$(f,g) = C(X) \Longleftrightarrow Z(f) \cap Z(g) = \emptyset$

Question

Theorem:

a: If $I$is an ideal in $C(X)$ , then the family $Z[I] = \{ Z(f) : f \in I \} $ is a $z$-filter on $X$.

b: if $\mathbf{F}$ is a $z$-filter on $X$, then the family $Z^{-1} [ \mathbf{F} ] = \{ f : Z(f) \in \mathbf{F} \}$ is an ideal in $C(X)$.

Remark:

$(f,g) \neq C(X) $ if only if $Z(f) $meet $Z(g)$, hence if only if $f^{2}+g^{2}$ is not a unit of $C(X)$.

Can you give me more information about the above result? Why can we say" $(f,g) \neq C(X) \Longleftrightarrow Z(f) \cap Z(g) \neq \emptyset$ or

$(f,g) = C(X) \Longleftrightarrow Z(f) \cap Z(g) = \emptyset$

Answer 1

I am assuming $C(X)$ denotes the set of real valued (continuous) functions on $X$.

First of all, notice that the two equivalences are equivalent, because $A$ is equivalent to $B$ if and only if $\neg A$ is equivalent to $\neg B$ .

Then if $(f,g) = C(X)$, there are $h,k$ such that $fh+gk =1$. If $x\in Z(f)\cap Z(g)$, then $0=1$. Thus $Z(f)\cap Z(g)=\emptyset$.

Conversely, if $Z(f)\cap Z(g)=\emptyset$, then $f^2 + g^2$ never vanishes (because we're considering real valued functions), so it's invertible in $C(X)$, and is in $(f,g)$ , so $(f,g) = C(X)$ .