Consider the ring \[\begin{align*} S=C[0,1]=\{f:[0,1]\rightarrow\mathbb R:f\text{ is continuous}\} .\end{align*}\]
with the usual operations of addition and multiplication of functions.
What are the invertible elements of \(S\)?
For \(a\in[0,1]\), define \(I_a=\{f\in S:f(a)=0\}\). Show that \(I_a\) is a maximal ideal of \(S\).
Show that the elements of any proper ideal of \(S\) have a common zero, i.e., if \(I\) is a proper ideal of \(S\), then there exists \(a\in[0,1]\) such that \(f(a)=0\) for all \(f\in I\). Conclude that every maximal ideal of \(S\) is of the form \(I_a\) for some \(a\in[0,1]\).
Hint: As \([0,1]\) is compact, every open cover of \([0,1]\) contains a finite subcover.
Solution
1. A function \(f\in S\) is invertible if and only if\[ f(x)\ne0\qquad\text{for every }x\in[0,1]. \]
If \(f\) is invertible, choose \(g\in S\) with \(fg=1\). Then for every \(x\), \[ f(x)g(x)=1, \] so \(f(x)\ne0\).
Conversely, suppose \(f\) is nowhere zero. Then \[ g(x)=\frac1{f(x)} \] is continuous on \([0,1]\), because reciprocal is continuous on \(\mathbb R\setminus\{0\}\) and \(f([0,1])\subseteq\mathbb R\setminus\{0\}\). Hence \(g\in S\) and \(fg=1\).
2. For each \(a\in[0,1]\), the set \(I_a\) is a maximal ideal of \(S\).
Consider evaluation at \(a\), \[ \operatorname{ev}_a:S\longrightarrow\mathbb R, \qquad \operatorname{ev}_a(f)=f(a). \] This is a surjective ring homomorphism, since every real number is the value at \(a\) of the corresponding constant function. Its kernel is exactly \[ \ker(\operatorname{ev}_a)=I_a. \] Therefore the first isomorphism theorem gives \[ S/I_a\cong\mathbb R. \] Because \(\mathbb R\) is a field, \(I_a\) is maximal.
3. Let \(I\subsetneq S\) be a proper ideal. Then the elements of \(I\) have a common zero.
Suppose not. Then for every \(x\in[0,1]\) there exists \(f_x\in I\) such that \[ f_x(x)\ne0. \] By continuity of \(f_x\), there is an open neighborhood \(U_x\) of \(x\) on which \(f_x\) is nowhere zero. The family \(\{U_x:x\in[0,1]\}\) is an open cover of \([0,1]\). By compactness, choose finitely many points \(x_1,\dots,x_n\) such that \[ [0,1]=U_{x_1}\cup\cdots\cup U_{x_n}. \] Set \[ h=f_{x_1}^2+\cdots+f_{x_n}^2. \] Since each \(f_{x_i}\in I\) and \(I\) is an ideal, \(h\in I\).
For every \(y\in[0,1]\), some \(U_{x_i}\) contains \(y\), so \(f_{x_i}(y)\ne0\). Hence \[ h(y)=\sum_{i=1}^n f_{x_i}(y)^2>0. \] Thus \(h\) is nowhere zero, so by 1 it is a unit of \(S\). Since \(h\in I\), the ideal \(I\) contains a unit, hence \(I=S\), contradicting that \(I\) is proper. Therefore there exists \(a\in[0,1]\) with \[ f(a)=0\qquad\text{for every }f\in I. \]
4. Every maximal ideal of \(S\) is \(I_a\) for some \(a\in[0,1]\).
Let \(M\) be maximal. By 3, there exists \(a\in[0,1]\) such that every \(f\in M\) vanishes at \(a\). Hence \[ M\subseteq I_a. \] By 2, \(I_a\) is a proper maximal ideal. Since \(M\) is maximal and is contained in the proper ideal \(I_a\), we must have \[ M=I_a. \]