Let $M$ be a finitely presented $R$ module and $0 \to N' \to N \to M \to 0 $ be an exact sequence with $N$ a finitely generated $R$ module. Prove that $N'$ is also a finitely generated $R$ module.  By $M$ finitely presented $R$ module we mean there is a ses $0 \to K \to F \to M \to 0$ where $F$ is free of finite rank and $K$ is finitely generated $R$ module. I proved the case when $N$ is free of finite rank but can't show it in this case. Any help will be appreciated. Thanks.
Let $R$ be a ring (unital, not necessarily commutative), $M$ a finitely presented left $R$ module. Suppose $m_1,\ldots,m_n\in M$ generate $M$. This determines a surjection $f:R^n\to M$.     Must the kernel of $f$ be finitely generated?   This is true by definition if the generating set $m_1,\ldots,m_n$ is the one coming from the finite presentation of $M$.
I was able to post in chat with less than 20 rep by joining 20 sites I joined 20 other sites (out of the many) and even with 1 rep was able to post in any of the chatrooms. For example: In ELL, I have just 1 rep and have posted messages in its chat.  ell.stackexchange.com/users/18923/lamart This seems like a bug because I should have needed 20 rep to post. . EDIT: From , I got the idea to make this edit. To reproduce this bug/error:  Create a new SE account. Use it to make an account here. You will get 1 default rep. Choose a random SE site and use this account to get 1 rep on that random site. Repeat the above step 20 times. Now you have 1 rep on 20 network sites but the net SE rep is 20. Go to the ELL chat and ping me. (important step)  From my comment: I have now got 40 rep since people voted this question up. Initially I had net 20 rep gained by creating 20 accounts on 20 different SE sites. Those default 1rep added up to get me to 20rep and I was able to chat. The link says one needs 20 rep in any of the SE site to chat in &quot;all the other&quot; category chatrooms..... but I did not have 20 rep on ELL at that moment. Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$? Annoying Viber notification I understand that the expressions on both sides of an equal sign are the same entity, and I know that when you modify one side, the other must be changed because it is referring to the same thing. What I do not understand is why making a new equation (adding or taking away from an expression) allows one to know what an unknown represents. What about equations lets us do this? Question on how to normalize regression coefficient Not sure if normalize is the correct word to use here, but I will try my best to illustrate what I am trying to ask. The estimator used here is least squares.  Suppose you have $y=\beta_0+\beta_1x_1$, you can center it around the mean by $y=\beta_0'+\beta_1x_1'$ where $\beta_0'=\beta_0+\beta_1\bar x_1$ and $x_1'=x-\bar x$, so that $\beta_0'$ no longer has any influence on estimating $\beta_1$.  By this I mean $\hat\beta_1$ in $y=\beta_1x_1'$ is equivalent to $\hat\beta_1$ in $y=\beta_0+\beta_1x_1$. We have reduced equation for easier least square calculation.  How do you apply this method in general? Now I have the model $y=\beta_1e^{x_1t}+\beta_2e^{x_2t}$, I am trying to reduce it to $y=\beta_1x'$. XeLaTeX font rendering tend to be slightly bolder compared to PDFlatex Group of order $8p$ is solvable, for any prime $p$ Consider the following question:     Show that a group $G$ of order $8p$ is solvable, for any prime $p$.   I am kind of stuck, but here are my first attempts:  I chose the series of subroups $G&gt;H_8&gt;H_4&gt;H_2&gt;H_1=\{e\}$ where $H_k$ has order $k$. All of these subgroups exist due to Sylow's Theorems. We have the quotients $H_8/H_4\cong H_4/H_2 \cong H_2/H_1 \cong \mathbb Z/2\mathbb Z$, so $G&gt;H_8\rhd H_4\rhd H_2\rhd H_1=\{e\}$. Only the factor $G/H_8 \cong \mathbb Z/p\mathbb Z$ is causing me a headache, because although its of prime order (and hence abelian) I don't know whether $H_8$ is normal in $G$ (unless $p=2$).  If $p\ne 2$, then the number $k$ of Sylow $2$-subgroups is $1$ mod $2$. Since $k$ divides $|G|$ we might have $k=1$ or $k=p$. If we had $k=1$ then $H_8$ would be the only Sylow $2$-subgroup (and hence normal in $G$). But how could we show this?  Or is there even an easier way to approach this problem?  $\ $  edit: I am going to try a different approach: The case $p=2$ is clear. When $p=3$ or $p=7$ the group has order 24 or 56 and since the smallest simple non-abelian group has order 60, the group must also be solvable in these exceptional cases (as Mariano pointed out).  In any other case we get a Sylow $p$-subgroup $H_p$ of order $p$, which is is normal in $G$. The quotient $G/H_p$ has order $8=2^3$; and prime power order implies solvable. So $G/H_p$ is solvable and $H_p$ is also solvable. And since the composition factors of $G$ are those of $H_p$ together with those of $G/H_p$, we conclude that $G$ is solvable iff $H_p$ and $G/H_p$ are solvable, which is the case. Hence $G$ is solvable if $|G|=8p$. Upper bound for $\sqrt[n]{n!}$ Light fixture with a switch that has an outlet I hope this explains it better: in this room there is a switch with one set of wires and only the black one went through the hole in it, and another black one in the other hole, and is hooked to the outlet by being wrapped around the screw terminals. The ground is cut short and not grounded.  Now on to the outlet: the wire from the switch is connected to the top screw - but it's on the side that says "white".   There are two more sets of wires that connect to the outlet as follows:   the two black ones are on the the other screw terminal on the bottom (one on the "white" side)  the grounds are grounded to the screw terminal below that  the white ones are connected by the one from the switch to the bottom (black side), with one of the white ones from the other set on the screw terminal, and the one from the other cable is on the top screw terminal on the "white" side.   Now to the light: there are the black and white that come out of the box, and that is the set that has the ground cut short.  The original light fixture was a single 660w/250v rated socket.  The one my friend put up was a fixture with two 660w /250v sockets.  At the light only black and white connected, not ground... It worked then made noise and this is where I'm at.  When I wire test them, the black one from the switch and the white ones from either sets nothing...  but all the other ones light up.....! Survey database design: associate an answer to a user I'm doing the conceptual model for a survey database.  The goal is store the answers given by users (it's going to be an Android app).  I have three entities: user, question and option.  A question will have one or more options (for example: How many employees do you have? 1-40, 40-1000, +1000).  Options will have a text (1-40) and a value (the value selected by user).  User will select one (or more) of these options.  My conceptual design is:    I don't know how to associate an answer with an user.  How can I represent that relation? Do I have another entity to represent option value?  This model will store questions and pre-made answers (offered answers) and allows them to be re-used in different surveys.  I have to represent question like this one:    This question is related to this one: INSERT takes over 5 hours, when inserting more than specific number of rows When inserting less then about 1,350,000 rows to the table it all takes about 2 minutes, however when number of inserted rows is bigger, then the time needed to insert data grows to about 5 hours.  The problem is not connected with query or indexes, because everything has been working fine for a long time and nothing has changed in the structure of query, tables or indexes.   Problem appeared for the first time about 2 weeks ago and it appears repeatedly on days, when number of inserted rows is bigger than +-1,350,000. For example on one day number of inserted rows is 1,200,000 and the process takes 2 minutes, on the other day number of rows is 1,450,000 and it takes 5-6 hours to insert data.  I tried to rebuild indexes, but it has not helped. About number truncation of ticks display in ListPlot I have heard of some methods, but none of them have stuck. Personally I try to avoid complex types in C and try to break them into component typedef.  I'm now faced with maintaining some legacy code from a so called 'three star programmer', and I'm having a hard time reading some of the ***code[][].  How do you read complex C declarations? Convergence/Divergence of infinite series $\sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$ $$ \sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$$  Does it converge or diverge?   Can we have a rigorous proof that is not probabilistic?   For reference, this question is supposedly a mix of real analysis and calculus. Calculationg base-resistor between microcontroller and transistor I'm curious, when connecting my 2N2222 transistor's base to a PWM output pin on my Arduino, I know the Arduino can deliver a maximum of 40mA current. The 2N2222 only requires around 5 to 10 mA on it's base to switch the collector and emitter. Now, when calculating the base-resistor under "normal" conditions where I'm merely switching it directly from Vcc, the following applies:   However, when putting in a 500 ohm or 1K resistor in (1K for 5mA) directly it works great. But putting it in between the Arduino and the transistor, the transistor doesn't switch.  I suspect thus, that the resistance is too high of the base-resistor. Therefore, we may very well not need a base-resistor.  But, according to spec, the Arduino can deliver 40mA. So, am I correct in assuming the folling: 1) I should calculate the perceived resistance that the Arduino is giving, if it goes high to 5Volt and only gives 40mA, and then 2) Subtract that from the 500ohm resistor needed to make the transistor operate optimally.  Example:    So, I'm theorising that I have to then put in a 375 ohm resistor in to make the total resistance between the Arduino and the base resistor 500 ohm.  Somehow this doesn't feel right. Keep in mind, I'm not an electronic engineer, so I might have the cat by the tail here :P Hide the term id on autocomplete widget Chen weibull distribution How to choose the default gcc and g++ version? I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help! Is a constant condition more costly than switching shaders? In general, branching in shaders is not a good idea. But now I have a shader with a condition that is constant with respect to the entire draw call. So the branch that is executed is always the same for one draw call.  Is such kind of branching still more costly than having multiple shaders without these branches and switch between them?