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Let $R \neq 0$ be a commutative, unital ring and $f: R \to \mathbb{Z}$ be a ring homomorphism.

Does this force $f$ to be surjective?

Attempt of Proof: By the universal property of $\mathbb{Z}$, there is a unique ring homomorphism $g: \mathbb{Z} \to R$. But then $f \circ g: \mathbb{Z} \to \mathbb{Z}$ is the unique ring homomorphism which is already given by $id: \mathbb{Z} \to \mathbb{Z}$. Thus $f \circ g = id$ and therefore $f$ must be surjective.

Does this proof hold? It seems weird that I can nowhere find this result...

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    $\begingroup$ Your proof is ok. (You‘re probably having a hard time to find this result because it is very immediate.) $\endgroup$ – Qi Zhu Jan 23 at 14:07
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The image of a ring homomorphism (preserving $1$) is a subring of the codomain.

Since the only subring of $\mathbb{Z}$ is $\mathbb{Z}$ itself, the statement is proved.

The same is true for ring homomorphisms to $\mathbb{Z}/n\mathbb{Z}$.

Your proof is good as well, but overkill.

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  • $\begingroup$ On the other hand, op's proof could serve as a very nice proof that $\mathbb Z$ only has one subring! $\endgroup$ – Vercassivelaunos Jan 23 at 14:32
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    $\begingroup$ @Vercassivelaunos I'm afraid it would be circular. The universal property of $\mathbb{Z}$ being an initial object in the category of rings relies on it having no proper subring. $\endgroup$ – egreg Jan 23 at 14:44
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    $\begingroup$ Depends on your definition. If you define $\mathbb Z$ as the initial ring, then it having no proper subrings is a theorem. $\endgroup$ – Vercassivelaunos Jan 23 at 15:28
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You certainly assume that a ring homomorphism between unital rings preserves the multiplicative identity.

However, all what we need to show that $f(R) =\mathbb Z$ is the fact that $1 \in f(R)$ - which is an immediate consequence of $f$ being a ring homomorphism. Beyond this fact the multiplicative structure of $R$ is completely irrelevant, your result is true also for non-commutative unital rings $R$. In fact the essence is this:

A group homomorphisms $\phi : G \to \mathbb Z$ is surjective if and only $1 \in \phi(G)$.

One half is trivial. For the other part, let $g \in G$ such that $\phi(g) = 1$. Then clearly $\phi(g^n) = n $.

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If $f$ is a ring homomorphism, then for the unit elements $f(1_R) = 1_Z$.

For completeness define of multiples of ring elements (unit element is only require):

For each natural number $n$, write $n\cdot 1_R = 1_R+\ldots+1_R$ $n$-times and for each negative integer $n = -m$, where $m>0$, write $n\cdot 1_R = -(m\cdot 1_R)$.

Then for each integer $n$, $f(n\cdot 1_R) = n\cdot f(1_R) = n\cdot 1_Z$. It follows immediately that the mapping is onto.

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