Proposition 6

Denote by $\mathscr{B}_A$ and $\mathscr{B}_B$ the sets of two-sided ideals of A and two-sided ideals of B.

a) There exists an isomorphism of ordered sets $f$ from $\mathscr{B}_A$ to $\mathscr{B}_B$ determined by the following property: if $\mathfrak{a}$ is a two-sided ideal of A and $\mathfrak{b}$ a two-sided ideal of B, then the relation $f(\mathfrak{a}) =\mathfrak{b}$ is equivalent to $\mathfrak{a}P = P\mathfrak{b}$.

b) Suppose that the ring A is commutative, so that A can be identified with the center of B (VIII, p. 102, Corollary 1). The isomorphism $f:\mathscr{B}_A\rightarrow$ $\mathscr{B}_B$ sends an ideal $\mathfrak{a}$ of A to the two-sided ideal $B\mathfrak{a}$ of B, and we have $\mathfrak{a}= A\cap B\mathfrak{a}$.

Assertion a) follows from Example 2.

Now, suppose that A is commutative, and identify A with the center of B. Let $\mathfrak{a}$ be an ideal of A. Then $B\mathfrak{a}$ is a two-sided ideal of B; we have PB$\mathfrak{a}=\mathfrak{a}P$ and therefore $f(\mathfrak{a}) = B\mathfrak{a}$. Let $\mathfrak{a}'$ be the ideal $A\cap B\mathfrak{a}$ of A; it is contained in $B\mathfrak{a}$ and contains $\mathfrak{a}$, so $B\mathfrak{a}'$ is equal to $B\mathfrak{a}$. Since $f$ is bijective, it follows that $\mathfrak{a}'=\mathfrak{a}$.