Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I try to solve exercise 2.19 on page 65 in "Algebraic Geometry I" by U.Görtz/T.Wedhorn. The exercise reads:

Let $(X,O_X)$ be a locally ringed space, and $f\in O_X(X)$. Define $X_f:=\{ x\in X; f(x) \neq 0\}$. Show that $X_f$ is an open subset of $X$. What is $X_f$ if $X$ is an affine scheme?

With regard to the first question I don't know what to do. Isn't it necessary to have more information on the topology to prove this?

If we take $X$ as an affine scheme we can assume $X=\text{Spec}(A)$ for a commutative ring $A$. Then $X_f = D(f) = V(f)^c$ is clearly open. Is that right?

share|improve this question

1 Answer 1

up vote 6 down vote accepted

What does the expression $f (x)$ itself mean? I will take it to mean the residue of the germ $f_x$ modulo the maximal ideal of the local ring $\mathscr{O}_{X,x}$.

So suppose $x$ is a point such that $f(x) \ne 0$. That means $f_x$ is not in the maximal ideal of $\mathscr{O}_{X,x}$, and so $f_x$ is invertible. Then, there must exist an open neighbourhood $U$ of $x$ and an element $g$ in $\mathscr{O}_X (U)$, such that $f_x g_x = 1$ in $\mathscr{O}_{X,x}$. Consider the element $f |_U g$. Since $f_x g_x = 1$, there must exist an open neighbourbood $V$ of $x$ such that $V \subseteq U$ and $f |_V g |_V = 1$ (by the construction of directed colimits), i.e. $f |_V$ is invertible in $\mathscr{O}_X (V)$. It is now clear that for all $y$ in $V$, the germ $f_y$ is invertible in $\mathscr{O}_{X,y}$, and so $f (y) \ne 0$ for all $y$ in $V$. Thus $$D(f) = \{ x \in X : f (x) \ne 0 \}$$ is open in $X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.