# Does the Dorroh Extension Theorem simplify ring theory to the study of rings with identity?

I came across a theorem called Dorroh Extension Theorem while reading a textbook on ring theory. What the theorem essentially says is that any ring $R$ can be embedded in a ring $R^{\prime}$ with identity, i.e. there exists a subring $S^{\prime}$ of $R^{\prime}$ such that $R\cong S^{\prime}$. What I cannot understand is the following question.

Why does not this theorem simplify ring theory to the study of rings with identity?

The fastest answer that comes to my mind is that a subring of a ring is not necessarily a ring with identity. But is this the only reason? I'll be grateful for any help provided.

Edit: Proof of the Dorroh Extension theorem.

Consider $R\times\mathbb{Z}$. Define the operations as $(a,m)+(b,n)=(a+b,m+n)$ and $(a,m)(b,n)=(ab+an+mb, mn)$. Then $R\times \mathbb{Z}$ is a ring with identity $(0,1)$. And $R\times\{0\}$ is a subring of $R\times\mathbb{Z}$. Moreover $f:R\to R\times\{0\}$ given by $f(a)=(a,0)$ is an isomorphism. Hence the theorem "Any ring can be embedded in a ring with identity".

• Please give a reference where Dorroh's extension theorem is explicited. – Jean Marie May 21 '17 at 4:56
• @JeanMarie At this point, Janitha357 would presumably just be googling a resource which you can do just as well. The only thing Janitha357 can provide beyond that is verifying that it does correspond to what is in the book. Alternatively, you could ask Janitha357 to provide the exact statement of the theorem. – Derek Elkins May 21 '17 at 5:12
• @JeanMarie I will edit my question so that you can see the exact statement of the theorem and it's proof. – Janitha357 May 21 '17 at 5:21
• @Janitha357 : I think it's the same reason why Steinitz's theorem doesn't simplify the study of fields to algebraically closed fields. Indeed, any ring can be embedded in a unitary ring, but there are properties of non unitary rings that are special to them and that you can't see when looking at the bigger unitary field in which it is embedded. See this answer I gave that shows that some properties are drastically different in non unital rings : math.stackexchange.com/questions/2194671/… – Max May 21 '17 at 10:08
• Check this answer: math.stackexchange.com/a/16171/133781 – Xam May 21 '17 at 17:32