Algebraic properties of zero point of Continuous function

Consider the Algebra of continuous function $f:\mathbb{R}\to \mathbb{R}$$f:\mathbb R\to\mathbb R$

We know that $a(f):=f(a)$$a(f):=f(a)$ is a Algebra homomorphism

And ${\mathcal{I}}_a=\{f\in C|a(f)=0\}$$\mathcal I_a=\{f\in C|a(f)=0\}$ is the kernel of the homomorphism

And actually, it is a maximal ideal because it is a surjection to $\mathbb{R}$$\mathbb R$

Thus $\begin{array}{c}\frac{C}{{\mathcal{I}}_{\mathcal{a}}}\cong \mathbb{R}\end{array}$$\frac{C}{\mathcal{I_a}}\cong \mathbb R\\$

Thus ${\mathcal{I}}_a$$\mathcal I_a$ is a maximal ideal

If there exists no empty open interval $(a,b)$$(a,b)$, and $\mathrm{\forall }x\in (a,b),f(x)=0$$\forall x\in(a,b),f(x)=0$, then $f(x)$$f(x)$ is a zero divisor

And if $f_1,f_2,f_3,...,f_n$$f_1,f_2,f_3,...,f_n$ have no common zero point, then the ideal $(f_1,f_2,f_3,...,f_n)=(1)=C$$(f_1,f_2,f_3,...,f_n)=(1)=C$

$\mathrm{Proof}.$$\mathrm{Proof.}$

Consider $(g)=(f_1^2+f_2^2+f_3^2,...,+f_n^2)\subseteq (f_1,f_2,f_3,...,f_n)$$(g)=(f_1^2+f_2^2+f_3^2,...,+f_n^2)\subseteq(f_1,f_2,f_3,...,f_n)$

Observe that $\mathrm{\forall }x\in \mathbb{R},g(x)>0$$\forall x\in\mathbb R,g(x)>0$, thus $g$$g$ is invertible, there exists $h,hg=gh=1$$h,hg=gh=1$

Thus $1\in (g)\subseteq (f_1,f_2,f_3,...,f_n)$$1\in (g)\subseteq (f_1,f_2,f_3,...,f_n)$

Therefore $(f_1,f_2,f_3,...,f_n)=(1)=C$$(f_1,f_2,f_3,...,f_n)=(1)=C$