homomorphism from a (semi)local ring to $\mathbb Z$

Question

One can easily construct homomorphisms from $\mathbb Z$ to a (semi)local ring $\mathbb Z/6\mathbb Z$, even a field, $\mathbb Q$.

How about the converse?
Is there a homomorphism from a (semi)local ring to $\mathbb Z$?

Thank you.

Answer 1

Commutative semilocal rings only have finitely many maximal ideals.

Suppose you had a homomorphism $\phi$ from a semilocal ring $R$ to $\mathbb Z$. The identity has to map to either $0$ or $1$.

In the case $\phi(1)=1$, then $\phi$ must be surjective (as any abelian group homomorphism into $\mathbb Z$ with $1$ in its image would have to be.) This would imply that $\mathbb Z$, as a homomorphic image of $R$, would also have finitely many maximal ideals, which is an absurdity.

In the other case $\phi(1)=0$ you would indeed have a (very boring) homomorphism from $R$ to $\mathbb Z$.

Answer 2

In the case of commutative rings (as you tagged commutative-algebra), your question can be answered negatively:
Let $A$ be a semi-local ring. Since we assumed $A$ to be commutative, it has finitely many maximal ideals $\mathfrak m_1,\dots,\mathfrak m_n$. Suppose we had a ring homomorphism $\varphi\colon A\to\mathbb Z$. In particular $\varphi(1_A)=1_{\mathbb Z}$ (by a common definition) and we see $\varphi$ is surjective. Its kernel is the preimage of the prime ideal $\{0\}\subseteq\mathbb Z$, hence we have $\ker(\varphi)=\mathfrak p$ for some prime ideal $\mathfrak p\subseteq A$. Now this gives us an isomorphism $$\overline{\varphi}\colon A/\ker(\varphi)\xrightarrow{\cong}\mathbb Z$$ and we note for the Jacobson ideal $$\{0\}=J(\mathbb Z)\cong J(A/\ker(\varphi))=J(A)/\ker(\varphi)=\bigcap_{i=1}^n\mathfrak m_i/\ker(\varphi)$$ This means $\ker(\varphi)=\bigcap_{i=1}^n\mathfrak m_i$. But then $A/\bigcap_{i=1}^n\mathfrak m_i$ has only finitely many maximal ideals in contrast to $\mathbb Z$ having infinitely many, which is a contradiction to the isomorphism above.