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.

Question

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?

Answer 1

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 $\mathfrak{m} \in V(f)$.

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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.)