# Hausdorff spaces are never irreducible

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

# 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 $S_n$.

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 $3u^2 - v^2 - w^2 =0 \in \mathbb{Z}_3$and 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.