Hilbert Nullstellensatz and ring of continuous functions

Question

Is there any relation between Hilbert's Nullstellensatz and the fact that the maximal ideals in $\mathcal C([0,1])$ correspond to a point in $[0,1]$ (which can be generalized to compact hausdorff spaces)?

Answer 1

I assume that by Hilbert's Nullstellensatz you mean here that there is a bijection between the maximal ideals of $k[T_1,\dotsc,T_d]$ and the points in $\mathbb{A}^n(k)$, where $k$ is an algebraically closed field. Well I don't think that we can prove one of these theorems from the other one, but we can put them into the same framework as follows.

Let $X$ be a locally ringed space. Then there is a canonical morphism $i_X : X \to \mathrm{Spec}(\Gamma(X,\mathcal{O}_X))$, see EGA I, §1.6. It maps a point $x \in X$ to the prime ideal $\{f \in \Gamma(X,\mathcal{O}_X) : f_x \in \mathfrak{m}_x\}$.

If $X$ is an affine scheme, then $i_X$ is an isomorphism. If $X$ is an affine variety in the classical sense (only closed points and everything is defined over an algebraically closed field), then $i_X$ restricts to a homeomorphism $X \cong \mathrm{Spm}(\Gamma(X,\mathcal{O}_X))$ (the maximal ideals); this is Hilbert's Nullstellensatz.

If $X$ is a topological space equipped with its sheaf of continuous functions to $\mathbb{R}$, then $i_X(x)=\{f : X \to \mathbb{R} : f(x)=0\}$ is a maximal ideal of $X$ (the quotient is $\mathbb{R}$). If $X$ is compact Hausdorff, then $i_X$ restricts to a homeomorphism $X \cong \mathrm{Spm}(C(X,\mathbb{R}))$. I doubt that we can reduce this non-trivial result to something more algebraic. For example, injectivity uses Urysohn's Lemma, and for surjectivity we need a special property of $\mathbb{R}$, namely that $u_1^2 + \dotsc + u_n^2=0$ has only the trivial solution $u_1=\dotsc=u_n=0$. Let me recall the proof: If $\mathfrak{m}$ is a maximal ideal and $i_X(x) \neq \mathfrak{m}$ for all $x$, we find $f_x \in \mathfrak{m}$ with $f_x(x) \neq 0$. Since $X$ is compact, there are finitely many $x$ such that $X$ is covered by the open subsets $\{f_x \neq 0\}$. But then $\sum_x f_x^2 \in \mathfrak{m}$ vanishes nowhere, i.e. it is a unit, contradiction.