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Let $X=\operatorname{Spec}A$ be an affine scheme. In the book of Hartshone, he claimed that $X$ is integral if and only if $A$ is an integral domain. If $X$ is integral then we can deduce easily that $A$ is an integral domain.

How can I show that if $A$ is an integral domain then $X$ is integral ? Please give me some hint. Thanks.

Bowei Tang
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Arsenaler
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    This is just a matter of applying all of the relevant definitions. Where do you get stuck? – Qiaochu Yuan Jun 17 '13 at 08:14
  • @Quiaochu Yuan: The definition of Hartshone for an integral scheme is $O_{X}(U)$ is an integral domain for any open subset $U$ of $X$. So how can I apply it? – Arsenaler Jun 17 '13 at 08:19
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    What is $O_X(Spec A)$? (and what does the definition imply about that) – John C Jun 17 '13 at 08:25
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    @JohnC: Did you read my question ? If $X$ is integral then $O_{X}(X)=O_{X}(SpecA)=A$, therefore $A$ is an integral domain. What I can not prove is that if we have $A$ is an integral domain, then how can we prove that $X$ is integral. – Arsenaler Jun 17 '13 at 08:30
  • @QiaochuYuan: Thanks, I always think about some others tricks and forgets the definitions :| – Arsenaler Jun 17 '13 at 08:50
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    With Hartshorne, Proposition II.3.1 (which says that 'integral' $\Leftrightarrow$ 'irreducible'+'reduced') this is relatively straightforward. – Nils Matthes Jun 17 '13 at 08:53
  • It is too strong Nils Matthes. On the other hand, Hartshorne did mentioned the equivalent condition for integral property before that proposition. – Arsenaler Jun 17 '13 at 08:56
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    @Arsenaler: yes, from the strictly logical point of view, this is possibly too strong. However, I believe that Proposition II.3.1 gives the "correct" way of thinking about integrality of schemes. Why do I think so? Because integrality is neither a global notion (cannot be checked on global sections) nor a local notion (cannot be checked on stalks). But said proposition allows you to split integrality up into a global notion (irreducibility), and a local notion (reducedness). This yields a neat way of thinking about integrality, which may make life easier. – Nils Matthes Jun 17 '13 at 09:46
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    For example, the very desirable statement that a scheme $X$ is integral, if and only if all local rings are integral domains is not true. But if all local rings are integral domains (or even only reduced local rings), then the only possible obstacle to integrality of $X$ is the purely topological notion of irreducibility of the underlying topological space of $X$. – Nils Matthes Jun 17 '13 at 09:48
  • You can also think of it as "local rings are domains" + connected, I think. – TTS Jun 17 '13 at 17:17
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    @TTS: This is not true. A scheme $X$ is integral if and only if all of the following conditions hold: 1) $X$ is nonempty; 2) $X$ is connected; 3) the stalks of the structure sheaf of $X$ are integral; 4) the set of irreducible components of $X$ is locally finite. None of these conditions can be omitted. – Fred Rohrer Jul 03 '13 at 06:19
  • @Fred Ah, right, I forgot a Noetherian hypothesis. Whoops! – TTS Jul 03 '13 at 09:24

3 Answers3

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Suppose $A$ is integral and let $U$ be a non-empty open subset of $\operatorname{Spec} A$. Let $f, g \in \mathcal{O}_{\operatorname{Spec} A}(U)$. By definition of $\mathcal{O}_{\operatorname{Spec} A}(U)$ we have an open covering $U = \bigcup V_i$, such that $f|_{V_i} = f_i/s_i$ and $g|_{V_i} = g_i/t_i$. Now, $f|_{V_i}=0$ is equivalent to $f(Q) = 0$ for some $Q \in V_i$: if $f(Q) = 0$ for some $Q \in V_i$, then there exists $s \not\in Q$ such that $s f_i = 0$, and so $f_i=0 \Rightarrow f|_{V_i}=0$. Similarly for $g$.

Next, suppose that $f, g$ are non-zero but $fg=0$. Then there exist $P \in V_i$ and $Q \in V_j$, for some $i,j$, such that $f(P) \neq 0$ and $g(Q) \neq 0$. Suppose that $V_i \cap V_j \neq \emptyset$. Then for $P' \in V_i \cap V_j$ we must have $f(P')g(P') = 0$. Hence there exists some $s' \not\in P'$ such that $s' f_i g_j = 0$, which implies either $f|_{V_i \cap V_j} = 0$ or $g|_{V_i \cap V_j} = 0$. But this is a contradiction in view of the remark of the previous paragraph.

Hence it must be the case that $V_i \cap V_j = \emptyset $ and so $\operatorname{Spec} A = V_i^c \cup V_j^c, \, \, \, (*)$. Since $A$ is an integral domain, $\operatorname{Spec} A$ is irreducible, which contradicts $(*)$, since both $V_i^c, V_j^c$ are proper subsets. Hence either $f=0$ or $g=0$.

Manos
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Let's recall the definition of integral scheme: A scheme $X$ over a commutative ring $R$ is called integral if for every nonempty affine open $U \subseteq X$, the $R$-algebra $C^{\operatorname{alg}}_R(U)$ is an integral domain.

Now let $A$ be an $R$-algebra which is an integral domain and $P_A$ the corresponding affine scheme. Let $U_I \subseteq P_A$ be some nonempty affine open, where $I \subseteq A$ is an ideal with $I \neq 0$. Since $U_I$ is affine, there is some $R$-algebra $B$ and an isomorphism $U_I \cong P_B$ under which the inclusion $U_I \subseteq P_A$ corresponds to a map of $R$-algebras $\varphi: A \to B$. Our goal is to show that $C^{\operatorname{alg}}_R(U_I) = C^{\operatorname{alg}}_R(P_B) = B$ is an integral domain.

Now we have, for any $R$-algebra $C$: $$ \begin{array}{cl} & \{\tilde{\psi}: B \to C\} \\ \cong & P_B(C) \\ \cong & U_I(C) \\ \cong & \{\psi: A \to C \mid \langle \psi(I) \rangle = \langle 1_C \rangle\} \\ \end{array} $$ where the total map is given by composition with $\varphi$.

In particular, this gives us the following:

  • $\langle \varphi(I) \rangle = \langle 1_B \rangle$

  • For any $R$-algebra $C$ and any map $\psi: A \to C$ with $\langle \psi(I) \rangle = \langle 1_C \rangle$, there exists a unique lift $\tilde{\psi}: B \to C$.

This is exactly the universal property considered in If $A \to B$ is universal with respect to sending $I$ to the unit ideal and $A$ is an integral domain, does $B$ also have to be an integral domain? . By that question, $B$ must be an integral domain, finishing the proof.

Elia Immanuel Auer
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I'm digging up this post a little late with a different answer :)

Since $A$ is integral, $(0)$ is a prime ideal and $\mathrm{Spec A} = \overline{\{(0)\}}$ is irreducible. Moreover, the localizations of an integral domain remain integral domains and it follows that $\mathrm{Spec A}$ is reduced.

Now in a reduced scheme $X$, for any non-empty open subset $U$, $f \in \mathcal O_X(U)$, $f = 0$ in the ring $\mathcal O_X(U)$ iff $f(x) = 0$ for all $x \in U$ (in the non reduced cased, this would only make $f$ a nilpotent element on affine opens). In the affine case where $U = \mathrm{Spec} R$, this means that $f =0$ if and only if $f \in \mathfrak{p}$ for all $\mathfrak{p} \in U$ i.e $f(\mathfrak{p}) = 0$ for all primes $\mathfrak p$ of $R$.

Take some non-empty open subset $U$ of $\mathrm{Spec A}$, and suppose that $f,g \in \mathcal O_X(U)$ are such that $fg =0$. Then by the above, we have that $f(\mathfrak{p}) = 0$ or $g(\mathfrak{p})=0$ for any $\mathfrak{p} \in U$.

It is a general fact about locally ringed spaces that $V_f = \{x \in U \mid f(x) = 0\} = \{\mathfrak{p} \in U \mid f(\mathfrak p) = 0 \}$ is a closed subset of $U$, and the same goes for $V_g$. We have decomposed the open subset $U$ as : $$U=V_f \cup V_g$$ Since $U$ is a non-empty open in the irreducible space $\mathrm{Spec A}$, it is itself irreducible this means that either $V_f = U$ or $V_g = U$. Say $V_f = U$, then $f (\mathfrak{p}) = 0 $ for all $\mathfrak{p} \in U$ which, by reducedness once again implies that $f = 0$ in $\mathcal O_X(U)$ and we are done.

supermartruc
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  • What is your definition of reduced scheme? – Elia Immanuel Auer Nov 30 '25 at 10:07
  • The usual one : $X$ is reduced if the stalks $\mathcal{O}_{X,x}$ are reduced rings for all points $x \in X$. Since reducedness is a local property for rings, it is the same as asking that $\mathcal{O}_X(U)$ be a reduced ring for any open subset $U$ of $X$. – supermartruc Nov 30 '25 at 10:24
  • Can you explain why reducedness of $U$ together with the fact that $\forall \mathfrak{p} \in U : f \in \mathfrak{p}$ implies that $f = 0$ in $\mathcal{O}_X(U)$? – Elia Immanuel Auer Nov 30 '25 at 10:37
  • This is exercice 5.2.A in Vakil's notes. Take $U$ to be some open subset of $X$. It can be covered by affine opens $U_i$. If $f \in \mathcal{O}_X(U)$, then $f= 0$ in $\mathcal{O}_X(U)$ iff all of its restrictions $f_i$ are zero in $\mathcal{O}_X(U_i)$ (by the sheaf property). Hence we can assume WLOG that $U = \mathrm{Spec R}$ is affine (and $R$ is reduced by assumption). Take some $f \in \mathcal{O}_X(U) = R$ and suppose that $f \in \mathfrak p$ for all primes in $R$. Then $f$ belongs to the nilradical of $R$, which is $(0)$ by reducedness of $R$, hence $f=0$. – supermartruc Nov 30 '25 at 10:54
  • Thanks for the explanation! – Elia Immanuel Auer Nov 30 '25 at 10:59
  • Another question: If $f \in \mathcal{O}_X(U)$, what do you mean by "$f \in \mathfrak{p}$"? $\mathfrak{p}$ is an ideal of $A$ and not of $\mathcal{O}_X(U)$, right? So how does the expression "$f \in \mathfrak{p}$" make sense? – Elia Immanuel Auer Nov 30 '25 at 10:59
  • I guess this is some abuse of notations, when $U = \mathrm{Spec} R$ then $f \in R$ and $f \in \mathfrak p$ makes sense. For a non-affine open $U$ of $\mathrm{Spec R}$ what I meant was "the evaluation of $f \in \mathcal{O}X(U)$ at $\mathfrak{p} \in U$ is $0$ i.e $f \in \mathfrak{m}{X,\mathfrak{p}}$ or $f(\mathfrak{p})= 0$ in the residue field at $\mathfrak{p}$. – supermartruc Nov 30 '25 at 11:05