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I'm currently reading through Hartshorne, and have come across a few things that have left me wondering.

(i) Somewhat pedantic, but also because I don't actually know the answer, (in Example 2.3.4) he looks at the affine plane over an algebraically closed field, defined as the scheme of $k[x,y]$, and discusses some of its properties. He says

Also, for each irreducible polynomial $f(x,y)$, there is a point $\eta$ whose closure consists of $\eta$ together with all closed points $(a,b)$ for which $f(a,b) = 0$. We say that $\eta$ is a generic point of the curve $f(x,y) = 0$.

Now, is $\eta$ unique? The way I read the first sentence, I feel there should be only one such point, but then he says $\eta$ is a generic point instead of the generic point, which really made me feel it was not unique when I read it.

(ii) The two definitions before Example 3.2.1 define certain morphisms to be locally of finite type, finite type or simply finite, but the very next example says a scheme (with no morphism between schemes in the example) is of finite type. What does this mean? Perhaps that the scheme itself admits a covering via affine subsets that are the Spec of rings?

(iii) Consider $R = \mathbb{C}[x]/(x^2)$. Then Spec $R$ has only one element, namely $(x)$. Hence Spec $R$ is irreducible. In a somewhat entertaining way however, Spec $R$ is not reduced. This follows quite easily from Example 3.0.1 in the book (in particular "$X$ is reduced if and only if nil $A = 0$", where nil $A$ is the nilradical of $A$, which in our case is $(x)$) but I'm fairly new to localization, so what hoping someone could make sure I'm not making a silly mistake in trying to show this directly. Is the localization of $R$ at $(x)$ simple $R$ again? (And hence the sheaf $\mathcal{O}((x)) = R$, which certainly has non-zero nilpotents.)

I should add, for the last one, I'm sure it was not Hartshorne that decided to use irreducible and reduced for distinct things, but I'm curious if anyone knows why the terminology came about this way? Perhaps a raw translation from french, where the words are slightly more distinct?

Thanks!

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Dear Alex, yes the words reduced and irreducible definitely come from medieval French (and the French words come from Latin) in general and probably also in their mathematical sense. However the translation is quite faithful and the mathematical usage is just as absurd in French: the two concepts have no mathematical relation whatsoever, which leads to the ridiculous terminology that a scheme like $Spec(k[x]/(x^2))$ is irreducible and yet has a (non-trivial) reduction, just as you judiciously noticed. C'est la vie... –  Georges Elencwajg Oct 26 '11 at 8:31
    
Fair enough. Thanks! –  Alex Oct 26 '11 at 12:18
    
@GeorgesElencwajg: Actually, when you say the scheme has a reduction, that's not a term I've come across. Does that mean that we may look at $(k[x]/(x^2))/(x) = k$ and say that Spec $k$ is a reduction of Spec$(k[x]/(x^2))$? (and as long as $k$ does not have nilpotents, a reduced reduction?) –  Alex Oct 26 '11 at 12:30
    
Dear @Alex, yes every scheme $X$ has a canonical reduction $X_{red}$ (which is of course a reduced scheme!) obtained by keeping the topological space $|X|$ of $X$ and dividing out the structure sheaf $\mathcal O_X$ by the quasi-coherent ideal given by its nilpotent elements. This is Définition (4.5.3) page 269 of EGA I (Springer Edition): "On appelle schéma réduit associé à un schéma $X$ et on note $X_{réd}$ l'unique sous-schéma réduit de $X$ ayant $X$ pour espace sous-jacent" [Beware that Grothendieck and Dieudonné here denote by $X$ both the scheme and its underlying topological space] –  Georges Elencwajg Oct 26 '11 at 21:40
    
(continued) The example in your comment is perfectly correct. Only, don't say "reduced reduction", since reductions are always reduced. (And, by the way, all my wishes of good health and success for your Ph.D.) –  Georges Elencwajg Oct 26 '11 at 21:51

1 Answer 1

up vote 4 down vote accepted

i) $\eta$ is unique: it must be the point described by the ideal $(f(x, y))$ by the Nullstellensatz. In complete generality, a generic point of a topological space $X$ is a point whose closure is all of $X$, and in this generality generic points are not unique.

ii) Every scheme admits a canonical morphism to its base scheme. For affine schemes without further qualification this is $\text{Spec } \mathbb{Z}$. This map comes from the spec-ification and gluing of canonical maps $\mathbb{Z} \to R$ giving canonical maps $\text{Spec } R \to \text{Spec } \mathbb{Z}$ for all affine schemes. For schemes over a base field $k$ the canonical maps are instead $\text{Spec } R \to \text{Spec } k$. Apply the definition to these morphisms.

iii) Yes. Every element of $\mathbb{C}[x]/(x^2)$ not in $(x)$ is already invertible.

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Thanks! extra characters so I can say thanks –  Alex Oct 26 '11 at 4:16

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