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I'm trying to solve Exercise 5.2.F from Vakil's notes:

Show a scheme $X$ is integral if and only if it is irreducible and reduced.

Where we say $X$ is reduced (integral) if $\mathscr O_X(U)$ is reduced (an integral domain) for all open subsets $U$ of $X$.

Clearly, if $X$ is integral, then each $\mathscr{O}_X(U)$ is a domain, hence reduced, so $X$ is reduced. I'm not sure how to see that $X$ is irreducible. It's obvious for an affine scheme, since if $\mathscr{O}_X(X)=:A$ is a domain then $\text{Spec } A$ is irreducible. Am I able to use this to tackle the general case? Like if $X=\cup_i U_i$ with each $U_i$ affine open, does each $U_i$ being irreducible imply that $X$ is irreducible? This doesn't seem like the right way to approach it.

I'm not sure, I just feel stuck. Any hints would be greatly appreciated (I'd prefer that over someone just giving me the answer).

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  • $\begingroup$ Do you want hints for the reverse direction as well? $\endgroup$ Commented Jan 4, 2017 at 3:46
  • $\begingroup$ @EricWofsey Yeah that'd be helpful. I couldn't even find a starting point for that direction.. $\endgroup$ Commented Jan 4, 2017 at 3:51
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    $\begingroup$ If $X = \cup U_i$ with $U_i$ affine open and irreducible, then $X$ is irreducible IF you assume it's connected. in general, it is not $\endgroup$
    – user6246
    Commented Jan 4, 2017 at 3:57
  • $\begingroup$ Be careful here. The affine scheme $\text{Spec }\mathbb{R}[x,y]/V(xy)$ is connected but not irreducible. $\endgroup$ Commented Mar 29, 2018 at 20:52

3 Answers 3

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Hint for the forward direction: If $X$ is not irreducible, that means there are two disjoint nonempty open subsets $U,V\subset X$. Consider $\mathscr{O}_X(U\cup V)$.

Hint for the reverse direction: If $X$ is irreducible and reduced, so is any open subscheme of $X$, so every nonempty affine open subscheme of $X$ is Spec of a domain. To show $\mathscr{O}_X(U)$ is a domain for arbitrary $U$, you can show that the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective for any nonempty affine open subset $V\subseteq U$.

A stronger hint for the reverse direction is hidden below:

To show the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective, suppose $f$ is in its kernel. Then $f$ vanishes on $V$. To show $f$ vanishes on $U$, it suffices to show $f$ vanishes on $W$ for every other affine open $W\subseteq U$. Now use the fact that $W$ is Spec of a domain and $f$ vanishes on $V\cap W$.

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  • $\begingroup$ Why does showing that the restriction is injective imply that $\mathcal{O}_X(U)$ is a domain? $\endgroup$
    – user525033
    Commented Mar 8, 2019 at 16:10
  • $\begingroup$ A subring of a domain is a domain. $\endgroup$ Commented Mar 8, 2019 at 16:16
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    $\begingroup$ Note that by this point in Vakil's notes we have not shown that $\operatorname{Spec}{A}$ irreducible and reduced implies that $A$ is a domain (only the reverse implication), so some argumentation will be required to fill in this step $\endgroup$ Commented Jan 17, 2020 at 10:26
  • $\begingroup$ @EricWofsey Can you give a further hint to considering $\mathcal O_X(U \cup V)$. I'm considering this but don't see where to go next. $\endgroup$
    – user5826
    Commented Mar 21, 2021 at 21:01
  • $\begingroup$ @user46372819: What does the sheaf condition tell you about $\mathcal{O}_X(U\cup V)$? $\endgroup$ Commented Mar 21, 2021 at 21:48
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If there were any disjoint open subsets $U,V$ then by the sheaf axioms $$\mathcal{O}_X(U \cup V) \cong \mathcal{O}_X(U) \times \mathcal{O}_X(V)$$ via restrictions $s \mapsto (s|_U, s|_V)$. This product has zero divisors of the form $(0,t)$ and $(t,0)$ unless one of $U$ and $V$ is empty. Contradiction since $\mathcal{O}_X(U \cup V)$ is integral.

This appears as Proposition 3.27 in Görtz, Wedhorn's Algebraic Geometry (which is a great reference for the basics of schemes).

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  • $\begingroup$ Could you teach me what is the axiom of sheaf you used here? $\endgroup$ Commented May 28, 2020 at 12:24
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In Vakil's note there is a possible hint for the reverse direction by using exercise 4.3.G

exercise 4.3.G:
(a) If f is a function on a locally ringed space X, show that the subset of X where f vanishes is closed.
(b) Show that if f is a function on a locally ringed space that vanishes nowhere, then f is invertible.

So if fg=0, you can find two closed set covering the scheme by (a). To prove the closed set is neither empty or X, you can use (b) and the fact that X is reduced.

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