If $X$ is a scheme, we define the sheaf $(\mathcal{O}_X)_{red}$ as the sheafification of the pre-sheaf defined by $U\mapsto \mathcal{O}_X(U)/\sqrt{0}$, where $\sqrt{0}$ is the nilradical of the ring $\mathcal{O}_X(U)$.

I'm trying to show that $(Y,\mathcal{O}_Y):=(X,(\mathcal{O}_X)_{red})$ is a reduced scheme.

I was able to prove that $\mathcal{O}_{Y,x}\simeq \mathcal{O}_{X,x}/\mathfrak{N}_x$, where $\mathfrak{N}_x$ is the nil radical of $\mathcal{O}_{X,x}$, so that $\mathcal{O}_{Y,x}$ is a local reduced ring.

But I'm having difficulty making the scheme structure of $Y$ explicit.

If $X=\bigcup_iU_i$ is a cover with $(U_i,\mathcal{O}_X\big|_{U_i})$ affine, I guess the natural thing is to show $\mathcal{O}_Y\big|_{U_i}\simeq \mathcal{O}_{\text{Spec}(\mathcal{O}_Y(U_i))}$.

But I don't know how to do that. The sheaf $\mathcal{O}_Y$ seems so abstract, I don't know how to handle it.

What's the best way to do this?

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    $\begingroup$ Probably it is obvious, but the affine case is extremely explicit: if $X = \operatorname{Spec}(A)$, then $X_{\text{red}}$ is the subscheme $\operatorname{Spec}(A/\sqrt{0})$. $\endgroup$ Apr 24, 2020 at 1:27
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    $\begingroup$ Hint: verify that the map $(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$ is a closed immersion. Now you can work locally with the concrete description provided by juan diego rojas above. $\endgroup$
    – KReiser
    Apr 24, 2020 at 1:31
  • $\begingroup$ @KReiser, since $Y\to X$ is the identity and I've already shown that $\mathcal{O}_{X,x}\to\mathcal{O}_{Y,x}$ is surjective for all $x$, this already gives us that $(Y,\mathcal{O}_Y)\to(X,\mathcal{O}_X)$ is a closed immersion, right? I can't see why this is helpful $\endgroup$
    – rmdmc89
    Apr 24, 2020 at 17:49
  • $\begingroup$ @rmdmc89 For $U_i$ affine, yes. $\endgroup$ Apr 24, 2020 at 17:54
  • $\begingroup$ This exactly shows the claim about $\mathcal{O}_Y|_{U_i}\cong \mathcal{O}_{\operatorname{Spec} \mathcal{O}_Y(U_i)}$ for $U_i$ affine open you were trying to show at the end of the post. $\endgroup$
    – KReiser
    Apr 24, 2020 at 18:01

1 Answer 1


You're well and truly stuck in the comments, so let me try to write you an answer which resolves these difficulties. I'll be following my hint that we should try to understand $Y\to X$ as a closed immersion in the first bit, with an explanation of an alternate approach a little later.

We'll try to address your concern that $Y$ is only a locally ringed space (remember that the morphisms of schemes are just the morphisms of the underlying locally ringed spaces, so $Y$ being a scheme is the only obstruction to $Y\to X$ being a morphism of schemes as long as you believe your map is a morphism of locally ringed spaces).

First, we recall the definition of a closed immersion of locally ringed spaces: this is a map of locally ringed spaces which is a homeomorphism onto a closed subset of the target, the induced map on structure sheaves is surjective with kernel $\mathcal{I}$, and the module $\mathcal{I}$ is locally generated by sections (aka every point has an open neighborhood so that, there exists a surjective map $\bigoplus_{i\in I}\mathcal{O}_X\to\mathcal{I}$ on this neighborhood).

The latter condition is easily verified in our case: the ideal sheaf $\mathcal{I}$ restricted to any open affine $\operatorname{Spec} R\subset X$ has the description $\widetilde{\sqrt{0}}$, which is clearly globally generated. Combining this with your already-proven $\mathcal{O}_{Y,x}\simeq \mathcal{O}_{X,x}/\mathfrak{N}_x$ and the fact the morphism of underlying topological spaces $Y\to X$ is the identity gives that $(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$ is a closed immersion of locally ringed spaces.

Now I claim that for any closed immersion of locally ringed spaces $f:Z\to X$ with $X$ a scheme, $Z$ is actually a scheme too. To start this proof, recall that a scheme is a locally ringed space where every point has an open neighborhood isomorphic as a locally ringed space to the spectrum of some commutative ring. It is enough to prove this statement for $X$ affine: for any point $f(z)=x\in X$, pick an open affine neighborhood $U$, and then if we prove that the preimage of $U$ is an affine scheme, then we've produced an open affine scheme containing $z$ and $Z$ will then satisfy the definition of a scheme.

To handle the affine case, let $X=\operatorname{Spec} R$. Now let $\mathcal{I}\subset \mathcal{O}_X$ be the sheaf of ideals corresponding to the kernel of $\mathcal{O}_X\to i_*\mathcal{O}_Z$. This is locally generated by sections as a sheaf of $\mathcal{O}_X$-modules, so the quotient sheaf is locally the cokernel of a map of free $\mathcal{O}_X$-modules, so $\mathcal{O}_X/\mathcal{I}$ is quasicoherent. So $\mathcal{I}$ is too, which means that it's the sheaf associated to some ideal $I\subset R$, and $Z\cong\operatorname{Spec} R/I$. So we're done!

Alternatively, if you can prove that the sheaf $I$ associated to the presheaf $U\mapsto \sqrt{0}\subset \mathcal{O}_X(U)$ is quasicoherent, you can define the closed subscheme $V(I)$ by taking the underlying topological space to be the support of $I$ with the structure sheaf $\mathcal{O}_X/I$. This gives you a canonical closed immersion $V(I)=(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$.

Once we know that we have a closed immersion of schemes $(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$ we can investigate the structure sheaf of $Y$ by looking at open affines on $X$. As closed immersions are affine morphisms and our closed immersion is a homeomorphism on the underlying topological spaces, we get that for any $U$ open affine in $X$, $U$ is also open affine in $Y$, so $\mathcal{O}_Y|_U=\mathcal{O}_{\operatorname{Spec} \mathcal{O}_Y(U)}$, and $\mathcal{O}_Y(U)=\mathcal{O}_X(U)/\sqrt{0}$. In particular, this means that $Y$ can be covered by reduced affine open subschemes, which is equivalent to every stalk $\mathcal{O}_{Y,y}$ reduced which is equivalent to the sections of $\mathcal{O}_Y$ on any open $U$ is reduced. So no matter what definition of reduced you have in mind, $Y$ is reduced.

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    $\begingroup$ For an alternate, full exposition, you may wish to consult the StacksProject result. $\endgroup$
    – KReiser
    Apr 24, 2020 at 23:22
  • $\begingroup$ what do you mean by $\widetilde{\sqrt{0}}$? $\endgroup$
    – rmdmc89
    Apr 24, 2020 at 23:57
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    $\begingroup$ The tilde is for the associated sheaf construction, which to any $R$-module $M$ gives a sheaf $\widetilde{M}$ on $\operatorname{Spec} R$ with global sections exactly $M$. The $\sqrt{0}$ is the nilradical. $\endgroup$
    – KReiser
    Apr 25, 2020 at 0:01
  • $\begingroup$ I still don't understand the definition of locally generated by sections. What map $\mathcal{O}_X\to\mathcal{I}$ are you refering to? And by surjective you mean $\mathcal{O}_{X,x}\to\mathcal{I}_x$ surjective for all $x$ in the neighbourhood, right? How would this map look like in $U=\text{Spec}(R)$ $\endgroup$
    – rmdmc89
    Apr 26, 2020 at 18:28
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    $\begingroup$ Proving that a presheaf is a sheaf is a dirty business I avoid unless it is absolutely necessary. It is not necessary in this case (and it's probably even false in this situation for some choice of $X$)! You yourself have already sheafified in your desription of this problem, which is additional evidence against it being necessary. $\endgroup$
    – KReiser
    Apr 26, 2020 at 19:53

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