(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 .**

**Answer** :

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.

- The map is a set bijection because
- is a group homomorphism – easy check. [This is why we need instead of
- is a 1-1 function (or an embedding) because the map is identity if and only if which happens if and only if

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.

**Counter example**

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.

Thus .

Q : Find a quadratic extension of where has no nontrivial zero.

Non-solutions :

- Over , is a solution.
- Over , is a solution.

Solution :

Sketch of how to go about it :

- Finding a zero of is equivalent to finding a zero of . In fact, more generally, finding zero of is equivalent to finding a zero of over any field . A nice explanation of this fact is via the theory of quaternions. Our form is a really nice form, it is the norm form of the quaternions called which has generators : and relations : .
- A quaternion algebra is either a skew field (all elements have inverses) or it is , the 2×2 matrices.
- The norm is multiplicative, so it is a good test to check whether our quaternion is a skew field or not. The norm form has a nontrivial zero if and only if the quaternion is a matrix algebra.
- Another if and only if test for whether the quaternion is a matrix algebra over is to check that the form has a nontrivial zero.
- So from now on, we can assume our form is .
**Hasse-Minkowski’s theorem**: If q is a quadratic form over K, a number field, then q has a nontrivial zero if and only if it has a nontrivial zero over , all the completions of K.- For example, is the field of -adic numbers, which is the completion of at prime . In fact, here we only need the trivial direction of this theorem, namely the fact that if there is a rational solution, there is a local solution.

- So we try to find a and a completion such that our form has no solution over .
- In fact, let us try to imitate the proof of why this form has no solution over : If it does, it should have a nontrivial solution over (the 3 adics), so by clearing denominators, the solution will be in (the 3-adic integers).
- Let where are in and not all 0 simultaneously. Since is a DVR with parameter 3, this says that , so divide by 3, the whole equation to get . Continue this process till you can assume there is some such that and 3 does not divide one of v and w.
- is the field of 3 elements which is also modulo its maximal ideal , so the previous solution gives a nontrivial solution , which is a contradiction (you can check manually in )
- So if we find a and ring of integers and a prime of such that the residue field of (completion of K at P) is , and 3 is not ramified at P, the same explanation as above shows that has no solution over K.
- For where d is 2 or 3 mod 4, , and I think we can directly see how any prime p in splits in K [look at factorization of in ]
- For , where P and Q are distinct primes in (because In fact ). So 3 does not get ramified in . Also we see that residue field of
- In fact, there is a formula to check whether an odd prime is ramified, split or inert in depending on whether is a square modulo or not.