Is the ideal $I = \{f\mid f (0) = 0\}$ in the ring $C [0, 1]$ of all continuous real valued functions on $[0, 1]$ a maximal ideal?

Solution 1:

As suggested by Sigur, you can show that $\varphi : \left\{ \begin{array}{ccc} C([0,1]) & \to & \mathbb{R} \\ f & \mapsto & f(0) \end{array} \right.$ induces an isomorphism between $C([0,1])/I$ and $\mathbb{R}$. Therefore, $C([0,1])/I$ is a field and $I$ is a maximal ideal.

Otherwise, you can prove it directly from the definition: Let $J$ be an ideal of $C([0,1])$ such that $I \subsetneq J$. In particular, there exists $g \in J \backslash I$ (ie. $g(0) \neq 0$). For every $f \in C([0,1])$:

$$f= \underset{ \in I \subset J}{\underbrace{\left( f- \frac{f(0)}{g(0)}g \right)}}+ \underset{\in J}{\underbrace{\frac{f(0)}{g(0)}g}} \ \in J$$

Therefore, $J=C([0,1])$ and $I$ is a maximal ideal.

Solution 2:

Suppose $I\subset J$ and that inclusion is proper. Then $\exists f\in J$ with $f(0)\not= 0$. Clearly $f(x)-f(0)\in I\subset J$. Thus $f(x)-\Big(f(x)-f(0)\Big)\in J$ and so $f(0)\in J$, which is to say $J$ contains $1.\:\:$