I have a statement that says that if $R$ is a simple ring, then $Z(R)$ must be a field.
Since $R$ is a ring we get most of the field axioms (associativity and commutativity of addition, multiplication$\text{--}$since it's the center$\text{--}$, presence of the identities) immediately without need for explanation.
But I don't see how we can get the invertibility of multiplication.
On
Let $z\ne 0$ be an element of $Z(R)$. We know such a $z$ exists because $1\in Z(R)$.
Consider the ideal $I$ generated by $z$. Since $z\ne 0$ and $R$ is simple $I = R$. Therefore, there exists $s\in R$ such that $zs = 1$. Since $z\in Z(R)$ $zs = sz$ so $s=z^{-1}$.
It remains to show that $z^{-1}\in Z(R)$.
For any $r\in R$, $z^{-1}r = z^{-1}r(zz^{-1})=(z^{-1}z)rz^{-1} = rz^{-1}$.
Every central element is invertible. Suppose $a \in Z(R)$ is not. Then image of multiplication by $a$ does not contain $1$; but then it's a nontrivial two-sided ideal, because for all $r, r'$ $r'ar = ar'r$ and $arr'$ bot lie in $aR$.
If central element is invertible, then it's inverse also central. Take central $a$ and it's inverse $a^{-1}$. Then $a^{-1}r = a^{-1}(ra)a^{-1} = a^{-1}(ar)a^{-1} = ra^{-1}$.