cokernels in $\mathbf{Ring}$ of the inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$

Question

Aluffi III.2.12 asks about cokernels of the inclusion $\iota : \mathbb{Z} \hookrightarrow \mathbb{Q}$.

I've managed to prove that for any ring $R$: $\alpha \iota = 0$ (where $\alpha : \mathbb{Q} \rightarrow R$) implies $R$ is a zero-ring (since $\alpha$ maps the multiplicative identity to $0$). The reasoning in this question agrees with mine, and the conclusion in the answer says that the cokernel would be the zero ring.

My line of thought is slightly different, though: the zero ring (that $R$ must be) is the final object in $\mathbf{Ring}$, so any ring $K$ would do as a cokernel as long as there is a surjective $\pi : \mathbb{Q} \rightarrow K$ (for instance, $\mathbb{Q}$ itself being a possible candidate for $K$). Is it correct? How does it work with objects satisfying universal properties being all isomoprhic?

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Aluffi does not define the notion of cokernels for $\mathbf{Ring}$ explicitly. Instead, he informally refers to the one for $\mathbf{Ab}$, so I'm assuming the following definition: given a ring homomorphism $\varphi : R \rightarrow R'$ the $\text{coker} \varphi$ is a pair $(K, \pi)$ of a ring $K$ and a surjective homomorphism $\pi : R' \rightarrow K$ such that every $\alpha : R' \rightarrow S$ (for which $\alpha \varphi = 0$) factors uniquely through $\text{coker} \varphi$ and $\pi$.

Answer 1

There is something crucial missing in your definition, which is that you need to additionally require that $\pi\varphi=0$. When you add that requirement, it becomes immediate that a cokernel can only every be the zero ring, since the map $0$ is not even a ring homomorphism unless the codomain is the zero ring.

(Note that if you applied your definition to $\mathbf{Ab}$, it would be highly non-unique for the same reason that you observed for rings and would not at all coincide with the usual meaning of cokernels.)

Answer 2

I guess after thinking some more what a cokernel is I realized there is no ambiguity.

Since we're talking about universal properties, let's define more formally what category are we considering when looking for terminal objects. For the homomorphism $\varphi : R \rightarrow R'$ it looks reasonable to consider the category where the objects are homomorphisms $\alpha : R' \rightarrow S$ (and their corresponding targets $S$) such that $\alpha \varphi = 0$, and morphisms between $\alpha_1 : R' \rightarrow S_1, \alpha_2 : R' \rightarrow S_2$ are precisely the ring homomorphims $\sigma : S_1 \rightarrow S_2$ such that $\alpha_2 = \sigma \alpha_1$.

Thus, by the very same reasoning as in the question, every target $S$ of every morphism $\alpha$ comprising this category is actually a zero-ring. Thus we cannot take $K$ in the question to be $\mathbb{Q}$ or anything else besides zero-ring, and there is no ambiguity at all.