# Let $A$ be a commutative ring and $f \in A$. If $X_f=\operatorname{Spec}A \setminus V(f)$ show that $X_f = \operatorname{Spec}A$ iff $f$ is a unit.

Let $$A$$ be a commutative ring and $$f \in A$$. If $$X_f=\operatorname{Spec}A \setminus V(f)$$ show that $$X_f = \operatorname{Spec}A$$ if and only if $$f$$ is a unit.

Suppose that $$X_f = \operatorname{Spec}A$$, then taking complements $$V(f) = \emptyset \iff \forall \mathfrak p \subset A : f \notin \mathfrak p$$ i.e $$f$$ is not in the nilradical $$\mathfrak N = \bigcap_{\mathfrak p \in \operatorname{Spec}A} \mathfrak p$$.

How can I conclude that $$f$$ must be a unit? In Atiyah & MacDonald they have a Corollary

Every non-unit of $$A$$ is contained in a maximal ideal

but I cannot see if this can be used here. I know that every maximal ideal is prime, but the converse doesn't neccessarily hold.

I considered the arugment that if $$f$$ isn't in any prime, then it is not in any maximal ideal which by the corollary would imply that $$f$$ is a unit, but this would require the converse of "every maximal ideal is prime" to hold?

• When is $V(f)$ the empty set? Apr 29, 2022 at 8:02
• When $f$ isn't contained in any prime ideal of $A$? Apr 29, 2022 at 8:02

It does not require the converse of "maximal $$\Rightarrow$$ prime" to hold.

If $$f$$ is a unit, then $$f$$ cannot be contained in any proper ideal. In particular, $$f$$ is not contained in any prime ideal. Thus, $$V(f)$$ is empty.

Conversely, if $$f$$ is not a unit, then $$f$$ is contained in some maximal ideal $$\frak m$$, which is also a prime ideal. Thus, $$V(f)$$ is not empty as $$\mathfrak{m} \in V(f)$$.

Remark. You have written

$$\forall \mathfrak p \subset A : f \notin \mathfrak p$$ i.e $$f$$ is not in the nilradical $$\mathfrak N = \bigcap_{\mathfrak p \in \operatorname{Spec}A} \mathfrak p$$.

Note that the statements before and after the "i.e" are not equivalent. It is possible that an element is not nilpotent but is still contained in some prime ideal. (Example: $$2$$ in $$\Bbb Z$$.)

Rather, $$V(f) = \varnothing$$ is equivalent to $$f \notin \bigcup_{\mathfrak{p} \in \operatorname{Spec} A} \mathfrak{p}$$. (Union instead of intersection.)