Question

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

Answer 1

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.