Let $R$ and $S$ be rings, and consider the ring $R \oplus S$. Let $X=\{(x,0_S):X\in R\}$. Then, $X \cong R$.
I slightly know what I am doing, but I do not know how to do it.
I know I must prove for $\phi : X \to R$,
(i) $\phi (u+v)=\phi (u)+\phi (v)$ for all $u,v \in X$, and
(ii) $\phi (uv)=(\phi (u))(\phi (v))$ for all $u,v \in X$, and
(iii) The function is one-to-one, and
(iv) The function is onto.
Can anyone help explain to me how I would do this, please?
Update: This is what I have of the proof so far:
We will now to prove part (i) of this proof. That is we will prove $\phi (u+v)=\phi (u)+\phi (v)$ for all $u,v \in X$. We will first begin with $\phi (u+v)$ and substitute $u$ and $v$ in to get $\phi (u+v)=\phi ((a,0_S)+(b,0_S))=\phi ((a+b,0_S+0_S)= \phi ((a+b,0_S))=a+b$. We will now substitute $u$ and $v$ into $\phi (u)+\phi (v)$ to get $\phi (u)+\phi (v)=\phi ((a+0_S))+\phi ((b+0_S))=a+b$. Since we know $\phi (u+v)=a+b$ and $\phi (u)+\phi (v)=a+b$, we know $\phi (u+v)=\phi (u)+\phi (v)$ which proves part (i) of this proof.
We will now prove part (ii) of this proof. That is we will prove $\phi (uv)=(\phi (u))(\phi (v))$ for all $u,v \in X$. We will first begin with $\phi (uv)$ and substitute $u$ and $v$ in to get $\phi (uv)=\phi ((a+0_S)(b+0_S))=\phi ((ab,0_S0_S)= \phi ((ab,0_S))=ab$. We will now substitute $u$ and $v$ into $(\phi (u))(\phi (v))$ to get $(\phi (u))(\phi (v))=(\phi ((a+0_S)))(\phi ((b+0_S)))=ab$. Since we know $\phi (uv)=ab$ and $(\phi (u))(\phi (v))=ab$, we know $\phi (uv)=(\phi (u))(\phi (v))$ which proves part (ii) of this proof.
Final parts of the proof:
We will now prove part (iii) of this proof. That is, we will show the function is one-to-one. We will let $u,v \in X$ such that $\phi (u)=\phi (v)$. Then, there exists integers $a$ and $b$ such that $u=(a,0_S)$ and $v=(b,0_S)$. Thus, after substituting, we get $\phi ((a,0_S))=\phi ((b,0_S))$. By the definition of the function, we then get $a=b$. Therefore, since $a=b$, $(a,0_S)=(b,0_S)$. Then, by substitution, we know $u=v$ which proves the function is one-to-one and proves part (iii) of this proof.
We will now prove part (iv) of this proof. That is, we will show the function is onto. If $a \in R$, then, by the definition of the function, $\phi ((a,0_S))=a$. Thus, the function is onto which proves part (iv) of this proof.