Lemma 4

Let $A$ be a commutative ring and let $a$ be a finitely generated ideal of $A$ such that $a = a^2$. Then there exists an idempotent $e$ in $A$ such that $a = Ae$.

Let $(a_1, ..., a_r)$ be a generating system of the ideal $a$; since $a = a^2$, there exist elements $x_{ij}$ in $a$ such that $a_i = \sum_{j=1}^r x_{ij} a_j$ for $1 \leq i \leq r$. Write $M$ for the square matrix of order $r$ whose elements are $\delta_{ij} - x_{ij}$ and let $D$ be its determinant. There exists (III, p. 532, Formula (26)) a square matrix $N$ of order $r$ with elements in $A$ such that $N \cdot M = D \cdot I$, whence immediately $Da_j = 0$ for $1 \leq j \leq r$ and so finally $Da = 0$. Now the matrix $M$ is congruent to $I$, mod $a$, hence $D \equiv I$ mod $a$. Put $e = 1 - D$; then $e \in a$ and $ex = x$ for all $x \in a$. It follows that $e$ is an idempotent and $a$ is equal to $Ae$.

With these lemmas established, let us show by induction on the degree of $A$ that $A$ is diagonalizable if $K$ is algebraically closed and $\Omega_K(A) = 0$. Let $m$ be a maximal ideal of $A$ (I, p. 104). By Lemmas 3 and 4, there exists an idempotent $e$ such that $m = Ae$; we have seen that $A/m$ is of degree 1 over $K$. Hence $A$ is a direct sum of the ideals $a = (1 - e)A$ and $m$ and we have $[a : K] = 1$, hence $A$ is isomorphic to $K \times A/a$. Since $\Omega_K(A/a)$ is isomorphic to a quotient of $\Omega_K(A)$ (III, p. 573, Prop. 22), it is zero and the induction hypothesis shows that $A/a$ is diagonalizable. This then shows $A$ to be diagonalizable.