The Stacks project

Proof. The question is local on $X$, hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring. Then $U$ is quasi-compact (Properties, Lemma 28.5.3) hence $U = D(f_1) \cup \ldots \cup D(f_ n)$ (Algebra, Lemma 10.29.1). In this situation $U$ is scheme theoretically dense in $X$ if and only if $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective, see Morphisms, Example 29.7.4. Condition (3) translated into algebra means that for every associated prime $\mathfrak p$ of $A$ there exists an $i$ with $f_ i \not\in \mathfrak p$.

Assume (1), i.e., $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective. Say $a \in A$ has annihilator a prime $\mathfrak p$. Since $a$ maps to a nonzero element of $A_{f_ i}$ for some $i$ we see that $f_ i \not\in \mathfrak p$. Thus (3) holds.

Note that $U$ is dense in $\mathop{\mathrm{Spec}}(A)$ if and only if it contains all minimal prime ideals. Since the set of associated primes of $A$ is equal to the union of the set of minimal primes and the set of embedded associated primes (see Algebra, Proposition 10.63.6), we see that (2) and (3) are equivalent.

Assume (3). This means that every associated prime $\mathfrak p$ of $A$ corresponds to a prime of $A_{f_ i}$ for some $i$. Then $A \to A_{f_1} \times \ldots \times A_{f_ n}$ is injective because $A \to \prod _{\mathfrak p \in \text{Ass}(A)} A_\mathfrak p$ is injective by Algebra, Lemma 10.63.19. In this way we see that (3) implies (1). $\square$