Is it possible to turn the set $\mathbb{Hom}(R,S)$ of ring homomorphisms from $R$ to $S$ into a ring?

Question

Is it possible to turn the set $\mathbb{Hom}(R,S)$ of ring homomorphisms from $R$ to $S$ into a ring? Discuss.

What I have observed that if I define the multiplication in $\mathbb{Hom}(R,S)$ s.t $fg(x)=f(x)g(x)$ $\forall x\in R$. Now,

$$ \begin{align*} fg(x+y)=f(x+y)g(x+y) \Rightarrow f(x)g(x)+f(y)g(y)=f(x)g(x)+f(x)g(y)+f(y)g(x)+f(y)g(y) \Rightarrow f(x)g(y)=-f(y)g(x) \qquad\forall x,y\in R \end{align*} $$

So is it the sufficient condition?

Answer 1

If $R$ is a set and $S$ a ring then the set $Map\left(R,S\right)$ of $S$-valued functions has a ringstructure if sums and products of the functions are defined pointwise. If there is a ringstructure on $R$ as well and $Hom\left(R,S\right)$ denotes the set of ringhomomorphisms one could wonder whether $Hom\left(R,S\right)$ can be recognized as a subring of $Map\left(R,S\right)$. If we are dealing with rings with identity then for this $Hom\left(R,S\right)$ needs to contain the map that serves as identity of $Map\left(R,S\right)$ wich is prescribed by $r\mapsto1_{S}$ for each $r\in R$. However, this is not a ringhomomorphism, because it lacks the property of sending $0_{R}$ to $0_{S}$.

Answer 2

$\hom_{\mathrm{Ring}}(\mathbb{F}_2,\mathbb{Z})=\emptyset$.