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Idempotents lift through adically complete quotients

Statement

Let $R$ be a commutative $C$ -algebra, let $I \subseteq R$ be an ideal, and suppose that $R$ is $I$ -adically complete and separated, so that $R ≅ {\underset{\leftarrow}{l i m}}_{n \geq 1} R / I^{n}$ (Separated and complete filtered modules, The $I$ -adic completion of a module); over a complete local ring $R$ one may take $I$ its maximal ideal. Let $A$ be an associative unital $R$ -algebra which is finitely generated as an $R$ -module and complete for the $I$ -adic topology of the filtration $I^{n} A$ , $n \geq 0$ , so that $A \to {\underset{\leftarrow}{l i m}}_{n \geq 1} A / I^{n} A$ is an isomorphism (The $I$ -adic topology on a module). If $x \in A$ satisfies $x^{2} - x \in I A$ , then there exists $e \in A$ with $e^{2} = e, e \equiv x (m o d I A) .$ Consequently: (1) every idempotent of $A / I A$ is the image of an idempotent of $A$ ; (2) for every finite family of pairwise orthogonal idempotents ${\overset{ˉ}{x}}_{1}, \dots, {\overset{ˉ}{x}}_{k}$ of $A / I A$ there are pairwise orthogonal idempotents $e_{1}, \dots, e_{k} \in A$ with $e_{i} \equiv {\overset{ˉ}{x}}_{i} (m o d I A)$ for all $i$ , and if ${\overset{ˉ}{x}}_{1} + \dots + {\overset{ˉ}{x}}_{k} = 1$ they may be chosen with $e_{1} + \dots + e_{k} = 1$ . Neither assertion uses a choice principle.

Facts & Assumptions

Given: A commutative $C$ -algebra $R$ , an ideal $I \subseteq R$ with $R$ $I$ -adically complete and separated, an associative unital $R$ -algebra $A$ finitely generated over $R$ and complete for the filtration $I^{n} A$ , and an element $x \in A$ with $x^{2} - x \in I A$ .

[F1]

The $I$ -adic topology on a module has the neighbourhood basis $I^{n} M$ of $0$ , so its basic open sets are the cosets $x + I^{n} M$ (The $I$ -adic topology on a module).

[F2]

Completeness of $A$ for the filtration $I^{n} A$ says that $A \to {\underset{\leftarrow}{l i m}}_{n \geq 1} A / I^{n} A$ is an isomorphism, and it includes separatedness, that is, $⋂_{n \geq 0} I^{n} A = 0$ (Separated and complete filtered modules, The $I$ -adic completion of a module).

[L1]

For every $n$ the submodule $I^{n} A$ is a two-sided ideal of $A$ , and multiplication $A \times A \to A$ is continuous for the $I$ -adic topology: if $a^{'} \equiv a$ and $b^{'} \equiv b$ modulo $I^{n} A$ , then $a^{'} b^{'} - a b = (a^{'} - a) b^{'} + a (b^{'} - b) \in I^{n} A + I^{n} A = I^{n} A$ .

Proof

technique · direct

1.1F1F2L1

Multiplication of $A$ is continuous by [L1], and limits in $A$ are unique because $A$ is separated by [F2]. A sequence is Cauchy precisely when for every $n$ its terms eventually have a fixed residue modulo $I^{n} A$ ; completeness then gives its unique limit with those eventual residues. In particular a sequence with $s_{m} \in I^{m} A$ tends to zero, and a series with its $m$ -th term in $I^{m} A$ has Cauchy partial sums, whose limit agrees with each partial sum modulo the ideal containing its tail.

2.1step 1.1algebra

Let $c_{m} : = (\binom{- 1 / 2}{m}) \in C$ for $m \geq 0$ . If $ε \in I A$ then $ε^{m} \in I^{m} A$ for every $m$ , so the partial sums $S_{N} : = \sum_{m = 0}^{N} c_{m} ε^{m}$ satisfy $S_{N^{'}} - S_{N} \in I^{N + 1} A$ whenever $N^{'} \geq N$ : the series converges, and its limit $f (ε) : = \sum_{m \geq 0} c_{m} ε^{m}$ satisfies $f (ε) \equiv S_{N} (m o d I^{N + 1} A)$ for every $N$ by step 1.1. To justify the formal identity, put $f (u) = \sum_{m \geq 0} c_{m} u^{m}$ . The binomial coefficients satisfy $2 (m + 1) c_{m + 1} = - (2 m + 1) c_{m}$ , so $2 (1 + u) f^{'} (u) + f (u) = 0$ . Consequently the formal derivative of $(1 + u) f (u)^{2}$ is zero, and its constant term is $1$ ; over $C$ this gives $(1 + u) f (u)^{2} = 1$ . Thus the identity $(1 + u) (\sum_{m = 0}^{N} c_{m} u^{m})^{2} = 1 + \sum_{m > N} d_{m} u^{m}$ of truncated formal power series over $C$ shows $(1 + ε) S_{N}^{2} \equiv 1 (m o d I^{N + 1} A)$ ; passing to limits using the continuity of multiplication and the uniqueness of limits gives $(1 + ε) f (ε)^{2} = 1$ .

2.2F2step 1.1construct

Let $e \in A$ be an idempotent and put $B : = (1 - e) A (1 - e)$ , with unit $1 - e$ . Then $B$ is an associative $R$ -algebra with unit $1 - e$ and scalar map $r \mapsto r (1 - e)$ , finitely generated as an $R$ -module, and $I^{n} B = B \cap I^{n} A$ for every $n$ : the inclusion $I^{n} B \subseteq B \cap I^{n} A$ is clear, and for $b \in B \cap I^{n} A$ the computation $π (b) = b$ , where $π (a) : = (1 - e) a (1 - e)$ is the $R$ -linear idempotent projection onto $B$ , exhibits $b \in π (I^{n} A) \subseteq I^{n} B$ . The projection $π$ satisfies $π (I^{n} A) \subseteq I^{n} B \subseteq I^{n} A$ , hence is continuous, so it induces an idempotent endomorphism $Π$ of the completion $A ≅ {\underset{\leftarrow}{l i m}}_{n} A / I^{n} A$ of [F2]; its image is exactly ${\underset{\leftarrow}{l i m}}_{n} B / I^{n} B$ with the maps induced by $π$ . Since $B \to {\underset{\leftarrow}{l i m}}_{n} B / I^{n} B$ agrees with $Π$ on $B$ and $Π$ is the identity on $B$ , the map $B \to {\underset{\leftarrow}{l i m}}_{n} B / I^{n} B$ is an isomorphism: it is injective by the separatedness of $A$ , and surjective onto the image of $Π$ . Thus $B$ is $I$ -adically complete.

3.1step 2.1algebraconstruct

Put $y : = 2 x - 1$ and $ε : = y^{2} - 1 = 4 (x^{2} - x) \in I A$ . Since $ε$ is a polynomial in $y$ , the element $z : = y f (ε)$ of step 2.1 satisfies $z^{2} = y^{2} f (ε)^{2} = (1 + ε) f (ε)^{2} = 1.$ Hence $e : = \frac{1}{2} (1 + z)$ satisfies $e^{2} = \frac{1}{4} (1 + 2 z + z^{2}) = e$ , and $e - x = \frac{1}{2} (z - y) = \frac{1}{2} y (f (ε) - 1) \in A \cdot I A \subseteq I A$ , because $f (ε) - 1 = \sum_{m \geq 1} c_{m} ε^{m} \in I A$ . Any idempotent $\overset{ˉ}{x} \in A / I A$ has a representative $x \in A$ with $x^{2} - x \in I A$ , so the preceding construction produces an idempotent $e$ of $A$ with image $\overset{ˉ}{x}$ : assertion (1) holds.

4.1step 3.1step 2.2algebra∎

Assertion (2) follows by finite iteration. Given orthogonal idempotents ${\overset{ˉ}{x}}_{1}, \dots, {\overset{ˉ}{x}}_{k}$ in $A / I A$ , choose representatives $x_{1}, \dots, x_{k}$ ; they satisfy $x_{i}^{2} - x_{i} \in I A$ and $x_{i} x_{j} \in I A$ for $i \neq j$ . Suppose $e_{1}, \dots, e_{j} \in A$ are pairwise orthogonal idempotents with $e_{i} \equiv x_{i} (m o d I A)$ for $i \leq j$ , put $E_{j} : = e_{1} + \dots + e_{j}$ and $B_{j} : = (1 - E_{j}) A (1 - E_{j})$ , and set $X_{j} : = (1 - E_{j}) x_{j + 1} (1 - E_{j}) \in B_{j}$ . Expanding, $X_{j}^{2} - X_{j} = (1 - E_{j}) (x_{j + 1}^{2} - x_{j + 1} - x_{j + 1} E_{j} x_{j + 1}) (1 - E_{j})$ , and both $x_{j + 1}^{2} - x_{j + 1}$ and $x_{j + 1} E_{j} x_{j + 1} \equiv \sum_{i \leq j} x_{j + 1} x_{i} x_{j + 1} \equiv 0$ lie in $I A$ , so $X_{j}^{2} - X_{j} \in (1 - E_{j}) I A (1 - E_{j}) \subseteq I B_{j}$ . As $B_{j}$ is complete by step 2.2, step 3.1 applied in $B_{j}$ supplies an idempotent $e_{j + 1} \in B_{j}$ with $e_{j + 1} \equiv X_{j} (m o d I B_{j})$ ; then $e_{j + 1}$ is orthogonal to $e_{1}, \dots, e_{j}$ , and $e_{j + 1} \equiv X_{j} \equiv x_{j + 1} (m o d I A)$ because $(1 - E_{j})$ is congruent to $1 - \sum_{i \leq j} x_{i}$ modulo $I A$ . Starting from $j = 0$ , where $B_{0} = A$ and $X_{0} = x_{1}$ , this yields orthogonal idempotents with the required congruences. If ${\overset{ˉ}{x}}_{1} + \dots + {\overset{ˉ}{x}}_{k} = 1$ , the last lift may be replaced by $e_{k} : = 1 - e_{1} - \dots - e_{k - 1}$ : it is idempotent, orthogonal to $e_{1}, \dots, e_{k - 1}$ , and congruent to $1 - \sum_{i < k} {\overset{ˉ}{x}}_{i} = {\overset{ˉ}{x}}_{k}$ modulo $I A$ .

Depends on

Used by

Triviality of finite free deformations of semisimple algebras over the power series ring Lemma

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ivan Losev, Lecture 8: Representations of GL_n(F_q) - Section 2.3, Step 5 of the proof of Theorem 2.6 ('result about lifting of idempotents'), PDF p. 5 (standard reference, not scraped)
Olivier Dudas and Jean Michel, Lectures on Finite Reductive Groups and Their Representations - Section 11.2 (flatness of Hecke algebras and Tits' deformation theorem), printed p. 47 (standard reference, not scraped)