bigcup_T$.

It is the first part that I just cannot work around. How should one construct such an argument?

Any clever person out there who could help me?

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The point is that any singleton can be the zero module over $R$. But being a module is more than just being a set, you also come with the necessary operations and actions by the members of $R$.

So a zero module is in fact $(\{0\},+_0,r)_{r\in R}$ tuple. But now using Replacement, we can simply map it to $\{0\}$.

Taking a different set, $x$, we can again define a zero module by using $\{x\}$ as the underlying set, etc.

But the point here is that if the collection of all $0$ modules was a set, then mapping a zero module to its underlying set is a definable set operation, and so by Replacement it will produce a set. But since any singleton can be such module, we would get a set which contains exactly all of the singletons. And you followed the rest quite well.