Show that a collection of functions $I$ is an ideal

Question

Let $R$ be the ring of all continuous function on $[0,1]$, and let $I$ be the collection of functions $f(x)$ in $R$ with $f(1/3)=f(1/2)=0$. Prove that $I$ is an ideal of $R$ but not a prime ideal. I know that when you want to show something is an Ideal we have to show that the difference of two elements is in the ideal and by taking an element from the ring and another element in the ideal the product will be in the ideal. But with this problem i am having a hard time showing that. any idea.

Answer 1

It's trival that it's an ideal, and it's not a prime one as $f : x\mapsto (3x-1)(2x-1)$ is in it, but $g : x\mapsto 2x-1$ and $h : x\mapsto 3x-1$ are not in it. So you have a product $f = gh$ in the ideal, with $g$ and $h$ not in it, violating the fact that the ideal could be prime.

Remark. To show that $I$ is an ideal of $A$, with $A$ let's say commuatative, which is the case here, you have to show that $f_1+f_2\in I$ as soon as $f_1,f_2\in I$, and that $f_1 f_2 \in I$ as soon $f_1\in A$ and $f_2 \in I$.

Answer 2

It's simple to show that it's an ideal:

Let $f$ and $g$ be two functions in $I$. Then

$$(f+g)(\frac{1}{3}) = f(\frac{1}{3}) + g(\frac{1}{3}) = 0+0 = 0$$

$$(f+g)(\frac{1}{2}) = f(\frac{1}{2}) + g(\frac{1}{2}) = 0+0 = 0$$

And if you have a function h (in I or not), then

$$(fh)(\frac{1}{3}) = f(\frac{1}{3})h(\frac{1}{3}) = 0\times h(\frac{1}{3}) = 0$$ $$(fh)(\frac{1}{2}) = f(\frac{1}{2})h(\frac{1}{2}) = 0\times h(\frac{1}{2}) = 0$$

Hence $fg$ is in $I$ and I is an ideal

To show it's not prime, see Robert Green answer