Question #16892

Let R be the (commutative) ring of all real-valued continuous functions on [0, 1]. Is R a semisimple ring?

Expert's answer

The answer is “no”. Indeed, assume *R *is semisimple. Consider the ideal *I *= *{f **∈** R *: *f*(0) = 0*}.*

Then*I *is a direct summand of *R*, and *I *= *Re *for some idempotent *e **∈** I*. But *e*(*x*)^{2} = *e*(*x*) =*⇒** e*(*x*) *∈** {*0*, *1*}*

for any*x **∈*[0*, *1]. Since *e*(0) = 0, continuity of *e *forces *e *to be the 0-function, so *I *= 0; a contradiction.

Then

for any

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