Let $K = \Bbb Q(\alpha)$ be an algebraic number field of degree $n$ where $\alpha$ is an algebraic integer, and suppose the minimal polynomial for $\alpha$ satisfies the Eisenstein's criterion with a prime $p$. Then if for some $f(x) \in \Bbb Z[x]$ of degree $n-1$, $f(\alpha) \equiv 0 \pmod{p\mathcal{O}_K}$ then $f(x) \in p\Bbb Z[x]$.
Eisenstein basically says $\alpha^n \equiv 0 \pmod{p\mathcal{O}_K}$, so consider $\alpha^{n-1}f(\alpha) \equiv 0 \pmod{p\mathcal{O}_K}$. This forces $\alpha^{n-1}f_0 \equiv 0 \pmod{p\mathcal{O}_K}$, where $f_0$ is the initial coefficient of $f$.
So $\alpha^{n-1} f_0 = p \beta$ for some $\beta \in \mathcal{O}_K$. Take norm to get $N_K(\alpha)^{n-1} f_0^n = p^n N_K(\beta)$.
We also know by Eisenstein that $N_K(\alpha)$ is divisible by $p$ exactly once, so $p$ has to divide $f_0$.
Continue similarly
Essentially it's saying if a polynomial is zero on $\mathcal{O}_K/p\mathcal{O}_K$ then it's zero as a polynomial over $\Bbb Z/p\Bbb Z$, I suppose.
Which is not true in general; consider simply $x^p - x$ as a polynomial which is zero on $\Bbb Z/p\Bbb Z$ but is not zero as a polynomial. The Eisenstein criterion is saying when this is true?
@LeakyNun Apparently this says, in the same setup, $p$ doesn't divide $|\mathcal{O}_K : \Bbb Z[\alpha]|$. This is because if $\beta$ is some element of $p$-torsion in $\mathcal{O}_K/\Bbb Z[\alpha]$, then $p\beta \in \Bbb Z[\alpha]$. Write $\beta = r_0 + r_1\alpha + \cdots + r_{n-1}\alpha^{n-1}$ for $r_i \in \Bbb Q$. One of the $r_i$'s is not an integer as $\beta \notin \Bbb Z[\alpha]$, yet $pr_i \in \Bbb Z$.
So $p$ divides the denominator of each $r_i$. Write $r_0 + r_1\alpha + \cdots + r_{n-1} \alpha^{n-1}$ as a single fraction, $(f_0 + f_1 \alpha + \cdots + f_{n-1} \alpha^{n-1})/c$ where $p$ divides $c$. This belongs to $\mathcal{O}_K$, so $f_0 + \cdots + f_{n-1}\alpha^{n-1} \in p\mathcal{O}_K$
But this forces each $f_i$ to be divisible by $p$ by earlier, which is contradiction, because we normalized the fraction.
We can use this to prove the ring of integers of $K = \Bbb Q(\sqrt[3]{2})$ is $A = \Bbb Z[\sqrt[3]{2}]$. Because $2^2 3^3 = 108 = \text{disc}(A) = |\mathcal{O}_K : A| \text{disc}(A)$, and neither $2$ nor $3$ divides $|\mathcal{O}_K : A|$ as the minimal polynomial here is $x^3 - 2$ which is Eisenstein at $2$ and a linear change $(x - 1)^3 - 2$ is Eisenstein at $3$.
Wikipedia says that nikola tesla considered women to be superior in everyway to men, but it also says that in his elderly years he claimed to love a injured pigeon "The way a man would love a woman" I mean, if he held women in such high regard then why was he nailing a trash dove
@LeakyNun So $x^4 + 10x^2 + 1$ is also an example. The splitting field is $\Bbb Q(\sqrt{-2}, \sqrt{-3})$ which has discriminant $2^6 \cdot 3^2$, so $2$ and $3$ are the only primes which could be ramified.
How would I show by hand that neither has full ramification index?
So mod $2$ that's $(x + 1)^4$ and mod $3$ that's $(x + 1)^2(x - 1)^2$
Let me think why this helps (don't tell)
Rambling out loud: If $2$ totally ramifies then $(2) = \mathfrak{p}^4$ in $\mathcal{O}_K$.
If it was a PID I would be able to say $\mathcal{O}_K/2\mathcal{O}_K$ has $4$-torsion
If $\alpha = \sqrt{-2} + \sqrt{-3}$, then $\alpha^4 +10\alpha^2 + 1 = 0$, so reducing mod $2\mathcal{O}_K$ on both sides says $(\alpha+1)^4 \in 2\mathcal{O}_K$ I guess
Although you have a similar trick if $\mathfrak{p} + \mathcal{F}_\theta = 1$ where $\theta$ is a primitive element for your number field and $\mathcal{F}_\theta$ is the conductor
Usually professors refuse to supervise a thesis just because they don't have time (too many students already or they just have stuff to do), apart from that they won't eat you @ÍgjøgnumMeg
@Alessandro yeah, he often supervises master theses though (I've heard) and I don't know how many students he currently has, also Mathein mentioned that he's a really nice guy
The stuff you guys discuss is much more advanced than I have ever learnt, and judging from what you are saying you are all very qualified, are my problems just too boring for people at your level to consider attempting? (I'm not going to be offended)
no but things like field theory and rings etc like I have been reading the material for the past few years but it's a long way off being second nature do you understand what I mean? Like I know that it's all very relevant somehow to the things I work on and the questions I post, but you guys seem to be doing this with ease, that's why I ask
like I have to put things in first order sometimes and im not sure if its a good habit to get into or its obsessive compulsive disorder, but it's generally how I understand something, like this morning I had to get Lagrange's theorem out in formal and I known its ridiculously long, but the statements on the wiki didn't make the point that it makes no sense unless you are doing arithmetic modulo p
like the upper bound on the number of incongruent solutions only holds when you restrict $x$ and $y$ to ${\{0,1,2,...,p-1}\}$
in other words
put it this way, if that MIT assignment was due this week, and I was in that class, i would attempt one of the warm ups, then go on a 3 month bender in which i am permanently banned from campus then be able to get a decent score on it 3 years later
ergh "throughout this whole thesis we assume the conjectured finiteness of the Tate-Shafarevich group since this is crucial to computing the coefficients of blah blah"
@Adam The stuff that I have seen you post is usually one of 1) too unrigorous to even interpret, 2) contrived and unmotivated, 3) really basic. I think you should just work through the exercises in an algebra textbook like Dummit and Foote?
Ok thanks Ted is there a chance you submit an answer to one for me?
or perhaps give an example where rigor was lacking, and provide an example of what you would expect to make the question answerable or even interpretable?
But sure I will look for the text book you mentioned and go through some exercises
I'm not sure your second point is worth any salt tho, kinda seems like the type of point motivated by sporadic thyroid
In my Linear Algebra class, my professor talked about something called "Slotovi slices" which are these matrices that generate all characteristic polynomials. Does anyone know where I can find more about this? I tried Googling "Slotovi slice," but found nothing. Also, I think I am probably misspelling "Slotovi" which may be the reason why I can't find anything on it.
Basically, the way my professor explained it is that a "Slotovi slice" is a set of $n\times n$ matrices such that each matrix in the set has a unique characteristic polynomial, and that for every polynomial of degree $n$ with leading coefficient of $1$, there is a matrix in the set with that polynomial as its characteristic polynomial.
Let $G$ be a locally compact group. The union of countably many open $\sigma$-compact subgroups of $G$ generates an open $\sigma$-compact subgroup.
For each $n \in \Bbb{N}$, let $L_n = \bigcup_{j=1} K_{n,j}$ be an open subgroup, where $K_{n,j}$ is compact. I was able to prove that $\langle \...
@TedE ah buddy your initial statement carries a requisite of having classified the content I have posted don't you have bongs to smoke with mark Walberg?
Let $G$ be a locally compact group. The union of countably many open $\sigma$-compact subgroups of $G$ generates an open $\sigma$-compact subgroup.
For each $n \in \Bbb{N}$, let $L_n = \bigcup_{j=1} K_{n,j}$ be an open subgroup, where $K_{n,j}$ is compact. I was able to prove that $\langle \...
