# on the generic points of a scheme

This question may be a little bit metaphysical:are there any important properties about the generic points on a scheme?Or rather,why do we introduce the concept of generic point?I am not very clear the importance of generic points in the study of schemes.So,I would appreciate it if someone would like to say something on generic points.

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I've read somewhere that the whole point of introducing schemes was to have generic points! :) The ancients had generic points but they were a façon de parler, and schemes made them first class citizens. The history is told in Dieudonné's book on the history of algebrac geometry, if I recall correctly. – Mariano Suárez-Alvarez Aug 11 '11 at 15:00
See Emerton's answer here (4th answer): mathoverflow.net/questions/28496/… – Matt Aug 11 '11 at 15:20
@Mariano I think you meant "façon de parLer" meaning "manner of speaking". The verb "parer" means (among other things) "to parry". – AFK Sep 2 '11 at 16:18
@user8882: indeed! Thanks for catching the typo. Poor Jean's name algo got mangled... typing on a touchscreen is a pain :( – Mariano Suárez-Alvarez Sep 2 '11 at 16:31

A quick remark:

Classical varieties correspond to algebras of finite type over a (algebraically closed) field. Their points correspond to maximal ideals. But there are lots of reasons why one needs to consider more general rings: the concept of specialisation requires fields of transcendance degree $>0$, arithmetic geometry requires algebras over rings of integers, deformation theory requires artinian rings like $k[t]/(t^n)$ etc... Unfortunatly under a general morphism of rings $\phi:A\to B$ the inverse image of a maximal ideal is NOT a maximal ideal. So you do not have a map $\phi^{-1}: Spm(B) \to Spm(A)$ between maximal spectra as you would for algebras of finite type over a field (thanks to the Nullstellensatz). But you do have a map $\phi^{-1}: Spec(B) \to Spec(A)$ between prime spectra. For a classical complex variety this means than you need to add exactly one point for every non maximal prime ideal, i.e. every irreductible closed subset other than the usual (closed) points. And for this small effort you get a theory that is much much more flexible.

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One way to think about schemes, which can be especially useful when first learning about them, is as a way of adding extra structure to complex algebraic varieties. This is done so that any complex commutative algebra $A$ has a 'spectrum' $Spec(A)$, which generalizes the correspondence between nilpotent-free, finitely generated, commutative $\mathbb{C}$-algebras and varieties.

The need for generic points comes from the need for $Spec(\mathbb{C}(x))$ to exist, and more generally $Spec(K)$ for $K$ a field over $\mathbb{C}$. I claim $Spec(K)$ should always be a point, together with the extra information of its structure field. This is because closed subobjects of $Spec(K)$ should correspond to radical ideals in $K$, but there is only one ideal in $K$ (the zero ideal).

So how does this lead to generic points? Consider $\mathbb{C}[x,y]$, and let $p\in \mathbb{C}[x,y]$ be an irreducible polynomial. Then the ring $A=\mathbb{C}[x,y]/p$ is a domain, and so it embeds in its fraction field, which is isomorphic to $\mathbb{C}(\zeta)$ for some $\zeta$. This gives a map of algebras $$f:\mathbb{C}[x,y]\rightarrow \mathbb{C}(\zeta)$$ whose kernel is the ideal generated by $p$.

What does this become, geometrically? We want a dual map on affine schemes $$F:Spec(\mathbb{C}(\zeta))\rightarrow Spec(\mathbb{C}[x,y])$$ As a variety, $Spec(\mathbb{C}[x,y])\simeq \mathbb{C}^2$. The ring $\mathbb{C}(\zeta)$ is a field, and so the underlying set of $Spec(\mathbb{C}(\zeta))$ is a point. This means that the underlying set map of $F$ sends a single point to something in $Spec(\mathbb{C}[x,y])$.

But now think about what this geometric map is doing. For any function $a$ on $Spec(\mathbb{C}[x,y])$, its pullback along $F$ is $f(a)$. This only vanishes if $p$ divides $a$, which means that $a$ vanishes along the entire zero set $Z(p)$. This means that the map $F$ must somehow map to a Zariski-dense subset of $Z(p)$, while still only mapping to a single point. We are left with no other options... we must add a single point whose Zariski-closure is $Z(p)$. Thus, generic points are born.

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Can you give some more details on'for any function a on Spec(C[x,y]),its pullback along F is f(a)'.Thank you very much! – user14242 Aug 12 '11 at 0:01
Depending on how you approach it, this is the definition of F. On the level of varieties, we start with a map of varieties F and get a pullback map on functions f. On the level of affine schemes, we start with a map on algebras f, and we get a map on schemes F whose pullback map is f. – Greg Muller Aug 12 '11 at 5:13