Proposition 3

a) Let $\widetilde{\mathfrak{a}}$ be a left ideal of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$. We set $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$. There exists an element $u$ of A such that $u-e$ belongs to $\widetilde{\mathfrak{a}}$. If $u$ is such an element, then it is a right unit of A modulo $\mathfrak{a}$, and we have $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$; in particular, the ideal $\mathfrak{a}$ is regular.

b) Conversely, let $\mathfrak{a}$ be a regular left ideal of A and $u$ a right unit of A modulo $\mathfrak{a}$. Set $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$. Then $\widetilde{\mathfrak{a}}$ is a left ideal of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$, and we have $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$.

c) If $k$ is a field, then the condition $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ is equivalent to saying that $\widetilde{\mathfrak{a}}$ is not contained in A.

Under the assumption of a), the element $e$ of $\widetilde{A}$ can be written as $(e-u)+u$ with $u\in A$ and $u-e\in \widetilde{\mathfrak{a}}$. Let $x=\lambda e+a$ be an element of $\widetilde{A}$, where $\lambda \in k$ and $a\in A$; if $x$ belongs to $\widetilde{\mathfrak{a}}$, then the element $y=a+\lambda u=x+\lambda (u-e)$ of A belongs to $\mathfrak{a}$, and we have $x=y-\lambda (u-e)$; this proves the equality $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$. Moreover, for every $a$ in A, the element $au-a=a(u-e)$ of A belongs to $\mathfrak{a}$, so $u$ is a right unit modulo $\mathfrak{a}$. This proves assertion a).

Let $\mathfrak{a}$ and $u$ be as in b). Since $\widetilde{A}$ is the direct sum of A and $k(u-e)$, we have $\widetilde{\mathfrak{a}}+ A =\widetilde{A}$ and $\widetilde{\mathfrak{a}}\cap A =\mathfrak{a}$. Every element of $\widetilde{\mathfrak{a}}$ is of the form $x+\lambda (u-e)$ with $x\in \mathfrak{a}$ and $\lambda \in k$. For every $a\in A$, we have

Since $u$ is a right unit of A modulo $\mathfrak{a}$, the sum $ax+\lambda (au-a)$ belongs to $\mathfrak{a}$, so that $\widetilde{\mathfrak{a}}$ is a left ideal of A. This proves b).

Finally, if $k$ is a field, then A is a hyperplane in $\widetilde{A}$, and $\widetilde{A}$ is the sum of $\widetilde{\mathfrak{a}}$ and A if and only if $\widetilde{\mathfrak{a}}$ is not contained in A.