A backup of a deleted PSQ : https://math.stackexchange.com/q/3955443/290189

OP : irfanmat

It has a detailed answer by Atticus Stonestrom. It’s pity that his post got deleted. As there’s no reason for undeletion, I’m posting it here so as to preserve the contents.

Question body

Is there a semi-simple and projective but not injective module? I will be glad if you help.

Response(s)

In the non-commutative case, the answer is yes. Consider $R$ the ring of upper triangular $2\times 2$ matrices over a field $F$, and denote by $e_{ij}$ the element of $R$ with the $ij$-th entry equal to $1$ and all other entries equal to $0$. We can decompose $R$ as a direct sum of left ideals $$Re_{11}\oplus (Re_{12}+Re_{22})=Re_{11}\oplus Re_{22},$$ so let $M=Re_{11}$. Then $M$ is clearly simple, and – as a direct summand of the free module $R$ – is also projective. However, $M$ is not injective; consider the map $f:Re_{11}\oplus Re_{12}\to M$ taking $e_{11}$ and $e_{12}$ to $e_{11}$. $Re_{11}\oplus Re_{12}$ is a left ideal of $R$, but there is no way to extend $f$ to a map $R\rightarrow M$, so this gives the desired example.

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I wanted to start a meta question, but I don’t see a point of that after viewing some related posts listed at the end of the next subsection.

Intended question

You may vote on your preferred way.

Related questions:

The Math.SE question $2\cos(2x) - 2\sin(x) = 0$ has attracted several answers from high-rep users.

I am expanding @rhombic’s comment into an answer.

$$ \begin{aligned} 2\cos(2x)-2\sin(x)&=0 \\ 2 - 4\sin^2(x)-2\sin(x)&=0 \\ 2\sin^2(x)+\sin(x) - 1&=0 \\ (2 \sin(x) - 1)(\sin(x) +1) &= 0 \\ \sin(x) = \frac12 \text{ or } \sin(x) &= -1 \\ x = \frac{\pi}{6}, \frac{5\pi}{6} \text{ (rejected) or } & \frac{3\pi}{2} \text{ (rejected)} \end{aligned} $$

Goal

To create a copy button for my Math.SE comment template in order to save the trouble of copying and pasting.

My first attempt

I put the boilerplate inside a Markdown codeblock to prevent them from getting interpreted by Hugo’s Markdown parser. Under each codeblock, I placed the copy button.

Comment boilerplate goes here ...

📝

Another comment boilerplate goes here ...

📝

…

My page’s original layout

$(document).ready(function() {
  $('.copyBtn').click(function() {
    copy($(this).prev().children())
j  });
});

function copy(selector) {
  var screenTop = $(document).scrollTop();
  var $temp = $("<div>");
  $("body").append($temp);
  $temp.attr("contenteditable", true)
       .html($(selector).html()).select()
       .on("focus", function() { document.execCommand('selectAll',false,null) })
       .focus();
  document.execCommand("copy");
  $temp.remove();
  $('html, body').scrollTop(screenTop);
}

static/js/copyBtn.js at Git tag copyBtn0

Update: The question has been reopened.

I intended to answer 김종현’s problem on Math.SE. However, the programs in the question body aren’t typeset in MathJax. As a result, I downvoted and closed this question because found it unclear. From the proposed dual, it seems that I shouldn’t interpret the primal as a linear program. Anyways, without further clarifications from OP, I found no reason to look at this further. Here’s my intended answer:

I intend to post this for a Borel-Cantelli lemma exercise on Math.SE.

The target event is ${\exists i_0 \in \Bbb{N} : \forall i \ge i_0, X_i = 1}$, whose complement is

$$ {\forall i_0 \in \Bbb{N} : \exists i \ge i_0, X_i > = 0} = \limsup_i {X_i = 0}. $$

To apply Borel-Cantelli, one has to determine whether $\sum_i P(X_i = 0)<+\infty$.

My intended answer to a weak LLN problem on Math.SE.

Problem: Suppose $(X_n)$ is a sequence of r.v’s satisfying $P(X_n=\pm\ln (n))=\frac{1}{2}$ for each $n=1,2\dots$. I am trying to show that $(X_n)$ satisfies the weak LLN.

The idea is to show that $P(\overline{X_n}>\varepsilon)$ tends to 0, but I am unsure how to do so.

My solution: As in the accepted answer in OP’s previous question https://math.stackexchange.com/q/3021650/290189, I’ll assume the independence of $(X_n)$. By Chebylshev’s inequality,

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I intended to answer Maddle’s $p$-test question, but T. Bongers has beaten me by two minutes, so I posted my answer here to save my work.

The problem statement

This is the sum: $$\sum\limits_{n=3}^\infty\frac{1}{n\cdot\ln(n)\cdot\ln(\ln(n))^p}$$ How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all.

Unpublished solution

The integral test will do.

When $p \ge 1$, the improper integral diverges. When $p < 1$, it converges.