Prove that every rng with a left-side identity has a right-side identity? Given a rng $(R,+,\cdot)$ and $e \in R$ with $\forall a \in R: e \cdot a = a$, I would like to prove $a \cdot e = a$. What I have so far is the following:
Assume there exists $r \in R$ with $\forall a \in R: a \cdot r = a$. Plug in $e$ for $a$ to get $e \cdot r = e$. But by the given conditions, $e \cdot r = r$. Thus, $r=e$ and consequently $e \cdot r = a$.
Now the problem I have with this is that I have to assume there exists a right identity element. Any tips on how I could solve this problem?
Is the thing I'm trying to prove even true? I made this exercise for myself, but my feeling tells me it's true.
 A: Your argument shows that if both a left and a right identity exists, they must be equal (and hence must be an identity, and hence must be unique). But it's not true that the existence of a left identity implies the existence of a right identity. For example, let $S$ be the semigroup consisting of two elements $\{ a, b\}$ with multiplication table
$$a^2 = ba = a, b^2 = ab = b.$$
That is, every word in the generators evaluates to its rightmost letter. Then $a$ and $b$ are both left identities, but neither of them are right identities. The "semigroup rng" $\mathbb{Z}[S]$ similarly has two left identities, and hence it can't have any right identities (because that contradicts uniqueness). 
A: This is not true in general. Take the following example
$$R = \left\{\left(\begin{array}{cc} x & x \\ y & y\end{array}\right):a,b\in\mathbb{R}\right\}$$
with usual matrix addition and multiplication. This is a ring. (I'll leave you to show this.) However the right identities are those elements such that $x+y=1$ but these are not left identities. To see this, consider the matrix $\left(\begin{array}{cc} 0 & 0 \\ 1 & 1\end{array}\right)$.
To address the spirit of the question: consider $(a\cdot e)\cdot e$. Then by associativity, this is the same as $a\cdot(e\cdot e) = a\cdot e$. So for elements of the form $a\cdot e$, $e$ acts as a right identity. However the sticking point is that the image of $R$ under right multiplication by $e$ may not be all of $R$. If it were, then indeed $e$ would be a right identity.
