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?
Asked
Active
Viewed 2,748 times
11
-
7Why do you ask? What do you think? What have you tried? – hmakholm left over Monica Dec 30 '12 at 13:51
-
yes,infact I={f:f(c)=0} are the all possible maximal ideals – Koushik Dec 30 '12 at 13:51
-
6Hint: Try to determine the quotient $C[0,1]/I$ using the map $\phi: g\mapsto g(0)$. What is $\ker \phi$? What is ${\rm Im}~\phi$? – Sigur Dec 30 '12 at 13:56
-
where is the compactness of [0,1] used? – Koushik Dec 30 '12 at 14:04
-
Why do you think compactness would be used, @K.Ghosh? – Thomas Andrews Dec 30 '12 at 14:05
-
in C(0,1) does this hold? i don't know – Koushik Dec 30 '12 at 14:10
-
6@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. – Dec 30 '12 at 14:14
-
This question is contained in http://math.stackexchange.com/questions/375400/maximal-ideals-in-the-ring-of-real-functions-on-0-1 – May 26 '13 at 13:25
-
Related: Proof that ideals in $C[0,1]$ are of the form $M_c$ that should not involve Zorn's Lemma – PinkyWay Dec 10 '23 at 11:55
2 Answers
10
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.
Seirios
- 33,157
