# 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.

## 1 Answer

The proof is fine.
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 $\xi : R \rightarrow R''$.

The composition function is defined and exists on the underlying sets. What you have done is to show that if its factors are homomorphisms then it is one also. There is really no question of uniqueness.