Let the set $S=\{f \in C[a,b] : f(x)=0 \forall x \in [c,d]\}$ in $C[a,b]$

Question

Let $a,b,c,d$ be a real number such that $a<c<d<b$ .consider the ring $C[a,b]$ with pointwise addition and multiplication Let the set

$$S=\{f \in C[a,b] : f(x)=0 \forall x \in [c,d]\}$$ then we have to show that does $S$ is ideal but not prime ideal.

solution i tried-To check the given set is ideal or not i have to check that

$1$. for every $f,g \in S$, $f-g$ also in

$2$. for $f \in S$ and $g\in C[a,b]$ $fg \in S$

and for prime ideal there are two ways either quotient is an ideal of

if $fg \in S$ the either $f \in S $ or $g \in S$

I only know the defination but can't get how to apply them

Please help

Answer 1

Well: check that $f-g$ fulfills the condition of being in $S$: if $x \in [c,d]$ we know that $f(x)=0$ and $g(x)=0$ as $f,g \in S$. So also $(f-g)(x)=f(x)-g(x)=0-0=0$ and as this is true for all $x \in [c,d]$, $f-g \in S$ by definition.

The proof for 2. is similar, using that $a \cdot 0= 0$ for any $a$.

Answer 2

$S$ is not a prime ideal. Let $c<t<d$ and define $f(x)=0$ for $x \leq t$, $x-t$ for $x \geq t$. Let $g(x)=0$ for $x \geq t$, $x-t$ for $x \leq t$. Then $fg \in S$ but neither $f$ nor $g$ is in $S$.