Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be the ring of continuous functions from $[0,1]$ to the real numbers. Fix $c \in [0,1]$ and let $M_c$ = ker $E_c$ where $E_c$ denotes evaluation at $c$, a ring homomorphism from $R$ to the real numbers. That is, $E_c(f) = f(c)$ for $f \in R$. What is a nice clean way to show $M_c$ is not finitely generated? I figure that one way is to suppose that \begin{equation} A = \{ f_1, f_2,\dots, f_n \} \end{equation} is a minimal generating set and try to come up with $f_{n+1}$ which cannot possibly be an R-linear combination of the $f_k$, but that feels like a lot of guesswork. (And I didn't succeed.) Thanks for the help.

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Your idea does work, and unfortunately there does not seem to be a way around the guesswork. Suppose $\{ f_1, \ldots, f_n \}$ is a set of continuous functions that generate the ideal of functions vanishing at $c$. Then there is a continuous function $$f = \left| f_1 \right| + \cdots + \left| f_n \right|$$ and it is clear that $f (x) = 0$ if and only if $x = c$. (Indeed, if $f (x) = 0$, then non-negativity means $f_1 (x) = \cdots = f_n (x) = 0$, so if $x \ne c$ that is tantamount to saying that any continuous function that vanishes at $c$ also vanishes at $x$, which is absurd for the closed interval $[0, 1]$.) Let $g = \sqrt{f}$. This is also a continuous function vanishing at $c$, so $$g = h_1 f_1 + \cdots + h_n f_n$$ for some continuous functions $h_1, \ldots, h_n$. Let $$h = \left| h_1 \right| + \cdots + \left| h_n \right|$$ and observe that $g (x) \le h (x) f (x)$ for all $x$. But $f (x) = g (x)^2$, so that means $h (x) \ge 1 / g(x)$ for all $x \ne c$, and since $g (c) = 0$, $h$ cannot be continuous at $c$, contradicting the fact that $h_1, \ldots, h_n$ are continuous.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.