2
$\begingroup$

When we usually talk about prime ideals "lying over" one another, we have a tower of rings $A \subseteq B$ with $\mathfrak{p} \subseteq A$ and $\mathfrak{q} \subseteq B$ prime ideals of $A$ and $B$ respectively. Then we can say that $\mathfrak{q}$ lies over $\mathfrak{p}$ if $$ A \cap \mathfrak{q} = \mathfrak{p}. $$ This is standard and I am comfortable with that. The issue is when texts have a morphism of rings $\phi: A \longrightarrow B$ and talk about primes of $B$ lying over primes of $A$. What exactly does this mean? The confusion stems from the fact that the extension of a prime ideal in $A$ is not generally a prime ideal in $B$, so there doesn't seem to be an obvious interpretation to me. In the case that $\phi$ is surjective, then it makes sense for primes containing the kernel of $\phi$, but in general, how does one interpret the notion of a prime of $B$ lying over a prime ideal when all we have is a general morphism of rings $\phi: A \longrightarrow B$?

$\endgroup$
4
  • 2
    $\begingroup$ The inverse image of a prime is a prime, and in the special case of the inclusion the inverse image is just the intersection. $\endgroup$
    – user363120
    Commented Sep 25, 2017 at 10:20
  • $\begingroup$ @user363120, so in the case of a morphism $\phi: A \longrightarrow B$, for $\mathfrak{q}$ to lie over $\mathfrak{p}$ simply means that $\phi^{-1}(\mathfrak{q}) = \mathfrak{p}$? That makes sense in terms of the geometric notion of a point lying over another point in the affine schemes I suppose. $\endgroup$
    – Luke
    Commented Sep 25, 2017 at 10:45
  • 1
    $\begingroup$ Note that the inclusion $A \subseteq B$ is a homomorphism as well, so it is a very natural extension of terminology. $\endgroup$
    – Arthur
    Commented Sep 25, 2017 at 10:47
  • $\begingroup$ @Arthur, yeah of course, that makes total sense. Thanks to both of you. It seems so obvious now but I couldn't actually find this written explicitly anywhere. $\endgroup$
    – Luke
    Commented Sep 25, 2017 at 10:48

1 Answer 1

0
$\begingroup$

$\def\frp{\mathfrak{p}} \def\frq{\mathfrak{q}} \def\spm{\operatorname{Spm}}$Given an algebraically closed field $k$, by a “classical $k$-(pre)variety,” I will mean what Serre calls in FAC a “(pre)variety.” For a modern definition, see Definition 5.1 and Example 3.5(a) here.

Let $\varphi:B\to A$ be a ring homomorphism. If $\frp\subset A$ is a prime ideal, then so is $$ \label{eq}\tag{1} \varphi(\frp)^{-1}=\frq. $$ We are about to give a geometric explanation of what does it mean for primes $\frp\subset A$ and $\frq\subset B$ to satisfy \eqref{eq}. For this, we consider the case where $A, B$ are finitely-generated reduced algebras over an algebraically closed field $k$, and $\varphi$ is a $k$-algebra homomorphism (the resulting category of such maps $\varphi$ is equivalent to affine classical $k$-varieties. A proof may be found in the link above: fully faithfulness is 4.8, and essential surjectivity is 4.14).

Let $f=\spm\varphi:X\to Y$ be the associated morphism to $\varphi$ of classical varieties. If $Z\subset X$ is any subset, it is not difficult to see that $\varphi^{-1}(I(Z))=I(f(Z))\;(=I(\overline{f(Z)}))$. In the case where $Z$ is closed irreducible, then this is exactly \eqref{eq}, by the one-to-one correspondence between (irreducible) closed subsets of an affine classical variety and the (prime) radical ideals of its coordinate ring (see 1.18(b) and 2.9 from last link).

Note that the continuous image of a closed set $Z$ needs not be closed, but if it is irreducible, then so is its image (continuity preserves irreducibility), and, in turn, also is the closure of the image (irreducibility is stable under closures), $\overline{f(Z)}$. We can make then the following

Conclusion. On finitely-generated reduced algebras over an algebraically closed field, what on the POV of affine schemes is just “mapping a point,” $f(x)=y$ (i.e., equation \eqref{eq}), on the POV of classical affine varieties corresponds to performing the mapping $Z\subset X\mapsto\overline{f(Z)}$, for $Z$ closed irreducible.

What can we say about this in the case of general schemes? For any space $X$, there is a function from points of $X$ to irreducible closed subsets of $X$ which maps $x$ to $\overline{\{x\}}$. A space is sober if this map is a bijection. Equivalently, every irreducible closed subset has a unique generic point. For a continuous map between sober spaces $f:X\to Y$, one has $f(x)=y$ if and only if the generic point of $\overline{f\left(\overline{\{x\}}\right)}$ is $y$. This is a consequence of the proof of the existence of “soberifications” (see 02AN, or, alternatively, EGA I, Springer 1971 ed., $\mathbf{0}_\mathrm{I}$ (2.9.2)).

If a scheme is regarded as a “generalized variety,” then points in a scheme are to be interpreted as “irreducible generalized subvarieties” of the scheme. This interpretation is supported on the fact that schemes are sober, so the map $x\mapsto\overline{\{x\}}$ is a one-to-one correspondence between points in the scheme and these “irreducible generalized subvarieties.” For more a detailed discussion on these ideas, see Appendix C here.

If $\overline{k}=k$, the category of classical $k$-prevarieties is equivalent to reduced $k$-schemes of finite type. The equivalence, to the right direction, is given by soberification of spaces; and the quasi-inverse, to the left direction, is given by restriction to the subset of closed points. The above conclusion generalizes, in formally the same way, to interpretation, inside the POV of classical $k$-prevarieties, of the equation $f(x)=y$ on the POV of reduced $k$-schemes of finite type.

For a morphism between arbitrary schemes, the equation $f(x)=y$ should be thought intuitively about as a generalization of this idea from the classical setting.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .