No, I have never heard of that theorem.
My confusion comes from this phrase: "...which determine the zero function...". Determine in what sense, and the zero function on which domain? These are two unknowns which can change the question completely, I cannot keep guessing what they mean. If you know what they mean, please elaborate. Otherwise, you should clarify with your instructor/book.
I only have two more problems in this chapter then I can move on to the next chapter. I hope that the next chapter is better than this one. Well I have to go to work tomorrow!! Thanks for trying to help if you figure it out that would be great
I just took a look at your book, and it seemed we can just substitute elements of $\displaystyle F$ whenever we want. In this case, our job is quite easy actually. To see that
$\displaystyle \mathcal{S}=\{p(x)\in F[x]|p(a)=0 \quad \forall a\in F\}$
is an ideal, we want to show that it is closed under addition by elements in $\displaystyle \mathcal{S}$ and closed under multiplication by elements of $\displaystyle F[x]$. For the first bit, suppose $\displaystyle p(x),q(x)\in \mathcal{S}$. Then $\displaystyle (p+q)(a)=p(a)+q(a)=0+0=0$ for all $\displaystyle a\in F$. Thus $\displaystyle (p+q)(x)\in \mathcal{S}$. For the second assertion, suppose $\displaystyle p(x)\in \mathcal{S},r(x)\in F[x]$. Then $\displaystyle r(a)p(a)=r(a)\cdot 0 = 0$ for all $\displaystyle a\in F$. Therefore $\displaystyle r(x)p(x)\in \mathcal{S}$.