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 $A$ be the banach algebra of continuously differentiable complex functions on $[0,1]$ with pointwise multiplication normed by

$ ||f||=||f||_{\infty}+||f'||_{\infty}. $

I have to show that the ideal

$ J= \{f\in A : f(p)=f'(p)=0\} $

is closed.

Is the following proof correct?

Let $f_n$ be a sequence in $J$ such that $\text{lim}_{n \rightarrow \infty} f_n = g \in A$. Then $f_n(p)=f'_n(p)=0$ for all $n$ from which it follows that $\text{lim}_{n \rightarrow \infty} f_n(p)=0$ and $\text{lim}_{n \rightarrow \infty} f_n'(p)=0$ and thus $g\in J$. So J is closed since it contains its limit points.

Then for the second part I have to show that $A/J$ is a two-dimensional algebra which has a one-dimensional radical.

Here is my attempt.

Suppose $f\in A$ then $f$ is equal to a function $h,i,j,k\in A$ such that either

$$h(p)=h'(p)=0, \, i(p)\neq 0,i'(p)=0, \, j(p)= 0,j'(p)\neq 0 \text{ or } k(p)\neq 0,k'(p)\neq0.$$

Then $i+J\neq j+J$, but an element of $h+J$ or $k+J$ is in $i+j+J$ so $A/J$ contains two cosets and is therefore two-dimensional.

Since the kernel of any complex homomorphism on $A/J$ contains $J$ the radical is equal to $J$, but why would this be one-dimensional?

share|improve this question
    
Yes, this is correct, assuming you have already proved that it is an ideal. –  Prahlad Vaidyanathan Dec 3 '13 at 14:51
    
Thank Prahlad. Yes I did already proved that is is an ideal. I also added a second part. –  simon Dec 3 '13 at 15:54
    
Perhaps you can recover $J$ as the kernel of the map $$A \mapsto \mathbb{C}\oplus \mathbb{C} ; \qquad f \mapsto (f(p), f'(p)) $$ or something to that effect. –  Prahlad Vaidyanathan Dec 3 '13 at 16:03
    
But then I have just one map. Why should this imply that the intersection of all maximal ideals are $J$? –  simon Dec 3 '13 at 16:07
    
I misinterpreted. You mean to prove that $A/J$ is two-dimensional. So my reasoning is wrong? About the two cosets. –  simon Dec 3 '13 at 17:14

1 Answer 1

What Prahlad means to say is that you can recover J as the kernel of the (continuous!) ring homomorphism $A \to C[x]/x^2$ given by $f \mapsto f(p)+f'(p)x$. (The map Prahlad wrote down is not a ring homomorphism when $C \oplus C$ is given the product ring structure.)

share|improve this answer

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.