the center of a simple ring is either $0$ or a field The definition for a simple ring:  

A ring $R $ is said to be simple if $R^{2} \neq 0$ and $0$ and $ R $ are the only ideals
  of $R$.

The definition for center of a ring:    

The center of $R$ is the subset $C(R) = \{x\in R \mid xr = rx , \forall r\in R\}$. 

my question is: 
is the center of a simple ring  either $0$ or a field. In particular, a
commutative simple ring is thus necessarily a field.
 A: You are correct, a commutative simple ring is a field, but you need to show it has an identity.
Proof: Pick $x\neq 0$ from $R$ so that $xR\neq 0$. This is true since $R^2\neq 0$.
Since $xR$ is an ideal in $R$ and $xR\neq 0$, by simplicity, $xR=R.$
In particular, there is an $e\in R$ so that $xe=x$. But then for each $r\in R=xR$, we have $r=xr'$ for some $r'\in R$, and then $re=xr'e=xer'=xr'=r.$ So $e$ is a multiplicative identity of $R$.
Then the rest of your proof follows. For any non-zero $x\in R$, $x\in xR$, so $xR$ is a non-zero ideal, so $xR=R$ and hence $e\in xR$, so $x$ has a multiplicative inverse.
You can only conclude that $C(R)$ is a field if you know that $C(R)$ is simple. It is not generally true that if $S$ is a subring of a simple ring $R$, then $S$ is simple. (For example, $R=\mathbb Q, S=\mathbb Z$.) So you'd need some proof that if $R$ is simple then $C(R)$ is simple. 
I haven't found that proof, but I haven't found a counter-example, either.
If $x\in C(R)$ and $xC(R)\neq 0$ and $xC(R)\neq C(R)$, then we have that $xR\supseteq xC(R)\neq 0$, so $xR=R$ by $R$'s simplicity. If $xC(R)\neq C(R)$, then some element of $y\in C(R)$ can be written as $y=xr$ for some $r\in R\setminus C(R).$ Now, given $a\in R$. you have that $a=xa'$ for some $a'$ so that:
$$ar = xb'r=b'xr=b'y=yb'=rx'b=ra$$
So we get that $y\in C(R)$, which is a contradiction.
We are left with the case where $C(R)$ is not simple because $C(R)^2=0$. I wonder if it is possible to find such a case. Is it perhaps possible that $C(R)=0$? Certainly not when $R$ has an identity. When $R$ is a simple ring with identity, then $C(R)$ is definitely a field.
A: I know this is an old post, but recently I was wondering the same thing.
By Thomas Andrews' arguments, it remains to show that if a ring $R$ is simple and non-unital, its centre is zero. To obtain a contradiction, assume there exists a non-zero central element $z\in R$. As $R^2 \neq \{0\}$, note that $zR$ is a non-zero ideal of $R$. Otherwise we would have $zR=Rz=\{0\}$, so $\mathbb{Z}z:=\{nz: \, n \in \mathbb{Z}\}$ would be a non-zero ideal of $R$. As $R$ is simple, this forces $\mathbb{Z}z=R$. But then $R^2=\{0\}$; a contradiction.
Hence, as $zR$ is a non-zero ideal of $R$, by simplicity of $R$ we have $zR=R$, so there is $e \in R$ such that $ze=z$. Then, since $zR=R$ and $z$ is central, it follows that $e$ is an identity in $R$; a contradiction.
