# Homomorphism of Rings from Composition of 2 Ring homomorphisms.

I would like to know if this proposition is true or not.

$\textbf{Proposition:}$ Suppose there exist a ring homomorphism $\varphi:R \rightarrow R'$ and another ring homomorphism $\phi:R'\rightarrow R''$, where all $R,R',R''$ are rings. Then, there exist a ring homomorphism $\xi : R \rightarrow R''$ s.t. $\xi= \phi \circ \varphi$ is composition function.

Proof:

Let $r_1,r_2 \in R; \; r_{1}^{'},r_{2}^{'} \in R' \text{ and } r_{1}^{''},r_{2}^{''} \in R''$ s.t.

$\varphi (r_1)=r_{1}^{'} \; ; \varphi (r_2)=r_{2}^{'} \; ; \phi (r_{1}^{'})=r_{1}^{''} \; ; \phi (r_{2}^{'})=r_{2}^{''}$

Then, $$\begin{split} \xi(r_1+r_2) &= \phi ( \varphi (r_1+r_2) ) = \phi ( r_{1}^{'} + r_{2}^{'} ) = r_{1}^{''} + r_{2}^{''} = \phi ( \varphi (r_1) )+\phi ( \varphi (r_2) ) =\xi(r_1)+\xi(r_2) \end{split}$$

for the multiplication, $$\begin{split} \xi(r_1r_2) = \phi ( \varphi (r_1r_2) ) = \phi ( r_{1}^{'} r_{2}^{'} ) = r_{1}^{''} r_{2}^{''} = \phi ( \varphi (r_1) ) \phi ( \varphi (r_2) ) =\xi(r_1) \xi(r_2) \end{split}$$

and for the unit element in $R$,

Then, $$\begin{split} \xi(1_{R}) &= \phi ( \varphi (1_{R}) ) = \phi ( 1_{R'} ) = 1_{R''} \end{split}$$

Hence, $\xi$ is a ring homomprphism.

Doubt:

1) Is the proof right?

2) If this is right, is this (i.e. $\xi$) the unique ring homomorphism? I have no idea of how to prove uniqueness result if it's true.

Your question on uniqueness exhibits a slight misunderstanding. Actually the second sentence of the proposition is written awkwardly. It should read: Then the composition function $\xi= \phi \circ \varphi$ is a homomorphism $\xi : R \rightarrow R''$.