# Ideals in $C[0,1]$

$(a)$ not necessary, If I take $S=(1/2,1/3)$ still $Z(S)$ is an ideal.

$(b)$ I know that maximal ideals in $C[0,1]$ comes in this fashion(I dont know the proff rigoriously) i.e $C_a=\{f\in C[0,1]:f(a)=0\}$ so $S$ may be finite or countable set? so I gues $b$ is true statement and for the same reason $c$ is also true. but I will be happy if some one explain me a bit about $b$ and $c$. Thank you.

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Could you provide us with a reference? –  Hans Giebenrath Dec 18 '12 at 9:06
it is not from a book just a question paper of past year exam paper –  Bunuelian Trick Dec 18 '12 at 9:12
To show that $C_a$ is maximal define $\phi:C[0,1]\longrightarrow\mathbb R$ with $\phi(f)=f(a)$ and then use the first isomorphism theorem.
For the converse define $\phi:C[0,1]\longrightarrow\mathbb C(S), \ f \mapsto f|_S$ to show that $C[0,1]\ /Z(S)$ is isomorphic to $C(S)$.
Then use that $m \leq R$ is maximal $\iff R/m$ is a field.
$S$ can be uncountable or countably infinite to be a maximal ideal? –  Bunuelian Trick Dec 18 '12 at 7:03
@Kuttus: $Z(S)$ is maximal iff $S$ is an one point set. –  P.. Dec 18 '12 at 10:22