is it like two parties both having almost enough to win the elections, or is it a "we can't really have a coalition of $n$ parties because $n$ is too large" thing
@Ultradark There is no such function. y=x^a grows slower than y=x for a<1 and a can be made as close to 1 as you want
Unrelated: I'm pretty sure I saw a comment on MathOverflow about how there are probability distributions on the integers where the probability of choosing any integer is zero, but (IIRC) the proof is nonconstructive. Am I imagining things or is this real?
@EsolangingFruit I don't follow. If probability of choosing every integer is zero, by countable additivity probability of $\Bbb N$ is zero.
Maybe you only want finite-additivity? Then it's not as clear to me anymore
It should be possible for sure
Choose some ultrafilter on $\Bbb N$ and assign $A$ to be of probability $1$ if upto finitely many elements $A$ is in the ultrafilter, something like that. Every finite set has probability zero because $\emptyset$ is not in the ultrafilter.
by "upto finitely any elements" I mean all the sets obtained from $A$ by adding/removing finitely any elements is contained in the ultrafilter
idk about the tech- tbh but then again my musical diet was different from the people who listened to Death albums when they came out so we have different standards for "tech"
on the topic of kvlt throwback to when i tried to tell someone to listen to a track from Bergtatt but couldn't actually pronounce the name and just said "google ulver capitel 3"
i recall looking up the lyrics to some (i was unaware that it was dsbm) album as a teenager and it was just a very detailed description of how the singer wanted to off himself
@Balarka yeah, as in, one studies the group $Z_\mathfrak{P} := \lbrace \sigma \in \operatorname{Gal}(L/K) : \sigma(\mathfrak{P}) = \mathfrak{P}\rbrace$ (i.e. the stabiliser of the prime under the action of the Galois group)
and this is obv a subgroup of the Galois group (as a stabiliser) so there's a fixed field, and it turns out ALL of the splitting of a prime occurs between $L$ and this fixed field, and nothing happens between that fixed field and $K$
There's also a surjection $Z_\mathfrak{P} \to \operatorname{Gal}(\Bbb F_\mathfrak{P}/\Bbb F_\mathfrak{p})$ (where the automorphisms of $\Bbb F_\mathfrak{P}$ are induced by those of $L/K$)
If $L/K$ is an extension then we have a corresponding map $\text{Spec} O_L \to \text{Spec} O_K$ and fiber over the prime $\mathfrak{p}$ is $\text{Spec}(O_L \otimes \kappa(\mathfrak{p}))$ where $\kappa(\mathfrak{p})$ is the residue field at $\mathfrak{p}$
$\text{Gal}(L/K)$ acts on $\text{Spec}(L \otimes_K K^{alg})$, the geometric fiber of $\text{Spec} K \to \text{Spec} L$, so maybe there's a way to push this forward to an action on $\text{Spec}(O_L \otimes_{O_K} \kappa(\mathfrak{p}))$
Let $K$ be a number field and let $\mathfrak{a}$ be a non-zero ideal of $\mathcal{O}_K$ such that $\mathfrak{a}^m = (a)$ for some $a \in \mathcal{O}_K$.
Does there exist a finite extension $L/K$ such that $\mathfrak{a}\mathcal{O}_L$ is principal?
The answer is yes; one can take $L = K(\sqr...
@ÍgjøgnumMeg I have never really understood this story to be honest. I understand that if $q \subset O_L$ is a prime lying above $p \subset O_K$ then since $pO_L$ is a product of a bunch of stuff including $q$, applying $\sigma \in \text{Gal}(L/K)$ we get $pO_L = \sigma(pO_L)$ is a product of a bunch of stuff including $\sigma(q)$ and by unique factorization of ideals you get $\sigma(q)$ also lies above $p$, so $\text{Gal}(L/K)$ acts naturally on the primes lying above $p$.
fair, we're actually just reaching this point in our lectures atm, I've seen it before but this is my first proper treatment of it so if I can help I will but if not then I'll have learned smth too hahaha
No I definitely think you can help, but let me see if I can formulate a question first
Look at the residue field $\kappa(p)$ of $p$ in $O_K$ and take a prime lying above, say $q$, and look at the corresponding residue field $\kappa(q)$ in $O_L$. What can we say about the extension $\kappa(q)/\kappa(p)$?
How does $\text{Gal}(L/K)$ relate to $\text{Gal}(\kappa(q)/\kappa(p))$?
well the groups $Z_q$ that I described above are subgroups of $\operatorname{Gal}(L/K)$ and there's a surjection $Z_q \to \operatorname{Gal}(\kappa(q)/\kappa(p))$ (note that this extension of residue fields is defo Galois in the number field case because your residue class field extensions are just finite field extensions)
The question you can ask is what happens at the extremes?
Well by the fundamental equation (and the fact that quadratic extensions are Galois) you get $efg = 2$ and $(2)$ is ramified so $2\cdot f \cdot g = 2$ so $fg = 1$
The kernel of that map is called the inertia group of q, that guy also has a fixed field that sits between the fixed field of the decomposition group and the extension $L$
so you've got like $L/L^{T_\mathfrak{P}}/L^{Z_\mathfrak{P}}/K$
all of the splitting of $\mathfrak{p}$ comes from $L^{Z_\mathfrak{P}}/K$, all of the inertial degree comes from $L^{T_\mathfrak{P}}/L^{Z_\mathfrak{P}}$, and all of the ramification comes from $L/L^{T_\mathfrak{P}}$
Weird stuff. How do you get this surjection map in the first place? $Z_q$ is the set of Galois automorphisms of $L$ which fixes $q$, so I suppose it naturally acts on the local ring $(\mathcal{O}_L)_q$, hence on the residue field
Hm, but I am annoyed, because it feels what I would do is try to "analytically continue" the automorphism of $\kappa(q)$ to all of $O_L$. Maybe this is what the proof does, but hidden in some retarded algebraic language
Which means I have to struggle through the algebra to see it
OK. If $q$ is the unique prime in $O_K$ lying above $p$ in $O_L$ then $Z_q = \text{Gal}(K/L)$. Choose a primitive element $\alpha$ for $\kappa(q)$ over $\kappa(p)$, and essentially observe $\text{Gal}(L/K)$ acts transitively on the roots of the minimal polynomial for $\alpha$ over $K$.
Which gives all the automorphisms for $\text{Gal}(\kappa(q)/\kappa(p))$ as modulo $q$ that minimal polynomial is the minimal polynomial for $\alpha$ over $\kappa(q)$
Alright I have a dumb computational question now: $(p, \frac{1}{2}\sqrt{p})$ is prime in $\mathcal{O} := \Bbb Z[\frac{1 + \sqrt{p}}{2}]$ because $$\mathcal{O}/(p, \frac{1}{2}\sqrt{p}) \cong \Bbb Z[X]/(X^2 - X + \frac{1 - p}{4}, p, \frac{1}{2}X) \cong \Bbb F_p[X]/(X^2 - X + \frac{1}{4}, \frac{1}{2}X) \cong \Bbb F_p$$ I think
$X^2 - X + \frac{1}{4} \equiv (X-\frac{1}{2})^2 \bmod p$
Well $(p, \omega - c)$ is the ideal above $p$ where $c$ is the constant term in the factorsation and $\omega$ is a generator for the ring of integers lol
To get a real ($\sigma$-additive) measure you need what's called an $\omega$-closed filter, but it's a theorem of $\mathsf{ZFC}$ that there is no $\omega$-closed filter on $\Bbb N$ and it's independent of $\mathsf{ZFC}$ whether there is an $\omega$-closed filter on any set
(If there is one then there's a measurable cardinal)
Free ultrafilters are precisely the ultrafilters which contain all cofinite sets. So for any finite set $A\in\mathcal U$ $\Leftrightarrow A\setminus F\mathcal U$.
If you take an ultrafilter and throw away all the subsets $A \subset \mathcal{U}$ such that there is some finite subset $P$ and $Q$ for which $(A \setminus P) \cup Q \in \mathcal{U}$, does that give you an ultrafilter? I think so.
So I am wondering what exactly goes wrong in the construction. Choose an ultrafilter $\mathcal{U}$, define $\mu(A) = 1$ if $A \in \mathcal{U}$ and whenever $B =^* A$, $B \in \mathcal{U}$ as well, using your notation.
And you can say exactly the same thing if you take the dual ideal instead of finite sets. (I.e., the same is true if you say $X\setminus P,X\setminus Q\in\mathcal U$ instead of saying that $P$, $Q$ are finite.
Ok, I'm listening. (Sorry for the digression.) We have $\mu(A)=1$ for the sets from ultrafilter. What next?
Let $K$ be a number field and let $\mathfrak{a}$ be a non-zero ideal of $\mathcal{O}_K$ such that $\mathfrak{a}^m = (a)$ for some $a \in \mathcal{O}_K$.
Does there exist a finite extension $L/K$ such that $\mathfrak{a}\mathcal{O}_L$ is principal?
The answer is yes; one can take $L = K(\sqr...
