$R$ is a division Ring. $R-Z(R)$ is finite. How to prove $R$ is finite $R$ is a division Ring. $Z(R)$ is the center of ring and $R-Z(R)$ is finite. How should I prove that $R$ is finite?
 A: Since $R \setminus Z(R)$ is finite, we need to show that $Z(R)$ is finite.
Now
$Z(R) = \{z \in R \mid zr = rz, \; \forall r \in R \}; \tag 1$
as such, $Z(R)$ is a field.  It is clear that $Z(R)$ is a subring of $R$; if $0 \ne z \in Z(R)$, then since for $r \in R$,
$zr = rz, \tag 2$
we have
$rz^{-1} = z^{-1}r, \tag 3$
which shows $z^{-1} \in Z(R)$; thus $Z(R)$ is a field, a subfield of the division ring $R$.  
Now let $r \in R \setminus Z(R)$; we have, for $0 \ne z \in Z(R)$, 
$zr \in R \setminus Z(R); \tag 4$
for if $zr \in Z(R)$, then
$r = z^{-1}(zr) \in Z(R), \tag 5$
a contradiction.  Furthermrore, for $r \in R \setminus Z(R)$, the map $\theta_r:Z(R) \setminus \{0\} \to R \setminus Z(R)$ given by
$\theta_r(z) = zr \tag 6$
is injective, since
$z_1r = z_2r \Longrightarrow z_1 = (z_1r)r^{-1} = (z_2r)r^{-1} = z_2; \tag 7$
now since $R \setminus Z(R)$ is finite, the set $\theta_r(Z(R) \setminus \{0\}) \subset R \setminus Z(R)$ is also finite; since $\theta_r$ is injective, we must have $Z(R) \setminus \{0\}$, and hence $Z(R)$, are finite as well.  And that's that.
Note Added in Edit, Saturday 21 October 2017 9:23 AM PST:  First off, lauds to Milo Brandt and Lord Shark, who advanced the notion of a proof based on the additive cosets $r + Z(R)$; quite elegant.  What I saw, when first reading this problem, was that, since $Z(R)$ is a sub-field of $R$, we can, in a manner analogous to field extensions, conceive of $R$ as a vector space over $Z(R)$; then the whole rigamarole about $\theta_r(z) = zr$ is based on the notion that $\theta_r(Z(R))$ is a one-dimensinal vector subspace of $R$; the rest is just a simple counting argument.  End of Note.
A: Note that $Z(R)$ is an additive subgroup of $(R,+$). Consider the quotient map $\pi:R\rightarrow (R,+)/Z(R)$. The fibers of this map are all equally large and $Z(R)$ is the kernel. Assuming that $R$ is not a field, we have that $R\setminus Z(R)$ is not empty. Thus, $R\setminus Z(R)$ is a non-empty union of fibers of $\pi$, each of which has size $Z(R)$. Thus, $Z(R)$ must be finite, so $R$ is too.
