Corollary 1

The radical of A is equal to the radical of the opposite ring $A^o$, that is, to the intersection of the maximal right ideals of A.

For every $x\in \mathfrak{R}$(A), $1 +x$ is invertible in the ring A, hence in the ring $A^o$. Since $\mathfrak{R}(A)$ is a two-sided ideal of $A^o$, we have $\mathfrak{R}(A)\subset \mathfrak{R}(A^o)$. The equality follows by interchanging the roles of A and $A^o$.