I have read questions related to $\mathbb{Z}$-module and exact sequence of it.  I have a question about generalising this result (or finding counter example)  Let $R$ be an integral domain and $0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$ be an exact sequence of finitely generated $R$-module.  Then, is $rank(M)=rank(M_1)+rank(M_2)$ true?  (Definition of rank: $rank(M)$ denotes the cardinality of the largest linearly independent subset of $M$)  I have read statements online and I found that "flat modules" satisfy this result (though I have not learnt tahings about maps and tensor product so I cannot effectively understand all proofs), but there are no general result about integral domains.  Hence, I would like to ask this question here and seek for help.    Thank you for your kind help!  (I have stuck in this for half a week and I am not making any progress. The current state of my  work is that:  I have (kind of?) proved that $Rank(M)\geq Rank(M_1)+Rank(M_2)$ by putting their linear independent sets together and prove that it is still linearly independent.    I have proved that the statement is true if $Rank(M)-Rank(M_1)=0$ or $1$.  The proof of $0$ is by torsion module, and the proof of $1$ is proved by hardcoding (but it cannot be generalised to $2$ or larger.  )
Suppose that:  $0 \rightarrow M_{1} \rightarrow M \rightarrow M_{2} \rightarrow 0$  Is an exact sequence of $R$-modules (for $R$ commutative integral domain). Show that $\mathrm{rk}(M) = \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$    I have shown the inequality $\mathrm{rk}(M) \geq \mathrm{rk}(M_{1})+\mathrm{rk}(M_{2})$, however for the opposite direction I'm lost. I tried to take $N$ as a linear independent subset and tried to decompose it into $N_{1}$ (the elements in $\mathrm{Ker}(g)$) and $N_{2}$ (the elements not in $\mathrm{Ker}(g)$), but I haven't got far, since $N_{1}$ gives us a linearly independent set in $M_{1}$, however $N_{2}$ can give us a linearly dependent set in $M_{2}$.   We define $\mathrm{rk}(M)$ in this case as the supremum over all cardinalities of the linearly independent subsets of $M$ (where the only finite linear combination equaling $0$ implies trivial coefficients for $r \in R$).  Would this be the proper way to show the result? I think I'm missing something important.
Is $\sin^2(z) + \cos^2(z)=1$ still true for $z \in \Bbb{C}$? Is $$\sin^2(z) + \cos^2(z)=1$$ still true for all $z \in \Bbb{C}$? I've tried rewriting it using complex definitions of $\sin$ and $\cos$, and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold? Young Adult Fantasy about children living with an elderly woman and learning magic related to their skills I have heard that most spam filters use a reverse DNS lookup on the originating IP address of the sending SMTP server (or something like that) as part of the verification process.  I want to make sure my emails will pass this check.  How exactly can I verify that my emails are OK?  In researching it, I've noticed that if I look at the header of a valid email message coming from Gmail, I will see something like this:  Received: from mail-pz0-f185.google.com (mail-pz0-f185.google.com [209.85.222.185])         by mx.google.com with ESMTP id 10si2346996pxi.164.2009.07.22.22.26.19;         Wed, 22 Jul 2009 22:26:19 -0700 (PDT)   When I do an nslookup on 209.85.222.18, it responds with mail-pz0-f185.google.com, which is what I would expect.  If it responded with a different domain, then I assume that would indicate a problem and it would fail the check.  Is this what the reverse DNS lookup is all about or am I barking up the wrong tree? Synchronize/change Area51 profile image I just visited  and noticed that my profile picture is not the same as on all other sites.  According to  the image will be synchronized when synchronizing the complete profile. I did that but the image is still the same.  Then I wanted to change the image in the Area51 profile directly - but I can only change to a gravatar image and can't use the new and fancy Stack Exchange image.  Is that a bug? Converting point layer to raster grid that shows frequency of points per cell uisng QGIS I have a point layer that shows the distribution of a species.   Is it possible in QGIS to convert this point layer to a raster grid in which the value of each cell corresponds with the amount of points within that cell?   Until now, I only managed to attach attribute values to the grid cells. I have already added a column in my attribute table that has the value '1' for each point, hoping that there is a way to use a sum of attributes for each grid cell. I'm using soul to letterspace headings in a Hungarian document and compiling it with lualatex.  For some reason soul drops some of the accented letters at the end of a word, but not all. In the example below, only ő and ű are affected:  \documentclass{minimal} \usepackage{fontspec} \setmainfont[Ligatures = TeX]{TeX Gyre Pagella} \usepackage{soul} \begin{document} \noindent \so{áll llá ÁLL LLÁ}\\ \so{íll llí ÍLL LLÍ}\\ \so{űll llű ŰLL LLŰ}\\ \so{őll llő ŐLL LLŐ}\\ \so{üll llü ÜLL LLÜ}\\ \so{öll llö ÖLL LLÖ}\\ \so{úll llú ÚLL LLÚ}\\ \so{óll lló ÓLL LLÓ}\\ \so{éll llé ÉLL LLÉ} \end{document} OS: UBUNTU 18.04  I had installed Oracle Java 8 earlier. But decided to switch to open JDK 11. it's successfully installed. java -version openjdk version "11.0.2" 2019-01-15 OpenJDK Runtime Environment (build 11.0.2+9-Ubuntu-3ubuntu118.04.3) OpenJDK 64-Bit Server VM (build 11.0.2+9-Ubuntu-3ubuntu118.04.3, mixed mode, sharing)  Then I set the JAVA_HOME sudo nano /etc/environment  File changes: PATH="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/$ JAVA_HOME="/usr/lib/jvm/java-11-openjdk-amd64/"  Then executed: source /etc/environment  echo $JAVA_HOME /usr/lib/jvm/java-11-openjdk-amd64/  But the moment I exit the terminal and then launch the terminal again: I get this output:  ▶ echo $JAVA_HOME /usr/lib/jvm/java-8-oracle  which you can see is not what is expected. Please help me out Java Array, Finding Duplicates Grass tools not working in QGIS 2.14.3-ESSEN, OSGEO4W Expectation of random variables ratio How to combine various armatures? How to calculate Camera "correct" position and distance to Frame all my Scene? Conditional probability and venn diagram I know that there are questions asking like "intersection of a infinite collection of sets" and I can understand that the answer for that one is a null set, but I got a question here, in which all sets are finite and nonempty. Please take a look at the pic below.  It's a True/False question.  What I'm confused about is that the "infinity" sign in the interception. Since all sets are finite, does it mean that there are many sets in the chain are the same? (because the symbol used is for subsets not for proper subsets). Please give me some hints how I should think about this question.  Thank you Reason for the discreteness arising in quantum mechanics? What is the most essential reason that actually leads to the quantization. I am reading the book on quantum mechanics by Griffiths. The quanta in the infinite potential well for e.g. arise due to the boundary conditions, and the quanta in harmonic oscillator arise due to the commutation relations of the ladder operators, which give energy eigenvalues differing by a multiple of $\hbar$. But what actually is the reason for the discreteness in quantum theory? Which postulate is responsible for that. I tried going backwards, but for me it somehow seems to come magically out of the mathematics. I've never really got how chmod worked up until today. I followed a tutorial that explained a big deal to me.   For example, I've read that you've got three different permission groups:   owner (u) group (g) everyone (o)   Based on these three groups, I now know that:   If the file is owned by the user, the user permissions determine the access. If the group of the file is the same as the user's group, the group permission determine the access. If the user is not the file owner, and is not in the group, then the other permission is used.   I've also learned that you've got the following permissions:   read (r) write (w) execute (x)   I created a directory to test my newly acquired knowledge:  mkdir test   Then I did some tests:  chmod u+rwx test/ # drwx------ chmod g+rx test/ # drwxr-x--- chmod u-x test/ # drw-r-x---   After fooling around for some time I think I finally got the hang of chmod and the way you set permission using this command.    But...  I still have a few questions:   What does the d at the start stand for? What's the name and use of the containing slot and what other values can it hold? How can I set and unset it? What is the value for this d? (As you only have 7=4+2+1 7=4+2+1 7=4+2+1) Why do people sometimes use 0777 instead of 777 to set their permissions?   But as I shouldn't be asking multiple questions, I'll try to ask it in one question.  In UNIX based system such as all Linux distributions, concerning the permissions, what does the first part (d) stand for and what's the use for this part of the permissions? I have a requirement where a certain field can be edited without the validations firing for a particular user. Now I tried to write rule like LastModifiedId != '005000000000000AAA' but the validation was still firing. Then I realized that until the record saves, the LastModifiedId does not change and the validations still fire, as the LastModifiedId would not be updated to the exempted User.   Is there any workaround or a way to check the Id of the User who is currently editing the record? Show that if $a \equiv b \pmod n$, $\gcd(a,n)=\gcd(b,n)$ I've been wondering, what is the general form for the integral of $\sin(\cos x)$? I tried integrating it myself and couldn't, tried with Wolfram Alpha and it couldn't either. If you think about it conceptually, it should have a general form as it's graph if just like sin and cos, which both have general forms. If it doesn't have a general form, I'd like to know why. Thanks! SQL server Read Only permission automatically for new databases