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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?

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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

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.)

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