IMC / 2000 / Problems / Day 1, P5

IMC 2000 · Day 1 · P5

hard

Let $R$ be a ring of characteristic zero (not necessarily commutative). Let $e$ , $f$ and $g$ be idempotent elements of $R$ satisfying $e + f + g = 0$ . Show that $e = f = g = 0$ .

( $R$ is of characteristic zero means that, if $a \in R$ and $n$ is a positive integer, then $na \ne 0$ unless $a = 0$ . An idempotent $x$ is an element satisfying $x = x^2$ .)

Solution (official)

Suppose that $e + f + g = 0$ for given idempotents $e, f, g \in R$ . Then $g = g^2 = \bigl( -(e+f) \bigr)^2 = e + (ef + fe) + f = (ef + fe) - g,$ i.e. $ef + fe = 2g$ , whence the additive commutator $[e, f] = ef - fe = [e, ef + fe] = 2 [e, g] = 2 [e, -e-f] = -2 [e, f],$ i.e. $ef = fe$ (since $R$ has zero characteristic). Thus $ef + fe = 2g$ becomes $ef = g$ , so that $e + f + ef = 0$ . On multiplying by $e$ , this yields $e + 2ef = 0$ , and similarly $f + 2ef = 0$ , so that $f = -2ef = e$ , hence $e = f = g$ by symmetry. Hence, finaly, $3e = e + f + g = 0$ , i.e. $e = f = g = 0$ .

For part (i) just omit some of this.

How the field did

contestants scored

114

average (of 20)

7.78

solved (≥ 80%)

25.4%

near-0 (≤ 10%)

38.6%

discrimination

0.59

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

IMC 2000 · Day 1 · P5

How the field did

Score distribution (field cohort)

Similar problems