Let \(R = C[0, 1]\) be the ring of continuous real-valued functions on the interval \([0, 1]\). Let I be an ideal of \(R\).
Show that if \(f \in I, a \in [0, 1]\) are such that \(f (a) \neq 0\), then there exists \(g \in I\) such that \(g(x) \geq 0\) for all \(x \in [0, 1]\), and \(g(x) > 0\) for all \(x\) in some open neighborhood of \(a\).
If \(I \neq R\), show that the set \(Z(I) = \{x \in [0, 1] \suchthat f(x) = 0 \text{ for all } f \in I\}\) is nonempty.
Show that if \(I\) is maximal, then there exists \(x_0 \in [0, 1]\) such that \(I = \{ f \in R \suchthat f (x_0 ) = 0\}\).
Cool problem, but pretty specific topological tricks needed.
Solution
(of a)
- Suppose \(c\da f(a)\neq 0\), noting that \(c\) may not be positive.
- By continuity, pick \(\eps\) small enough so that \(\abs{x-a}<\eps \implies \abs{f(x) - f(a)} < c/2\). Since we’re on the interval, we have \(f(x) \in (f(a) - c/2, f(a) + c/2) = (c/2, 3c/2)\) which is a ball of radius \(c/2\) about \(c\), which thus doesn’t intersect \(0\).
- So \(f(x) \neq 0\) on this ball, and \(g \da f^2 > 0\) on it. Note that ideals are closed under products, so \(g\in I\)
- Moreover \(f^2(x) \geq 0\) since squares are non-negative, so \(g\geq 0\) on \([0, 1]\).
(of b)
- By contrapositive, suppose \(V(I)= \emptyset\), we’ll show \(I\) contains a unit and thus \(I=R\).
- For each fixed \(x\in [0, 1]\), since \(V(I)\) is empty there is some \(f_x\) such that \(f_x(x) \neq 0\).
- By (a), there is some \(g_x\) with \(g_x(x) > 0\) on a neighborhood \(U_x\ni x\) and \(g_x \geq 0\) everywhere.
- Ranging over all \(x\) yields a collection \(\ts{(g_x, U_x) \st x\in [0, 1]}\) where \(\ts{U_x}\covers [0, 1]\).
- By compactness there is a finite subcover, yielding a finite collection \(\ts{(g_k, U_k)}_{k=1}^n\) for some \(n\).
- Define the candidate unit as \[ G(x) \da {1\over \sum_{k=1}^n g_k(x)} .\]
- This is well-defined: fix an \(x\), then the denominator is zero at \(x\) iff \(g_k(x) = 0\) for all \(k\). But since the \(U_k\) form an open cover, \(x\in U_\ell\) for some \(\ell\), and \(g_\ell > 0\) on \(U_\ell\).
- Since ideals are closed under sums, \(H\da {1\over G} \da \sum g_k \in I\). But \(H\) is a unit: by definition \(HG = \id\), so \(H\) has a multiplicative inverse \(G\) in \(R\), hence \(H\) is a unit.
(of c)
- If \(I\normal R\) is maximal, \(I\neq R\), and so by (b) we have \(V(I) \neq \emptyset\).
- So there is some \(x_0\in[0, 1]\) with \(f(x_0) = 0\) for all \(f\in I\).
- Define \(\mfm_{x_0} \da \ts{f\in R \st
f(x_0) = 0}\), which is an ideal: it is the kernel of the
evaluation map \(\operatorname{ev}_{x_0}: R
\to \RR\), \(f \mapsto f(x_0)\),
which is a ring homomorphism (evaluation at a point preserves sums and
products), and the kernel of a ring homomorphism is an ideal.
- This is a proper ideal, since constant nonzero functions are continuous and thus in \(R\), not not \(\mfm_{x_0}\).
- We thus have \(I \subseteq \mfm_{x_0}\), and by maximality they are equal.
I’m not super convinced by c!