Unitisation of a non-unital ring If I have a non-unital ring $R$ and to it I add an element to act as the unitary element for the ring. How can I be sure that the unitisation $U$ is a ring itself?
I am confused because for any element $r \in R$, $r+1$ isn't defined to equal anything.
Do I just say $r+1 = 1$ for all $r$ and is that a suitable answer?
 A: The key observation is that you cannot add just an element $1$ to the non-unital ring $R$. Indeed, if $r\in\mathbb R$, then $r+1\notin R$, as otherwise is $1+r=s\in R$, it would follow that $1=s-r\in R$. But we also cannot define $r+1=1$, since this implies $r=0$ because of the group property of addition. So since $r+1$ is neither an element of $R$ nor equals $1$, we need to make that another new element of the extended ring.
So we see that to make an unital ring, we have to add other elements besides $1$. But how many should we add, and which ones?
One immediate idea would be to add as few as possible. But to do that, we have to dive into the specifics of the ring.
For example, consider the non-unital ring of even integers. Clearly we can make that into an unital ring by just adding the odd integers, arriving at the integers. That is, we assign $1+1$ to a specific element of the ring, namely $2$.
But that doesn't work for any ring. For example, in the non-unital ring of multiples of $3$, there's no element that we could assign $1+1$ to, as $x=1+1$ implies $x+x=x\cdot x$, but there is no multiple of $3$ that fulfils the condition.
So the question is, is there a construction which works for any ring, but doesn't add any new elements that are not needed at least for some rings?
Looking at the non-unital rings that are multiples of an integer, we can easily see that for any $n\in \mathbb Z\setminus 0$, there are non-unital rings where we can't identify $n1$ with an element in $R$, therefore for the construction to work on all rings, we need to add all of them as new elements.
But then for any of the new elements $n1$, we can make the same argument as in the first paragraph to see that $n1+r\notin R$ either.
So for the construction to work with any non-unital ring, we have to add at least the elements
$$n1+r, n\in\mathbb Z\setminus\{0\}, r\in R.$$
Obviously, with $n=0$, we recover the elements of $R$, so in total we now have the elements
$$n1+r, n\in\mathbb Z, r\in R.$$
Now it turns out that these elements are sufficient to give a ring, no matter what non-unital ring we started with, therefore we don't need to add any further elements.
To see that this set is also closed under multiplication (that is, we don't need additional elements to fulfil the multiplicative laws of rings), we remember that one of those laws is the distributive law. In particular,
$$(n1+r)(m1+s) = (n1)(m1) + (n1)s + r(m1) + rs$$
Here we have to remember that by definition
$$nx = \begin{cases}
\underbrace{x+\ldots+x}_{n\text{ terms}} & n\ge 0\\
-(\underbrace{x+\ldots+x}_{-n\text{ terms}}) & n<0
\end{cases}$$
and thus using the distributive law and the fact that $1$ by assumtion is the identity of the extended ring, it can easily determined that
$$(n1)(m1) = (nm)1, \quad (n1)s = ns, \quad r(m1) = mr$$
and also that all of $ns$, $mr$ and, of course, $rs$ are elements of $R$, which implies that also $ns+mr+rs\in R$. Therefore we get that
$$(n1+r)(m1+s) = (nm)1 + (ns+mr+rs)$$
is again of the same form $k1+t$ with $k\in\mathbb Z$ and $t\in R$, so we don't get any new elements by multiplication either.
A: Since $1\notin R$, it follows that $1+r\notin R$ for any $r\in R$. We then have to say that these are new elements of the augmented ring. But these are not the only new ones. Comprehensively, the elements of the ring are
$$n1+r$$
for an integer $n$ and $r\in R$. Multiplication and addition are defined in the obvious way.
I stress that this is only the case in the universal construction of a unital ring from a nonunital one. If $R$ is a subring of a ring with $1$, this in general will not be isomorphic to that ring. However, a subring of the ring containing $R$ will be a homomorphic image of the universal unital ring mapping $R$ to itself. 
