Suppose that $R$ is the ring of real valued continuous functions defined on $[0,1]$. If $M$ is a maximal ideal of $R$, then prove that $M$ will be of following form $$M=\{f \in R\mid f(x)=0\}$$ for a fixed $x\in[0,1]$.
Obviously the ideal given by $M=\{f \in R\mid f(x)=0\}$ is an maximal ideal for a fixed choice of $x$ in $[0,1]$.
But starting from a random maximal ideal how can I get to that form?
Let $J$ be a proper ideal of $C[0,1]$.
Since $1 \notin J$, it follows that each $f \in J$ has at least one zero.
Claim the elements of $J$ have a common zero.
Suppose otherwise.
For each $f \in J$, let $V_f = \{x \in [0,1]\mid f(x) \ne 0\}$.
Since each $f$ is continuous, it follows that $V_f$ is open in $[0,1]$, for all $f \in J$.
By the assumption that the elements of $J$ have no common zero, it follows that $\{V_f \mid f \in J\}$ is an open cover of $[0,1]$.
Since $[0,1]$ is compact, there must be a finite subcover $V_{f_1},...,V_{f_n}$, say.
Let $g = f_1^2 + \cdots + f_n^2$. Then $g \in J$, but $g$ has no zeros, contradiction.
Hence the elements of $J$ have a common zero, $a$ say.
Then $J$ is a subset of the maximal ideal $M_a = \{f \in C[0,1]\mid f(a) = 0\}$.
Therefore, if $J$ is maximal, we must have $J = M_a$.