Let $R$ be a sheaf of rings on a topological space $X$. Assume $R \neq 0$. Does then $R$ have a maximal ideal? So this is a spacified analogon of the theorem, that every nontrivial ring has a maximal ideal. Currently I try to develope this sort of spacified commutative algebra and algebraic geometry. If anyone knows some literature about it, please let me know.


So let's try to imitate the known proof for rings and use Zorn's Lemma. For that, we need that for every linear ordered set $(J\_k)\_{k \in K}$ of proper ideals in $R$, their sum $\sum\_{k \in K} J\_k$ is also a proper ideal. Note that if we replace $R \neq 0$ by $R\_x \neq 0$ for all $x \in X$ and the notion proper by "stalkwise proper", then everything works out fine since stalks and sum commute. However, global sections do not commute with (infinite) sums. Anyway, let's try to continue:


Assume $\sum\_{k \in K} J\_k = R$, that is, $1$ is a global section of the sum. Then there is an open covering $X = \cup\_{i \in I} U\_i$, such that $1 \in \sum\_{k \in K} J\_k(U\_i) = \cup\_{k \in K} J\_k(U\_i)$. Thus we get a function $I \to K, i \mapsto k\_i$, such that $1 \in J\_{k\_i}(U\_i)$. If this function has an upper bound, say $k$, then we get a contradiction $J\_k=R$. Thus the function is unbounded. And now?


I think that this already indicates that there will be counterexamples, but I'm not sure. Also note that everything is fine when $X$ is quasi compact.



Does every nontrivial sheaf of rings have a maximal ideal?
Take $X=\mathbf Z$ with topology $(k,\infty)\cap\mathbf Z$ for $-\infty\leq k\leq\infty$, so that sheaves on $X$ may be identified with sequences $\dots\to F\_k\to F\_{k+1}\to\dots$, $k\in\mathbf Z$. Now take for $R$ the constant sheaf with value $\mathbf Q$. All ideals have the form $\dots\to0\_{k-1}\to0\_k\to\mathbf Q\_{k+1}\to\mathbf Q\_{k+2}\to\dots$ for some $-\infty\leq k\leq\infty$, so there are no maximal ideals.


