Is $I:=\{ f \in A : \lim_{x \to \infty} f(x)=0 \}$ a prime ideal ? How does maximal ideals containing $I$ look like?

Question

Let $A$ be the ring of all real valued continuous functions on $\mathbb R$ under pointwise addition and multiplication , I have seen that $I:=\{ f \in A : \lim_{x \to \infty} f(x)=0 \}$ is a proper ideal of $A$ ; is $I$ a prime ideal of $A$ ? Let $M$ be a maximal ideal containing $I$ , does $\exists c \in \mathbb R : f(c)=0 , \forall f \in M $ ?

Answer 1

Do you know that you have non converging to zero functions whose product is equal to $0$ ? Example : $\mathbf{1}_{\mathbf{Q}}$ and $\mathbf{1}_{\mathbf{R}\backslash\mathbf{Q}}$ if you don't require continuity : their product is in $I$, but they are not in $I$. Other example for continuity : take $f$ to be the function equal to $0$ for negative $x$ and $x$ for positive $x$, and $g$ the function $x\mapsto f(-x)$... there product is zero, but they are not in $I$.

Answer 2

First, $I$ is not prime. Take two continuous functions, neither of which tend to zero (for example, because they oscillate) but which have disjoint support; then their product is zero and in particular is in $I$, but neither are in $I$.

Such a $c$ clearly does not exist, as the elements of $I$ themselves do not have this property.