Let $R$and $A$ be rings with $A \neq \{0\}$. Furthermore, assume that $R$ is unitary and commutative. Let $\eta: R \to Z(A)$ be a ring homomorphism where $Z(A)$ denotes the center of $A$.
Question: Is $\eta$ always injective?
Approach
I think the statement is true, and I tried the following: Let $r \in R$ such that $\eta(r)=0_A$. Assume that $r \neq 0_R$. If $R$ would be a field or, more generally, if $r$ has an inverse in $R$, we get $$1_A = \eta(r) \eta(r^{-1}) = 0_A$$ which would be a contradiction since we assumed $A \neq \{ 0\}$. But I cannot complete the proof for the case when $R$ is not a field.
Could you please help me with this problem? Thank you!
You can’t finish it because it’s false. Let $R$ be a unitary commutative ring that is not a field. Then $R$ has a nontrivial proper ideal $I$ (a unitary commutative ring with no nontrivial proper ideals is a field). Take $A=R/I$ and $\eta$ the canonical projection. Note that $Z(A)=A$.
If you want your $A$ to be strictly non-commutative (as opposed to “not necessarily commutative”, then note that if $S$ is any commutative unitary ring, then you can let $A=S\langle x,y\rangle$ be the ring of polynomials with coefficients in $S$ on the noncommuting variables $x$ and $y$. The center of this ring is the copy of $S$ inside it, and you can take any non-injective map $R\to S$ and compose it with the embedding $S\hookrightarrow A$ to get a non-injective map $R\to A$.