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According to Liu's "Algebraic Geometry and Arithmetic Curves", Proposition 2.4.2, if $X$ is any scheme, there is a unique closed subscheme $X_{red} \to X$ with the same underlying topological space, and this satisfies the property that each morphism $Y \to X$ from a reduced scheme $Y$ factors uniquely through $X_{red}$. Of course, it is also true that any morphism of schemes $X \to Y$ can be extended uniquely to a morphism $X_{red} \to Y$ by composition. Since "nice" schemes are usually required to be reduced, I wonder what the use of considering reduced schemes is. More in detail, I have the following questions:

  • Is the purpose of having non-reduced schemes solely to make definitions clearer and more consistent? (For instance, if $A$ is a non-reduced ring, we would need to change the definition of $\mathrm{Spec}A$ to make it a reduced scheme)
  • Is it true that one can, for all practical purposes, just assume that all schemes he's dealing with are reduced? This seems reasonable in view of the fact that morphisms to and from $X$ can be seen as morphisms to and from $X_{red}$. But is it possible to have two different morphisms from $X$ to a reduced scheme $Y$ that become the same when we compose them with the "inclusion" $X_{red} \to X$? Is this a major problem that hinders us from assuming that all schemes are reduced, or not?
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marked as duplicate by Watson, Lord Shark the Unknown, ArsenBerk, Jens, jgon Nov 13 '18 at 17:47

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  • $\begingroup$ You might want to read §4 "Some Applications of Nilpotents" from this document. $\endgroup$ – Watson Feb 18 '18 at 9:34
  • $\begingroup$ Related: mathoverflow.net/questions/55244 $\endgroup$ – Watson Feb 18 '18 at 9:35
  • $\begingroup$ @Watson Thank you, very useful links. $\endgroup$ – 57Jimmy Feb 18 '18 at 9:44
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    $\begingroup$ Non-reduced scheme occur naturally in the context of algebraic groups: $\tag*{}$ 1) if $k$ is a field of characteristic $p>0$, then $\mu_p = \mathrm{Spec}(k[X] / (X^p-1))$ is the non-reduced group scheme of the $p$-th roots of unity. $\tag*{}$ 2) More generally, the kernel of a morphism of smooth (hence reduced) algebraic groups need not be reduced (nor smooth) ; see example 1.47. $\endgroup$ – Watson Feb 18 '18 at 9:53
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    $\begingroup$ 3) The isomorphism theorems may fail if you don't allow nilpotents, see aside 5.40. $\endgroup$ – Watson Feb 18 '18 at 9:53