TITLE: About rings with a right multiplicative identity $1$ such that every element in $R\setminus\{0\}$ has a left-inverse
QUESTION [6 upvotes]: $R$ be a  ring with a right multiplicative identity $1$ such that for every $a \in R \setminus \{0\}$, $\exists x \in R$ such that $xa=1$ i.e. every element in $R \setminus \{0\}$ has a left-inverse. Then is it true that $R$ is a division ring ? 

(I know that left-identity and left-inverse implies group similarly right-identity and right inverse implies group, but am not sure about the other combination.) 
I have tried for every  $a,b \in R$ \ $\{0\}$ $a=a.1=a(x_b.b)=(a.x_b).b$ ; if for some $b$ , $a.x_b=0 $ then $a=0.b=0 \in R$ \ $\{0\}$ impossible ; so $a=yb$ has a solution for $y \in R$ \ $ \{0\}$ for every $a,b \in R$ ; but I cannot show solution for $a=b.x$ . Please help .

REPLY [6 votes]: (We must add to the assumptions that $1\ne0$, or the assertion would be false, because the zero ring is not a division ring.)
Let's first show that a right multiplicative identity is also a left identity.
Call $\rho$ a right multiplicative identity: $a\rho=a$, for every $a\in R$. If $\rho a-a\ne0$, there exists $x$ such that $x(\rho a-a)=\rho$. Now
$$
x(\rho a-a)=(x\rho)a-xa=xa-xa=0,
$$
a contradiction. So $\rho a=a$. Hence $\rho$ is also a left identity.
Now we can set $\rho=1$ and use it as a two-sided identity.
First of all, if $xa=1$, then $x\ne0$. Since $x\ne0$, it has a left inverse, call it $y$: $yx=1$.
Thus
$$
y=y1=y(xa)=(yx)a=1a=a
$$
and so $x$ is both a left and right inverse of $a$.