(Black + Gray) and (Black + White) form the whole space. That is, in an irreducible space, any two non-empty open sets intersect
Let denote the symmetric group on . A subgroup is said to be a transitive subgroup if the natural action on the set is transitive.
Q : Find all isomorphism classes of transitive subgroups of size n of .
Let us try to find one subgroup .., well, generated by the -cycle . Clearly it is transitive, and in fact, more is true, namely, the stabilizer of any element is the trivial permutation.
Can we think of more ?
Let us look at and look at [this is the cycle notation of permutations]. has order . Now if we can find a which somehow can make 1 go to n+1, then the subgroup generated by would be transitive.
In fact, we can find such that or rather . Just look at . Then =
The subgroup generated by is isomorphic to , so we see that is also a transitive subgroup of size 2n in
In fact, more is true !
Claim : Let be ANY subgroup of size . Then it can be embedded as a transitive subgroup of .
Proof : Let and let be the symmetric group on a set of size n, which we can assume to be G ! We have the following (usual) embedding of into got by the action of on itself by a modified left translation, namely
which sends , where is thought of as set bijections of G and itself.
So G can be identified as a subgroup of via . Now we check that it is a transitive subgroup !
Any can be moved to by the element because .
Infact, what is the stabilizer of any under this action of G ? It is the identity ! So this action of G is in fact regular (or transtive+free = simply transitive, namely given and , there is a unique element in G which takes one to the other)
Question : Let S be a projective surface and C, an irreducible curve on it. Show that C can’t be the divisor of a function.
Answer : A surface is a two dimensional scheme. I guess we assume S is defined over a field . So S being projective means that the natural map is a projective morphism, in that it factors through the following closed immersion :
Here and the map from this to is given by gluing maps from the usual affine cover where
degree 0 homogeneous elements of
To check : Thus
which has a natural map to .
Now the ring of functions regular on S is because is projective. (I guess that for this you need that S is projective over a field, or atleast a nice enough affine scheme below, like an integral domain ??)
What is the divisor of a function ?
If , the function field of , then the divisor of is where P runs over all codimension 1 points of . This is the construction of a principal weil divisor.
In more detail, if is a nice enough scheme (noetherian, separated, all local rings of dimension one are regular), then at any codimension 1 point P (which will correspond to a height one prime ideal of the ring of any affine open subset containing it), the local ring will be a noetherian regular one dimensional local ring and hence a DVR. This DVR defines a valuation on called .
Coming back to our problem, let be an irreducible curve. If possible, let it be . Thus we get (for an irreducible curve can’t have more than 1 generic point in it. Note that since S is two dimensional, closure of any codimension 1 point is 1-dimensional)
We will show that this is not possible.
Note that we can assume wlog that (if , just look at . Of course if , then it is not a curve !)
So this means , the maximal ideal of the local ring of S at P and in , the local ring of S at every other codimension 1 point Q.
Probably a stupid question : Is an integral projective scheme normal ? Ie, are the local rings integrally closed ?
Let us assume for a moment that our scheme S is normal. That is, all the local rings (whether at codimension 1 points or not are integrally closed in the function field). Let us look at any affine open set .
Claim : is also integrally closed.
Proof : where M runs over maximal ideals of
(Look at and look at , this is an ideal of R. If it is all of R, then and hence . If it is a proper ideal, then let it be inside maximal ideal . Since , for and hence . But and hence has to be in , a contradiction.)
We have assumed every local ring is normal, so are all integrally closed in . If such that it is integral over , then it is clearly integral over each (a bigger ring than ) and hence because of normality, in each . Thus .
Finally we have one more theorem :
Thm : If R is a noetherian integrally closed domain, then where P runs over height one prime ideals only.
Thus, our will be in since it is in each .
This means that is regular on each affine open set of S, and hence it is a regular function on S, and hence a constant in . In which case, it can’t have positive valuation at any codimension 1 point (for it is a unit in every local ring which are all k-algebras)
Hence we are done, modulo the question about whether S is a normal surface or not.
If we drop the assumption that S is projective, then this is false. For example, take .
What are the height one primes here ?
Well every prime ideal has an irreducible element in it. And the ideal generated by any irreducible element is a prime ideal because is a UFD.
But the Krull dimension of is 2, hence every height one prime looks like where is an irreducible polynomial in .
Let us look at . Since is irreducible, implies , thus is a unit in for every height one prime ideal except . At , x has valuation 1.
Q : Find a quadratic extension of where has no nontrivial zero.
Sketch of how to go about it :