For finite sets $A$ and $B$, it is clear that $A \subseteq B$ and $|A| \geq |B|$ implies $A = B$. While an obvious fact, it can sometimes be a nice shortcut in proofs.

Analogously, if $V$ and $W$ are finite-dimensional vector spaces such that $V \subseteq W$ and $dim\ V \geq dim\ W$ then $V = W$. This is an especially useful tool when $V$ is defined parametrically and $W$ is defined implicitly. Then you can easily prove $V \subseteq W$ by plugging the parametric expression for $V$ into the equations for $W$. Counting the dimensions can take more work, but you sometimes get lucky.

For a while I've been wondering how this extends to algebraic varieties.

Here's an attempted application to proving the spectral theorem. Fix the dimension $n$; all matrices will be $n \times n$. The theorem says that for every symmetric matrix $S$ there exists an orthogonal matrix $Q$ and a diagonal matrix $D$ such that

$$Q^T\ D\ Q = S.$$

Let $A$ and $B$ respectively denote the matrices of the form on the left-hand and right-hand side. $A$ is defined parametrically by a function $f$ from $D$ and $Q$, and $B$ is defined implicitly by the symmetry condition. We want to prove $A = B$. It is clear that $A \subseteq B$:

$$(Q^T\ D\ Q)^T = Q^T\ D^T\ (Q^T)^T = Q^T\ D\ Q,$$

so $Q^T\ D\ Q$ is symmetric.

The domain of $f$ has dimension $dim\ D + dim\ Q$ where $dim\ D = n$ and $dim\ Q = (n-1) + \cdots + 1$, while $S$'s space has dimension $n + (n-1) + \cdots + 1$, so the dimensions seem to match.

But what about $f$'s degree of injectivity? It isn't perfectly injective: if $D$ and $D'$ equal the identity matrix then $Q^T\ D\ Q = Q'^T\ D'\ Q'$ for any independent combination of $Q$ and $Q'$.

My question is thus twofold:

What is the right generalization of the theorem for vector spaces to algebraic varieties?


Can the attempted proof of the spectral theorem be salvaged with a genericity argument?

I'm happy with the extant proofs of the spectral theorem, so this is more curiosity than anything else.

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    $\begingroup$ My guess: if $X\subseteq Y$ is a pair of proper varieties of the same dimension, then $X=Y$. $\endgroup$ – Mariano Suárez-Álvarez Jul 30 '10 at 3:34
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    $\begingroup$ You want them to be irreducible too. Slightly more generally, if $X \subset Y$ are irreducible, $X$ is a closed subvariety, and $\dim X = \dim Y$, then $X = Y$. It's nice to be able to use this trick on affine varieties, like Per Vognsen is doing. $\endgroup$ – Ryan Reich Jul 30 '10 at 3:42
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    $\begingroup$ I hope you realize that over $\mathbb{C},$ symmetric matrices need not be diagonalizable! $\endgroup$ – Victor Protsak Jul 30 '10 at 5:55
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    $\begingroup$ Well, a straightforward argument based on dimension count can't work, because it should work over the complex numbers, where the result is false. One can show that the theorem is true for generic complex symmetric matrices, though, using a dimension count; to do the case of generic real symmetric matrices you have to add some calculations with differentials. Then conclude with a simple argument using the compactness of the real orthogonal group. I'll post the details later. $\endgroup$ – Angelo Jul 30 '10 at 8:25
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    $\begingroup$ Ignoring your intended application - if $f:X\to Y$ is a morphism of varieties it can be rather hard to show that $f$ is surjective. The reason is that in general the image $f(X)$ is not a subvariety - it is just a "constructible subset". E.g. consider $f:\mathbb{A}^2 \to \mathbb{A}^2$ given by $(x,y)\mapsto (x,xy)$. It is true that if $f$ is proper then $f(X)$ is a (closed) subvariety of $Y$, but this is a very strong condition. $\endgroup$ – Tony Scholl Jul 30 '10 at 8:41

I guess this answers at least part of the question, so I will post it as an answer. As in my comment above: the correct generalization is that if $X \subset Y$ is a closed subvariety, if $X$ and $Y$ are both irreducible, and if $\dim X = \dim Y$, then $X = Y$. At this level of generality this is an application of the noetherian property of finite-dimensional varieties over a field together with the definition of irreducible. From this you can deduce that even if $X$ is locally closed in $Y$, it must be dense (since $X$ has the same dimension as its closure). An even more general form of this theorem is that if you have merely a dominant map $X \to Y$ (that is, its image is dense) then the minimal dimension of a fiber is $\dim X - \dim Y$ and this occurs generically ($X$ and $Y$ should still be irreducible).

This is often combined with another theorem: if $X$ is irreducible and $X \to Y$ is surjective, then so is $Y$; conversely, if $Y$ is irreducible and the fibers of that map are irreducible of the same dimension, then so is $X$. Now you can perform the following trick beloved of algebraic geometers (e.g. anything in Harris' "Algebraic Geometry: A First Course", under the aegis of "incidence correspondences"; this theorem is 11.14): to compute the dimension of $Y$, cover it by $X$ and then have $X$ cover some other, irreducible $Z$ whose dimension is easy to compute. Then apply both the irreducibility and dimension theorems twice. Unfortunately, unless they are all projective varieties (which they often are in applications of this method) this is more of a way of computing dimension than of showing that maps are surjective (however, see for example the appendix to section 4 of Mumford's "Abelian Varieties" for an impressive example of this).

To show that $X \to Y$ is surjective, one general method that comes to mind is to show that it is dominant and has some kind of "closure" property. Either it could be actually closed, or perhaps you have a group acting on both $X$ and $Y$, transitively on $Y$, for which the map is equivariant. IANAAG (I am not an algebraic geometer) so this is probably hopelessly naive as far as general methods go.


Another generalization of the theorem for vector spaces could be the following.

Assume that $X \subset Y$ is a closed subvariety, with $Y$ connected. If

$\dim X = \dim Y $ and

$\textrm{mult}_xX = \textrm{mult}_x Y$ for all $x \in X$,

then $X=Y$.

In fact, assume $X \neq Y$. Since $Y$ is connected there exists $x \in X \cap \overline{Y \setminus X}$, and for such a $x$ we have

$\textrm{mult}_xY \geq \textrm{mult}_xX + \textrm{mult}_x \overline{Y \setminus X} \geq \textrm{mult}_xX+1$,

contradiction. This can be useful, since it does not require the irreducibility of $X$ and $Y$. However, the assumption about the connectedness of $Y$ cannot be dropped: consider the case where $Y$ is the disjoint union of two copies of $X$.

EDIT. In a first version of this answer I required the condition $\dim T_xX = \dim T_xY$ instead of $\textrm{mult}_xX = \textrm{mult}_x Y$. This actually did not work, see damiano's comment below.

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    $\begingroup$ I am not sure that this works: the dimension of the point x need not be larger in Y. An easy example is X an integral plane cubic with a node and Y the union of X and a line L that is tangent to X at one of the branches of X through the singular point. Thus L and X have in common just the singular point of X and the Zariski tangent spaces of X and Y always agree on the points of X. $\endgroup$ – damiano Jul 30 '10 at 17:57
  • $\begingroup$ @ damiano. Of course you are right, I wanted to write a condition involving dimensions of tangent spaces, but I overlooked a "degenerate" situation like the one you described. In order to fill this gap, I think one should consider multiplicities instead of tangent dimensions, namely replace the condition $\dim T_xX = \dim T_xY$ for all $x \in X$ with the condition $mult_xX=mult_xY$ for all $x \in X$. $\endgroup$ – Francesco Polizzi Jul 30 '10 at 21:22

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