Proposition 7

Suppose that the $(A\otimes_K\overline{A}^o)$-module $\overline{A}$ is projective.

a) Let $S_1$ and $S_2$ be subalgebras of A satisfying $A = S_1\oplus \mathfrak{r}= S_2\oplus \mathfrak{r}$. There exists a special automorphism of A transforming $S_1$ into $S_2$.

b) Suppose that $\pi$ has a K-linear section and that the radical $\mathfrak{r}$ of A is nilpotent. Then there exists a subalgebra S of A satisfying $A = S\oplus \mathfrak{r}$.

Let $S_1$ and $S_2$ be as in a). Let $s_1$ and $s_2$ be the elements of the set Σ corresponding to the subalgebras $S_1$ and $S_2$. Let $\varepsilon$ be the K-linear mapping from $A\otimes_KA$ to A given by $\varepsilon (a\otimes b) =ab$. By Proposition 5 of VIII, p. 236 and Remark 1 of VIII, p. 237, there exists an element $e=\sum^r_{i=1}\alpha_i\otimes \alpha '_i$ of $\overline{A}\otimes_KA$ satisfying $\sum^r_{i=1}\alpha_i\alpha '_i= 1$ and $\sum^r_{i=1}\alpha \alpha_i\otimes \alpha '_i=\sum^r_{i=1}\alpha_i\otimes \alpha '_i\alpha$ for every $\alpha \in A$. Set $x=\sum^r_{i=1}s_1(\alpha_i)s_2(\alpha '_i)$. We have $\pi (x) =\sum^r_{i=1}\alpha_i\alpha '_i= 1$ and therefore $x\in 1 +\mathfrak{r}$. Let $\alpha$ be an element of A. We have

The equality $x^{-1}S_1x= S_2$ follows, giving assertion a).

Under the assumptions of b), suppose furthermore that $\mathfrak{r}^2= 0$. In this case, the $(A$, A)-bimodule $\mathfrak{r}$ is annihilated by $\mathfrak{r}$, and we therefore view it as an $(A$, A)-bimodule. Choose a K-linear section $\sigma$ of $\pi$. We have

(19) $\alpha x=\sigma (\alpha )x$ and $x\alpha =x\sigma (\alpha )$

for $\alpha \in \overline{A}$ and $x\in \mathfrak{r}$. Set

for $\alpha , \beta \in A$. We have the relation $\pi (\varphi (\alpha , \beta )) =\alpha \beta -\alpha \beta = 0$ for $\alpha , \beta \in A$. Therefore, $\varphi$ defines an element of $C^2(A,\mathfrak{r})$. Let $\alpha , \beta , \gamma$ be elements of A; in view of (19), we have

By Theorem 3 of VIII, p. 242, the K-module $H^2(A,\mathfrak{r})$ is reduced to zero. Hence there exists an element $\psi$ of $C^1(A,\mathfrak{r})$ such that $\partial^1\psi =\varphi$, in other words, such that we have

(21) $\varphi (\alpha , \beta ) =\alpha \psi (\beta )-\psi (\alpha \beta ) +\psi (\alpha )\beta$ for $\alpha , \beta$ in $\overline{A}$.

We have $\psi (\alpha )\psi (\beta ) = 0$ because $\mathfrak{r}^2$ is zero; from (19) and (20), we then deduce (22) $(\sigma +\psi )(\alpha \beta ) = (\sigma +\psi )(\alpha )(\sigma +\psi )(\beta )$,

so that the K-linear section $\sigma +\psi$ of $\pi$ belongs to Σ. Its image is a subalgebra S of A such that $A = S +\mathfrak{r}$.

Let us now prove the existence of S in the general case. We reason by induction on the least integer $p\geqslant 1$ such that $\mathfrak{r}^p= 0$; the case $p= 1$ is trivial. Suppose $p\geqslant 2$, and set $A'= A/\mathfrak{r}^{p-1}$. The radical $\mathfrak{r}'$ of $A'$ is equal to $\mathfrak{r}/\mathfrak{r}^{p-1}$ (Proposition 5 of VIII, p. 155), so satisfies ${\mathfrak{r}'}^{p-1}= 0$, and the algebra $A'/\mathfrak{r}'$ is isomorphic to $\overline{A}= A/\mathfrak{r}$ and is therefore absolutely semisimple. By the induction hypothesis, there exists a subalgebra $S'$ of $A'$ such that $A'= S'\oplus \mathfrak{r}'$. Then $S'$ is of the form $A''/\mathfrak{r}^{p-1}$, where $A''$ is a subalgebra of A containing $\mathfrak{r}^{p-1}$, and we have

The algebra $A''/\mathfrak{r}^{p-1}$ is isomorphic to $A'/\mathfrak{r}'$; we have $(\mathfrak{r}^{p-1})^2= 0$, so $\mathfrak{r}^{p-1}$ is the radical of $A''$. By the case we just treated, there exists a subalgebra S of $A''$ such that $A''= S\oplus \mathfrak{r}^{p-1}$; we deduce the relation $A = S\oplus \mathfrak{r}$ from (23).