A Deeper Explanation of this Proof (or Alternative Proof) On the Maximality of Valuation Rings

I’m trying to see if there is a proof of the following theorem using Galois theory, or number theory of some kind. Specifically, I’m looking at (iii) implies (i) below.

Theorem: Let $ K$ be a field and let $ \mathcal{O}$ be a subring. Let $ l$ be an algebraically closed field, and let $ \phi : \mathcal{O} \rightarrow l$ be a map into $ l$ . Then, for any $ a \in K^\times$ , there is either an extension of $ \phi$ from $ \mathcal{O}[a]$ into $ l$ or an extension of $ \mathcal{O}[a^{-1}]$ into $ l$ .

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My intuition for the theorem is as follows. For any subring $ \mathcal{O}$ of $ K$ , we have a partially ordered abelian group $ K^\times / \mathcal{O}^\times$ , where the partial order on $ K^\times$ is given by declaring $ x \leq y$ when there is $ a \in \mathcal{O}$ such that $ x a = y$ , and the partial order on $ K^\times / \mathcal{O}^\times$ is given by declaring $ xU \leq yU$ when there exists $ a \in A$ such that $ xa \mathcal{O}^\times = y \mathcal{O}^\times$ .

This theorem is sort of saying that we can add at least one of the relations $ 0 \leq a$ and $ 0 \geq a$ in $ K^\times / \mathcal{O}^\times$ (mod out by the convex subgroup they generate), collapsing it so that it now corresponds to the ring $ \mathcal{O}[a]$ or $ \mathcal{O}[a^{-1}]$ . A corollary is that we can get $ K^\times / \mathcal{O}^\times$ to be totally ordered this way.

What is going on in this proof? Maybe it could be clarified with some higher number theory or Galois theory.

Alternative to displaying multiple plan columns

I’m working on a pricing page that has 6 different plans. What is the optimal way of displaying them considering the limited amount of space we have?

Mobile isn’t an issue since they will just be stacked, but on Desktop, our space is limited to around 1200px.

What are some suggestions you might have to display all the plans and make sure each one of them is discoverable?

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Null and alternative Variables

My task for the 2 following problems is to identify the null value and alternative value:

Question 1: Perform hypothesis test for population proportion

It is known that 29% of the months have rice production by Company1 above 307200 cwt (company1>307200.0). Is there sufficient evidence to suggest that the proportion is less than 29%? Test at 5% level of significance.

My answer:

Null = .29

alternative = .01

Question 2: Perform hypothesis test for population mean

It is claimed that average rice production by Company2 is 209,500 cwt (Company2 = 209500.0). Test this claim using a hypothesis test at 1% level of significance.

My answer:

Null = 209500.0

alternative = .01

I’m being told that these my answers are not correct. I have no idea why. Any help would be appreciated!

Advice on using this alternative method of finding the Laurent expansion of $\tfrac{1}{z^2(z-1)}$

Say we want to calculate the Laurent series of $ \tfrac{1}{z^2(z-1)}$ about $ z_0=1.$ Now I know that one way to do it is to say that $ f(z)=\tfrac{1}{z^2}(\tfrac{1}{z-1})$ and appy the geometric series expansion to the brackets term. But I wanted to try and do it a different way :

First we split f into partial fractions and compute the Laurent series separately.Now consider the Laurent expansion of $ \tfrac{1}{z^2}$

We know that $ 0$ is a pole of order 2 which implies that $ \forall n>2,a_{-n}=0$ . Therefore $ \tfrac{1}{z^2}$ haa series expansion $ \tfrac{a_{-2}}{(z-z_0)^2}+\tfrac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+…$

Now to calculate the the $ a_{-2}$ coefficient I applied the following trick

$ a_n=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{f(z)}{(z-z_0)^{n+1} }dz=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+3}}dx=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{z^{n+3}}dz$

$ a_{n-2}=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+1} }dz=\tfrac{f^n(z_0)}{n!}$ But as n $ f(z)=1$ this implies that $ n$ must be zero and so $ a_{-2}=1$

Now when I tried to use the same trick on $ a_{-1}$ It doesn’t work because now we can’t use Cauchy’s formula. Also when I tried to use u substitution by letting $ u=z-z_0$ it returns that $ a_{-1}=-\tfrac{1}{z}$ but I know this is not rue as I know from the method that I mentioned in the first paragraph that $ a_{-1}=1$ So does anyone have any suggestions on how I can find $ a_{-1}$ continuing with the method I’m trying to use ?

Looking for a free, non-email-oriented alternative to gnupg for bulk encryption/decryption of individual files [on hold]

I want to implement a system for bulk-encrypting individual files (as opposed to entire disks), and I would like to use a free, open-source tool to do this.

My first thought was to use gnupg for this, but after reading more about it I find that its design is too strongly geared towards the use-case of encrypting/decrypting emails, which is very different from the use-case I have in mind. As a result, as I read the documentation, I keep finding myself reading about considerations that are utterly irrelevant (e.g. web of trust, etc.) to what I want to do, and conversely, being unable to find information about the things I do want to do.

This is a clear sign that gnupg, whatever its merits for what it was designed to do, is just not the right tool for my problem.

What would be more a more appropriate tool?

My two main requirements are it be open source, and that it can be suitable to implement an unsupervised (i.e. non-interactive) encryption/decryption pipeline. In other words, I am looking for something I can write programs with, not something I use interactively, since the goal is to use the program to encrypt thousands of individual files.

multiple IF statements with between number ranges alternative

How to efectively deal with multiple IF-statements and with multiple “between number” ranges?
(e.g. C column based on B column)

9
with a rule set as follows:

01-10 11-20 21-50 51-100 101-150 151-200 201-300 301-500 501-800 801-1000 1001-1222 1223-1568 1269-1800 etc... 

IF()

=IF(AND(B2>=1;    B2<10);     "01-10";   IF(AND(B2>=11;   B2<20);     "11-20";   IF(AND(B2>=21;   B2<50);     "21-50";   IF(AND(B2>=51;   B2<100);    "51-100";  IF(AND(B2>=101;  B2<150);   "101-150";   IF(AND(B2>=151;  B2<200);   "151-200";   IF(AND(B2>=201;  B2<300);   "201-300";  IF(AND(B2>=301;  B2<500);   "301-500";   IF(AND(B2>=501;  B2<800);   "501-800";   IF(AND(B2>=801;  B2<1000);  "801-1000";  IF(AND(B2>=1001; B2<1222); "1001-1222";   IF(AND(B2>=1223; B2<1568); "1223-1568";   IF(AND(B2>=1569; B2<1800); "1569-1800";)))))))))))))

  • disadvantages of a standard IF statement nestings are:
    • long syntax
    • per single row dependency
    • tricky post-editing

IFS()

=IFERROR(ARRAYFORMULA(  IFS(B2:B>=1569; "1569-1800";      B2:B>=1223; "1223-1568";      B2:B>=1001; "1001-1222";       B2:B>=801;   "801-1000";      B2:B>=501;   "501-800";      B2:B>=301;   "301-500";      B2:B>=201;   "201-300";      B2:B>=151;   "151-200";      B2:B>=101;   "101-150";      B2:B>=51;     "51-100";      B2:B>=21;     "21-50";      B2:B>=11;     "11-20";      B2:B>=1;      "01-10"));)

  • disadvantages of IFS statement are:
    • long syntax
    • tricky post-editing
    • reverse order logic dependency