# Non-unital ring contains a unital ring as a subring?

It is well known that every non-unital ring can be embedded into a unital ring (e.g., Dorroh's adjunction). I am curious about the converse: every unital ring can be viewed as a subring of a non-unital ring?

If this converse is not correct, will this be still partially true? I am looking for a non-trivial example. One trivial example would be that $\{0\}$ as a subring of $2\mathbb{Z}$.

PS: by a ring here I mean a set that is an additive abelian group and a multiplicative semigroup, and satisfies the distributive laws.

• What do you mean by embedded? Do you mean that there is an injective ring homomorphism? Because any unital ring would have no homomorphism to a non-unital ring if morphisms must preserve $1.$ – Chickenmancer Aug 15 '18 at 4:49
• @Chickenmancer You may have a look at here for what I mean:math.stackexchange.com/questions/1113097/… – Easy Aug 15 '18 at 5:10
• @Chickenmancer I think we should note that a ring and its subring do not share the same "1" in general. – Easy Aug 15 '18 at 5:14
• math.stackexchange.com/questions/170953/… They do share the same 1 in general. – Chickenmancer Aug 15 '18 at 6:32
• @Chickenmancer Can you tell what are the "1"s in $\{0\}$ and $\mathbb{Z}$ respectively? – Easy Aug 15 '18 at 7:31

## 1 Answer

Take any ring $S$ without identity. If $R$ is any ring with identity, then $R\times S$ does not have identity. Is this what you seek?

• Nice to see that you're gonna hit 100k reputation. – Xam Aug 16 '18 at 4:53