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

  • 7
    $\begingroup$ Why do you ask? What do you think? What have you tried? $\endgroup$ Dec 30, 2012 at 13:51
  • $\begingroup$ yes,infact I={f:f(c)=0} are the all possible maximal ideals $\endgroup$
    – Koushik
    Dec 30, 2012 at 13:51
  • 6
    $\begingroup$ Hint: Try to determine the quotient $C[0,1]/I$ using the map $\phi: g\mapsto g(0)$. What is $\ker \phi$? What is ${\rm Im}~\phi$? $\endgroup$
    – Sigur
    Dec 30, 2012 at 13:56
  • $\begingroup$ where is the compactness of [0,1] used? $\endgroup$
    – Koushik
    Dec 30, 2012 at 14:04
  • 6
    $\begingroup$ @K.Ghosh, Compactness is needed for the statement that every max ideal is of the form $I_c = \{f: f(c) = 0\}$, not this one. $\endgroup$
    – user27126
    Dec 30, 2012 at 14:14

2 Answers 2


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.


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 $1.\:\:$


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