Proposition 4

a) Let $\mathfrak{a}$ be a regular maximal left ideal of A. There exists a unique left ideal $\widetilde{\mathfrak{a}}$ of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ and $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$. This ideal is maximal and does not contain A.

b) The mapping $\widetilde{\mathfrak{a}}\mapsto \widetilde{\mathfrak{a}}\cap A$ is a bijection from the set of maximal left ideals of $\widetilde{A}$ not containing A to the set of regular maximal left ideals of A.

Let $\mathfrak{a}$ be a regular maximal left ideal of A. By Proposition 3, a), the left ideals $\mathfrak{b}$ of $\widetilde{A}$ such that $\mathfrak{b}+ A =\widetilde{A}$ and $\mathfrak{b}\cap A =\mathfrak{a}$ are the ideals $\mathfrak{a}+k(u-e)$, where $u$ is a right unit modulo $\mathfrak{a}$. To prove the uniqueness of $\widetilde{\mathfrak{a}}$, it therefore suffices to prove that two right units $u$ and $u'$ of A modulo $\mathfrak{a}$ are congruent modulo $\mathfrak{a}$. Let us reason by contradiction and suppose that $u-u'$ does not belong to $\mathfrak{a}$. The formula $x(u-u') = (xu-x)-(xu'-x)$ shows that we have $A(u-u')\subset \mathfrak{a}$; it follows that $\mathfrak{a}+k(u-u')$ is a left ideal of A containing $\mathfrak{a}$ and distinct from $\mathfrak{a}$. Since $\mathfrak{a}$ is maximal, we therefore have $\mathfrak{a}+k(u-u') = A$, and so AA $\subset \mathfrak{a}$. For every $x\in A$, we have $x\equiv xu$ (mod $\mathfrak{a}$), and therefore $x\in \mathfrak{a}$ by the above, which contradicts the assumption $\mathfrak{a}\not= A$.

Hence there exists a unique left ideal $\widetilde{\mathfrak{a}}$ of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ and $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$. Let $\mathfrak{b}$ be a proper left ideal of $\widetilde{A}$ containing $\widetilde{\mathfrak{a}}$. Then $\mathfrak{b}\cap A$ is a proper left ideal of A containing $\mathfrak{a}$. It is therefore equal to $\mathfrak{a}$ because $\mathfrak{a}$ is maximal, which implies $\mathfrak{b}=\widetilde{\mathfrak{a}}$ by Lemma 2 of VIII, p. 4. This proves that $\widetilde{\mathfrak{a}}$ is a maximal ideal of $\widetilde{A}$; for such an ideal, the condition $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ means that $\widetilde{\mathfrak{a}}$ does not contain A.

It remains to prove that if $\widetilde{\mathfrak{a}}$ is a maximal left ideal of $\widetilde{A}$ that does not contain A, then the left ideal $\mathfrak{a}= A\cap \widetilde{\mathfrak{a}}$ of A is maximal. Let $\mathfrak{b}$ be a proper left ideal of A containing $\mathfrak{a}$. Let $u$ be a right unit modulo $\mathfrak{a}$ such that $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$ (Proposition 3). It is also a right unit modulo $\mathfrak{b}$; denote by $\widetilde{\mathfrak{b}}$ the ideal $\mathfrak{b}+k(u-e)$ of $\widetilde{A}$. By Proposition 3, b), we have $\mathfrak{b}= A\cap \widetilde{\mathfrak{b}}$. The ideal $\widetilde{\mathfrak{a}}$, equal to $\mathfrak{a}+k(u-e)$, is contained in $\widetilde{\mathfrak{b}}$ and is therefore equal to $\widetilde{\mathfrak{b}}$ because $\widetilde{\mathfrak{b}}$ is distinct from $\widetilde{A}$ and $\widetilde{\mathfrak{a}}$ is maximal. Consequently, we have $\mathfrak{a}=\mathfrak{b}$, and $\mathfrak{a}$ is maximal.

We leave it to the reader to translate Propositions 3 and 4 to right ideals.