Exercises

Hausdorff spaces are never irreducible

irreduciblenothausdorff

 

(Black + Gray) and (Black + White) form the whole space. That is, in an irreducible space, any two non-empty open sets intersect

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Transitive subgroups of the symmetric group

Let S_n denote the symmetric group on A=\{1,2,\ldots n\}. A subgroup G\subseteq S_n is said to be a transitive subgroup if the natural action on the set A is transitive.

Q : Find all isomorphism classes of transitive subgroups of size n of S_n.

Answer :

Let us try to find one subgroup .., well, \mathbb{Z}_n generated by the n-cycle (1 2 \ldots n). 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 S_{2n} and look at p=(1, 2, \ldots n)(n+1, \ldots, 2n) [this is the cycle notation of permutations]. p has order n. Now if we can find a q which somehow can make 1 go to n+1, then the subgroup generated by p,q would be transitive.

In fact, we can find q such that qp=pq or rather qpq^{-1}=p. Just look at q=(1, n+1)(2 , n+2)(3 , n+3)\ldots (n , 2n). Then qpq^{-1} = (1, 2 ,\ldots, n)(n+1, \ldots, 2n)=p

The subgroup generated by p,q is isomorphic to \mathbb{Z}_n\times \mathbb{Z}_2, so we see that \mathbb{Z}_{n}\times \mathbb{Z}_{2} is also a transitive subgroup of size 2n in S_{2n}

In fact, more is true !

Claim : Let G be ANY subgroup of size n. Then it can be embedded as a transitive subgroup of S_n.

Proof : Let G=\{g_1,\ldots,g_n\} and let S_n be the symmetric group on a set of size n, which we can assume to be G ! We have the following (usual) embedding of G into S_n got by the action of G on itself by a modified left translation, namely

f:G\to S_n which sends g\to [g_i\to g^{-1}g_i], where S_n is thought of as set bijections of G and itself.

  1. The map [g_i\to g^{-1}g_i] is a set bijection because g^{-1}g_i=g^{-1}g_j\iff g_i=g_j
  2. f is a group homomorphism – easy check. [This is why we need g_i\to g^{-1}g_i instead of g_i\to gg_i
  3. f is a 1-1 function (or an embedding) because the map [g_i\to g^{-1}g_i] is identity if and only if g^{-1}g_i=g_i\forall g_i which happens if and only if g=e

So G can be identified as a subgroup of S_n via f. Now we check that it is a transitive subgroup !

Any g_i can be moved to g_j by the element a=g_ig_j^{-1} because a\star g_i = a^{-1}g_i = g_jg_i^{-1}g_i=g_j.

Infact, what is the stabilizer of any g_i 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 g_i and g_j, there is a unique element in G which takes one to the other)

Can an irreducible curve in a surface be the divisor of a function

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 k. So S being projective means that the natural map S\to Spec(k) is a projective morphism, in that it factors through the following closed immersion i:

S\to^{i} \mathbb{P}^n_k\to Spec(k)

Here \mathbb{P}^n_k =Proj k[x_0,x_1,\ldots,x_n] and the map from this to Spec(k) is given by gluing maps from the usual affine cover D_{+}(x_i)\forall x_i\to Spec(k) where

D_+(f)=\{P\in Proj k[x_0,\ldots,x_n]| f\not\in P\} \cong Spec degree 0 homogeneous elements of k[x_0,\ldots,x_n]_{f}

To check : Thus D_{+}(x_i)\cong Spec k[\frac{x_0}{x_i},\ldots, \frac{x_n}{x_i}]

which has a natural map to Spec(k).

Now the ring of functions regular on S is k^* because S 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 f\in k(S)^*, the function field of S, then the divisor of f is div(f)=\sum v_{P}(f) where P runs over all codimension 1 points of S. This is the construction of a principal weil divisor.

In more detail, if S 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 k(S) called v_P.

Coming back to our problem, let C be an irreducible curve. If possible, let it be div(f). Thus we get div(f)=nP (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 n>0 (if n<0, just look at \frac{1}{f}. Of course if n=0, then it is not a curve !)

So this means f\in m_{P,S}, the maximal ideal of the local ring of S at P and in O_{Q,S}, 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 Spec(R).

Claim : R is also integrally closed.

Proof : R=\cap R_M where M runs over maximal ideals of R

(Look at a\in \cap R_M and look at I=\{x\in R|xa\in R\}, this is an ideal of R. If it is all of R, then 1\in I and hence a.1=a\in R. If it is a proper ideal, then let it be inside maximal ideal N. Since a\in R_N, a=b/c for c\not\in N and hence c\not\in I. But ca=b\in R and hence c has to be in I, a contradiction.)

We have assumed every local ring is normal, so R_M are all integrally closed in k(S)=K. If b\in K such that it is integral over R, then it is clearly integral over each R_M (a bigger ring than R) and hence because of normality,  in each R_M. Thus b\in \cap R_M=R.

Finally we have one more theorem :

Thm : If R is a noetherian integrally closed domain, then R=\cap R_P where P runs over height one prime ideals only.

Thus, our f will be in R since it is in each R_P.

This means that f is regular on each affine open set of S, and hence it is a regular function on S, and hence a constant in k. 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 S=\mathbb{A}^2=Spec k[x,y].

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 k[x,y] is a UFD.

But the Krull dimension of k[x,y] is 2, hence every height one prime looks like (t) where t is an irreducible polynomial in k[x,y].

Let us look at div(x). Since x is irreducible, x\in (t) implies (t)=(x), thus x is a unit in k[x,y]_P for every height one prime ideal P except Q=(x). At Q, x has valuation 1.

Thus div(x)=Q.

Finding zeros (or not) of quadratic forms over number fields

Q : Find a quadratic extension K of \mathbb{Q} where u^2-3v^2-3w^2+9z^2 has no nontrivial zero.

Non-solutions :

  1. Over \mathbb{Q}(\sqrt{2}), (0,\sqrt{2},1,1) is a solution.
  2. Over \mathbb{Q}(\sqrt{3}), (0,\sqrt{3},0,1) is a solution.

Solution : \mathbb{Q}(\sqrt{7})

Sketch of how to go about it :

  • Finding a zero of <1,-3,-3,9> is equivalent to finding a zero of <1,-3,-3> . In fact, more generally, finding zero of <1,-a,-b,ab> is equivalent to finding a zero of <1,-a,-b> over any field K. 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 (a,b) which has generators : i, j and relations : i^2 = 3 =a , j^2 = 3 = b , ij=-ji= k.
  • A quaternion algebra is either a skew field (all elements have inverses) or it is M_2(K), 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 (a,b) is a matrix algebra over K is to check that the form <1,-a,-b> has a nontrivial zero.
  • So from now on, we can assume our form is <1,-3,-3>.
  • 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 K_v, all the completions of K.
  • For example, \mathbb{Q}_p is the field of p-adic numbers, which is the completion of Q at prime p. 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.
  1.  So we try to find a K=\mathbb{Q}(\sqrt{d}) and a completion K_v such that our form q=<1,-3,-3> has no solution over K_v.
  2. In fact, let us try to imitate the proof of why this form has no solution over \mathbb{Q} : If it does, it should have a nontrivial solution over \mathbb{Q}_3 (the 3 adics), so by clearing denominators, the solution will be in \mathbb{Z}_3 (the 3-adic integers).
  3. Let a^2 - 3b^2 - 3c^2 =0 where a,b,c are in \mathbb{Z}_3 and not all 0 simultaneously. Since \mathbb{Z}_3 is a DVR with parameter 3, this says that 3|a, so divide by 3, the whole equation to get 3x^2 - b^2 - c^2 =0 . Continue this process till you can assume there is some u,v,w\in \mathbb{Z}_3 such that 3u^2 - v^2 - w^2 =0 \in \mathbb{Z}_3and 3 does not divide one of v and w.
  4. F_3 is the field of 3 elements which is also \mathbb{Z}_3 modulo its maximal ideal (3), so the previous solution gives a nontrivial solution v^2+w^2=0 \in F_3, which is a contradiction (you can check manually in F_3)
  5. So if we find a K= Q(\sqrt{d}) and ring of integers O_K and a prime P of O_K such that the residue field of K_P (completion of K at P) is F_3, and 3 is not ramified at P, the same explanation as above shows that u^2 - 3v^2 - 3 w^2 has no solution over K.
  6. For K = Q(\sqrt{d}) where d is 2 or 3 mod 4, O_K = Z[\sqrt{d}], and I think we can directly see how any prime p in \mathbb{Z} splits in K [look at factorization of x^2-d in F_p[x]]
  7. For Q(\sqrt{7}) , 3 = PQ where P and Q are distinct primes in O_K (because x^2-7 = x^2 - 1 = (x+1)(x-1) \in F_3[x] In fact P=(3,\sqrt{7}+1), Q=(3,\sqrt{7}-1)). So 3 does not get ramified in K_P. Also we see that residue field of K_P = \mathbb{Z}[\sqrt{d}]/P = F_3
  8. In fact, there is a formula to check whether an odd prime p is ramified, split or inert in \mathbb{Q}(\sqrt{d}) depending on whether d is a square modulo p or not.