Why isn't there necessarily a direct sum of rings? I've just seen on Wikipedia that we can't speak of direct sum of rings. Let $R$ and $S$ be rings. It says we can't have a direct sum of rings because the direct sum $R\times S$ doesn't receive a natural ring homomorphism. I don't understand what it means by this, and why we need this natural ring homomorphism to have direct sum of rings.
 A: Some of this confusion stems from confusing uses of the terms product, direct sum, and coproduct. Hopefully reading this answer by Martin Brandenburg will help clear things up.
The general categorical definitions of a direct sum (coproduct) of two objects $R$ and $S$ is the object $R \oplus S$ equipped with inclusion maps $i_R \colon R \hookrightarrow R\oplus S$ and $i_S \colon S \hookrightarrow R\oplus S$ such that for any other object $C$ into which we have inclusions $c_R \colon R \hookrightarrow C$ and $c_S \colon S \hookrightarrow C$, there's a unique map $\phi \colon R \oplus S \to C$ such that $c_R = \phi i_R$ and $c_S = \phi i_S$. So in a sense, the direct sum of $R$ and $S$ is the "smallest" thing into which both $R$ and $S$ include.
We may wish that the object $R \times S$ is the direct sum $R \oplus S$, with the inclusion maps being given by $i_R \colon r \mapsto (r,0)$ and $i_S \colon s \mapsto (0,s)$, but ring homomorphisms must send $1$ to $1$, and neither of these maps do this. So a direct sum object $R \oplus S$ with the property required in the last paragraph may not exist as a ring.
A: There is no natural ring homomorphism that maps $1$ to $(1,1)$ in the direct sum. 
However projections onto the factors  are natural ring homomorphisms, as they map $(1,1)$ onto $1$.
