D+F 7.4.9

Let $R$ be the ring of continuous functions on $[0,1]$ and let $I$ be the collection of functions $f(x)\in{R}$ s.t. $$f(\frac{1}{3})=f(\frac{1}{2})=0$$ Prove that $I$ is an ideal of $R$ but not a prime ideal.

$\textit{Proof}$ : To show that $I$ is an ideal, we need to show that $I \neq \emptyset$, $I$ is closed under subtraction and $I$ is closed under left and right multiplication by arbitrary elements of $R$.
$0$, the zero function is in $I$, so $I \neq \emptyset$
Let $f(x),g(x) \in{I}$. Then $f(\frac{1}{3})=f(\frac{1}{2})=g(\frac{1}{3})=g(\frac{1}{2})=0$. $$(f-g)(\frac{1}{3})=f(\frac{1}{3})-g(\frac{1}{3})=0-0=0$$ and $$(f-g)(\frac{1}{2})=f(\frac{1}{2})-g(\frac{1}{2})=0-0=0$$ Or also $$f(\frac{1}{2})-g(\frac{1}{3})=0-0=0$$ $$f(\frac{1}{3})-g(\frac{1}{2})=0-0=0$$ So, $I$ is closed under subtraction.
Now let $f(x)\in{I}$ and $h(x)\in{R}$. $$(fh)(\frac{1}{2})=f(\frac{1}{2}) \cdot h(\frac{1}{2})=0 \cdot a = 0$$ Similarly, $(hf)(\frac{1}{2})=0=(fh)(\frac{1}{3})=(hf)(\frac{1}{3})=0$.
So, $I$ is an ideal of $R$.

Since the ring of continuous functions is a commutative ring, we may have the notion of a prime ideal. An ideal $P$ is prime in the commutative ring $R$ if $P \neq R$ and $$fg \in{P} \Longrightarrow f\in{P} \quad or \quad g\in{P}$$ We would like to find $f,g \in{I}$ s.t. $fg\in{I}$ but $f \not \in{I}$ and $g \not \in {I}$.
Consider the linear functions $f(x)=x-\frac{1}{3}$ and $g(x)=x-\frac{1}{2}$.
$$f(\frac{1}{2}) \neq 0 \Longrightarrow f\not \in {I}$$ $$g(\frac{1}{3}) \neq 0 \Longrightarrow g \not \in{I}$$ However, $$(fg)(x)=(x-\frac{1}{3})(x-\frac{1}{2})$$ and so $$(fg)(\frac{1}{3})=(fg)(\frac{1}{2})=0 \Longrightarrow fg\in{I}$$ which amounts to a counterexample to $I$ being prime. So, $I$ is an ideal of $R$ but not a prime ideal. $\blacksquare$