Isomorphism of Ring of real valued continuous functions on [0,1]

Question

Let $R$ denotes the ring of real valued continuous functions on $[0,1]$. For each $x \in [0,1]$ we have $I_x=\{f\in R|f(x)=0 \}$ is a maximal ideal of $R$. I have seen the proof that $x\to I_x$ gives bijection between points in $[0,1]$ and maximal ideals of $R$. Because of this we a natural homomorphism $$\phi : R \to \prod_{x\in I_x}R/I_x$$ Now this homomorphism is clearly injective as $\cap_{x\in[0,1]} I_x=(0)$. Also, for $x\neq y$ it is easy to construct continuous functions $f$ and $g$ such that $f(x)=g(y)=0$ and $f+g=1$, which shows that $I_x$ and $I_y$ are pairwise coprime. Now, my question is that can I apply Chinese remainder Theorem to conclude that $\phi$ is in fact a ring isomorphism ? Then can we say that $R \cong \mathbb{R}^{[0,1]}$ as we have each $R/I_x \cong \mathbb{R}$ ?

Answer 1

An easy thing to prove about $C[0,1]$ (which I am using to denote the ring of continuous functions you define) is that the only idempotents are $0$ and $1$, and therefore it cannot be the product of even two nonzero rings.

As for the Chinese remainder theorem, a proof will usually go about showing that the individual coordinates (like $\{(a,0,0,0...)\mid a\in R\}$ and $\{(0,a,0,0,0...)\mid a\in R\}$ and so on) can be mapped onto, and therefore their finite sums can be mapped onto. When the index set is finite, that is the entire image, but when the index set is infinite, it is no longer the case, so the proof breaks down.