# Answer on Abstract Algebra Question for Irvin

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

Then

for any

*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.Need a fast expert's response?

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