# 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.

• Are you familiar with what "product" and "coproduct" mean in category theory? Did you see note #3, which was cited in the section you link to?
– anon
Mar 27, 2015 at 17:30
• The problem is possibly with the definiion of "ring". The last sentence of the Wikipedia section speaks of a "rng", so we are hinted that their defiinition of "ring" is what some other authors might call "ring with unity" or similar. - Another problem is that the WP article lacks a clearly formulated definition of "direct sum" in the first place ... Mar 27, 2015 at 17:33
• @anon yes I'm familiar with it. However I didn't understand why direct sum of rings must be the coproduct in the category of rings. Mar 27, 2015 at 17:37
• @HagenvonEitzen so if we work over the rings without unit we can define direct sum of rings in this context? Mar 27, 2015 at 17:40
• @user42912 You can define the coproduct of rings in any context. You can form the "direct sum" of rings in any context as well. The point is that the coproduct is not the "direct sum" in the category of rings. I'm not sure if the category of rngs suffers the same problem. I have a feeling it's not just the preservation of identity that causes problems... Mar 27, 2015 at 17:50

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.

• the problem is you are assuming that the coproduct's maps are defined as $r\into (r,0); s\into (0,s)$, it can be a good guess but the question asks, if we assume there is a coproduct as $R\times S$ with $f:R\to R\times S$ and $g: S\to R\times S$ which should not satisfy the universal property. Jun 14 at 11:15

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 $1$.