# What is the unitization of a perhaps-unital ring? Is it the same thing as a Dorroh extension?

What exactly is the unitization of a ring, described here? I tried reading the answer a few times and Googling around but I'm having trouble understanding what $$\mathbb{Z} \oplus R$$ really is and why.

My current best guess is that unitization is the Dorroh extension.

Let $$R$$ be a ring that is commutative but not necessarily unital.

I've read about the Dorroh extension which fixed a perhaps-unital ring $$R$$ by creating a new ring with an identity element in it, defined as follows:

Elements have the form $$(a, r) \in \mathbb{Z} \times R$$.

$$(a, r) - (b, s)$$ is defined as $$(a-b, r -s)$$.

$$(a, r)(b, s)$$ is defined as $$(ab, as+br+rs)$$ with $$as$$ and $$br$$ defined by repeated addition.

$$(1, 0)$$ is the identity.

This construction makes a lot of sense and I think of it as the way to add one to a non-unital ring.

I had to look up the details of the Dorroh construction to write this question, but I remember the trick because I thought it was cool.

This leads me to some confusion about this decade-old answer by Qiaochu Yuan describing unitization of a perhaps-commutative ring using the notation $$R \oplus \mathbb{Z}$$.

I'm having some trouble understanding what this notation means.

My first thought was that it just means a coproduct in the category of rings because $$\oplus$$ means a coproduct in the category of modules over a fixed ring. Then I realized that I don't actually know what the coproducts in the category of rings are, so I thought about something else.

My second thought was that $$\oplus$$ should define multiplication componentwise, since that's easy to do, call this the naive sum.

$$(a, b)(c, d) = (ac, bd)$$

However, this answer by Mike Pierce points out that this thing is actually not a coproduct in the category of rings-with-identity because the obvious maps into the naive sum do not preserve $$1$$ in the rings that they came from.

My third thought is that that can't be an issue here because $$R \oplus \mathbb{Z}$$ doesn't take place in the category of rings with unity because $$R$$ is not a ring with unity.

So, now I am thoroughly confused.

The Dorroh extension is not trivial, we're using the fact that every ring is also a $$\mathbb{Z}$$-algebra to define multiplication in the first place. I don't understand how $$\oplus$$, in any category, could pick out the Dorroh extension here (if it indeed that's what it does) or really anything other than the naive sum. And the naive sum won't have an identity element unless both of its inputs do.

Also, the naive sum doesn't lose any information in the category of non-unital rings. So if we have arrows going from $$A$$ and $$B$$ to whatever the true meaning of $$A \oplus B$$ is, we can make a pit stop at the naive sum of $$A$$ and $$B$$ along either path without losing anything, which is what we want out of a coproduct.

I think.

What is the unitization of a perhaps-unital ring?

I wrote it $$R \oplus \mathbb{Z}$$ to name the underlying abelian group, not to suggest that this was either the coproduct or the product with componentwise multiplication, neither of which is true as you've observed. The multiplication is uniquely determined by the condition that it restricts to the original multiplication on $$R$$ and that $$1 \in \mathbb{Z}$$ is the new unit, so I didn't think I needed to explicitly describe it.