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Here's my version of Tyler's argument that $\mathbb{Z}_p$ is a counterexample. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

 

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

 

If $e_i$ is in $A$, I define $f(e_i) = e_i$

 

If $e_i$ is not in $A$, I define $f(e_i) = 0$

 

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

 

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.

Here's my version of Tyler's argument that $\mathbb{Z}_p$ is a counterexample. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

If $e_i$ is in $A$, I define $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.

Here's my version of Tyler's argument. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

If $e_i$ is in $A$, I define $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.

Here's my version of Tyler's argument that $\mathbb{Z}_p$ is a counterexample. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

If $e_i$ is in $A$, I define $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.

Here's my version of Tyler's argument. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

If $e_i$ is in $A$, I define $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a domain.

Here's my version of Tyler's argument. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,

Let's suppose that $\mathbb Z_p$ were such a colimit. Then $\mathbb Z_p$ could be written as having a presentation as follows:

It would have a set of generators $e_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e_i = e_j$ (indexed by morphisms in the diagram).

Now, (and here's the change Tyler suggested later) one of the the $e_i$'s must be a $p$-adic unit - otherwise the image of the $e_i$'s would be contained in the proper subgroup $p\mathbb Z_p \subset \mathbb Z_p$. Pick such a generator $e$ and define $A = \mathbb Z_{(p)} e \subset \mathbb Z_p$ (I might be the only one not to realize this, but $\mathbb Z_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ not divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z_p$, so we might as well assume $e = 1$ and $A = \mathbb Z_{(p)}$. Then, as Tyler says,

Then I would be able to define a self-map $f$ of $\mathbb Z_p$ as follows:

If $e_i$ is in $A$, I define $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z_p / A$ both being torsion-free. But

• If $e_i,e_j \in A$ or $e_i,e_j \not \in A$, then the relation is trivial.
• If $n=0$, then the relation is respected.
• If $n \neq 0$, then either $e_i,e_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.

Then as Tyler concludes,

Therefore this gives a well-defined such map $f$.

However, any abelian group homomorphism $\mathbb Z_p \to \mathbb Z_p$ which is the identity on $\mathbb Z_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).

Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z_p$ is a UFD.