Prove the zero set of a proper ideal $I$ of the ring of continuous complex-valued function on a compact space $X$ is nonempty.
The above problem is from Lang' s Real and Functional analysis Chapter III, exercise 3. I know how to prove it when the ring of functions is real-valued:
Assume for the sake of contradiction that the zero set is empty, this means that given $x\in X, \exists f_x\in I$ such that $f_x(x)\neq 0$. Since $f_x$ is continuous, there exists open sets $O_x$ such that $f_x(O_x)$ is never equal to zero. By compactness of $X$, $\exists x_{1},...,x_n$ such that $O_{x_1},..., O_{x_n}$ cover $X$. Let $f=f_{x_1}^2+...+f_{x_n}^2$, then $f\in I$ and $f(x)>0$ for every $x\in X$. Thus $1=f(x)\times \frac{1}{f(x)}\in I$, implying $I=C(X)$, contradiction.
However, the above 'sum of squares' trick doesn't seem to apply to the complex case, since a sum of squares of complex numbers can be zero even if each summand is nonzero, such as $0=1^2+i^2$. How to proceed?
Thanks for @user10354138, I think I've got it. My proof is slightly different from his hint, but I think it yields more information.
Let $Re(I)$ be the set of all real parts of functions in $I$. Then it is easily seen that $Re(I)$ is an ideal of continuous real valued functions. Assume for the sake of contradiction that the zero set of $I$ is empty, then the zero set of $Re(I)$ must also be empty. Say if there is an $a\in X$ such that $Re [f(a)]$ is equal to zero for all $f\in I$, then $f(a)=c_f i$ for some real number $c_f$ depends on $f$. Then $if(x) \in I$ for all $f\in I $ and we have $if(a)=-c_f$. For $-c_f$ to be real and imaginary simultaneously, it must be the case that $c_f=0$ for all $f\in I$, which means $a$ is in the zero set of $I$, contrary to our assumption. That means $Re(I)$ is equal to the ring of all continuous real-valued functiosn on $X$ by our proof for the real case. The same argument applies to $Im(I)$. So $I=C(X)$, contradiction.