Part (iv) of exercise #2 for chapter 1 in Atiyah and Macdonald's book  asserts that if $f, g \in A[x]$ are primitive then $fg$ is primitive. We know that this is true when the coefficient ring $A$ is a UFD in which case it is the statement of Gauss' lemma. I have tried hard to show it for arbitrary ring as the problem says but did not manage to do that. So my question is: Though I do not doubt the author, is Gauss lemma really valid for arbitrary commutative rings with unity and can I have some hint?
I'm having troubles with one of the problems in the book Introduction to Commutative Algebra by Atiyah and MacDonald. It's on page 11, and is the last part of the second question.  Given $R$ a commutative ring with unit. Let $R[x]$ be the ring of polynomials in an indeterminate $x$ with coefficients in $R$. We say that a polynomial $f = \sum\limits_{i=0}^n r_ix^i \in R[x]$ (with coefficients $r_0, r_1, \ldots, r_n$) is primitive if $\langle r_0,r_1,...,r_n\rangle = R$, i.e., the ideal generated by the coefficients of $f$ is $R$. Let $f$ and $g$ be two polynomials in $R[x]$. Prove that $fg$ is primitive iff $f$ and $g$ are both primitive.  The $\Rightarrow$ part is easy. Say, $f = \sum\limits_{i=0}^n r_ix^i$, $g = \sum\limits_{i=0}^m s_ix^i$; then $fg = \sum\limits_{i=0}^{m+n} c_ix^i$, where $c_k = \sum\limits_{i + j = k}r_is_j$. Since $fg$ is primitive, there exists a set of $\{\alpha_i\} \subset R$, such that $\sum\limits_{i=0}^{m+n} \alpha_ic_i = 1$, to prove $f$ is primitive, I just need to write all $c_i$'s in terms of $r_i$'s, and $s_j$'s, then group all $r_i$ accordingly, rearranging  it a little bit, and everything is perfectly done. And the proof of the primitivity of $g$ is basically the same. The $\Leftarrow$ part is just so difficult. Say $f = \sum\limits_{i=0}^n r_ix^i$, $g = \sum\limits_{i=0}^m s_ix^i$ are both primitive, then there exists $\{\alpha_i\}; \{\beta_i\} \subset R$, such that $\sum\limits_{i=0}^{n} \alpha_ir_i = 1$, and $\sum\limits_{i=0}^{m} \beta_is_i = 1$. At first, I thought of multiplying the two together, but it just didn't work. So, I'm stuck since I cannot see any way other than multiplying the two sums together. I hope you guys can give me a small push on this. Thanks very much in advance, And have a good day.
Linear transformation of a binomial random variable? Let $X$ be a binomial random variable with parameters $n$ and $p$. Next, let: $$ Y = aX + b $$ I know that: $$ \mathbb{E}[Y] = \mathbb{E}[aX+b] = a\mathbb{E}[X] + b = anp + b \\ \text{Var}(Y) = \text{Var}(aX+b) = a^2 \text{Var}(X) = a^2np(1-p) $$ But how is $Y$ distributed? Turn off paragraph indention for a single line If shuffle-playing playlist &times;100 resulted in [10 13 10  3  2  2] different songs being repeated [1 2 3 4 5 6] times, what is the estimate for the total number of songs? (assuming shuffle play was completely random)  Update: (R code)  k &lt;- 50 # k number of songs on the disk indexed 1:k n &lt;- 100 # n number of random song selections m &lt;- 20 # m number of repeat experiments colnum &lt;- 10 mat &lt;- matrix(data=NA,nrow=m,ncol=colnum) df &lt;- as.data.frame(mat) for(i in 1:m){   played &lt;- 1+floor(k*runif(n)) # actual song indices (1:k) selected    freq &lt;- sapply(1:k,function(x){sum(played==x)})   # = number of times song with index x is being played   histo &lt;- sapply(1:colnum,function(x){sum(freq==x)});   for(j in 1:colnum){    df[i,j] &lt;- histo[j]   } } df   Resulting in: e.g. 20 distributions (V1=number of single plays, V2=number of double plays, etc):     V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 1  15 13 11  1  2  2  0  0  0   0 2  15 12 10  1  4  0  1  0  0   0 3  12 14  7  6  3  0  0  0  0   0 4  17 16  6  4  2  0  1  0  0   0 5  17 10 12  5  0  0  1  0  0   0 6  13 15 11  6  0  0  0  0  0   0 7  10 14  9  3  2  1  1  0  0   0 8  12 17  5  6  3  0  0  0  0   0 9   9 19  8  3  1  2  0  0  0   0 10 13  9 11  6  1  0  1  0  0   0 11 16  9 12  5  2  0  0  0  0   0 12 15  9 11  6  2  0  0  0  0   0 13 19  9  7  4  4  1  0  0  0   0 14 17 11  4  7  3  1  0  0  0   0 15 11 20  8  1  3  1  0  0  0   0 16 14 12 10  5  0  2  0  0  0   0 17  9 12  8  7  3  0  0  0  0   0 18 10 15  9  4  2  0  1  0  0   0 19 14 11 12  7  0  0  0  0  0   0 20 16 14 11  3  1  1  0  0  0   0   Now I need to get from here to the Poisson modelling--my R is a bit rusty (?lmer)...--Any help would be appreciated...  Attempted Poisson modelling: disappointing fit?!  plot(1:colnum,df[1,1:colnum],ylim=c(0,30),    type="l",xlab="repeats",ylab="count") for(i in 1:m){   clr &lt;- rainbow(m)[i]   lines(1:colnum,df[i,1:colnum],type="l",col=clr)   points(1:colnum,df[i,1:colnum],col=clr) }  df.lambda=data.frame(lambda=seq(1,5,0.1),ssq=c(NA));df.lambda for(ii in 1:dim(df.lambda)[1]){   l &lt;- df.lambda$lambda[ii]       ssq &lt;- 0       for(i in 1:20){         for(j in 1:10){           ssq &lt;- ssq + (df[i,j] - n*dpois(j,l))^2         }       }       print(ssq)       df.lambda$lambda[ii] &lt;- l   df.lambda$ssq[ii] &lt;- ssq     }     df.lambda     lambda.est &lt;- df.lambda$lambda[which.min(df.lambda$ssq)] lambda.est # 2.4 points(x &lt;- 1:10, n*dpois(1:10,lambda.est),type="l",lwd=2)  100*dpois(1:10,3) n/lambda.est   the estimated lambda stays around 2.3, with an n estimate of around 43; the fitted curve seems very discrepant, and seems to worsen with rising n !?  Doesn't this have to do with the fact, that our repeats are different from the 'classical' Poisson distributions: it's not just ONE event that repeats itself x number of times, but the sum of repeats of different items (songs)?! How can I tell whether a circle and a rectangle intersect in 2D Euclidean space? (i.e. classic 2D geometry) On which Stack Exchange site can I ask questions like        When will [some new feature] for [some product] be released?   Where can I download an update for [some product]?   I'm getting errors when accessing the customer portal for [some company] – how can I fix this?   Why was I banned for doing [some action] on [some platform]?   Why is [some company] discontinuing [some product]?   Why is [some company] giving away [some product] for free?   What's the price for [some product]?      or any other product support/customer service questions?  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In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes.    Can someone give a more down-to-earth description of what this means?  In addition, some examples of how "probability" and "likelihood" disagree would be nice. How to normalize my db schema for a quiz application? I am putting together a quiz application.   (I'm calling each quiz a "challenge")  I have a "challenge" table and a "challenge_question" table which is pretty straightforward.  challenge table: id name num_questions (I want to randomize and only display a certain number of Qs per each challenge)  challenge_question table: id challenge_id question (text of the question) required (some questions will be required to be displayed per each challenge) question_type (here I indicate what type of question: Mult. choice, T/F, Ordering, or Fill-in-the-blank)   Where I'm having trouble in normalization is with the types of question options:  Multiple choice, true/false, ordering, and fill-in-the-blank. I wonder, should I have a separate table for each type of challenge-question-option? Or is it possible to combine these into one somehow..  If I combined all possible options into a single table, it might look be something like this  challenge_options table: id challenge_question_id option (text field, can be used for both multiple choice and for fill-in-the-blank style questions) is_true (boolean field for true/false questions) is_correct (boolean, used to determine multiple choice answer) ordering_option_1  ordering_option_2 ordering_option_3 ordering_option_4 (these ordering fields would store the correct sequence of an ordering style of question)   This really strikes me as less than ideal.  Actually, I don't see how the ordering_options would work out correctly at all.  Wouldn't I need to have each of those be a separate option?  Surely there must be a better way.  Any suggestions/criticisms or alternate approaches? How are Wolverine's claws still in his body? Everyone knows that Wolverine's adamantium claws are super-sharp — sharper than even a razor — and virtually indestructible. What puzzled me the first time I saw Wolverine is: how do the claws stay in his body?  As it is shown, Wolverine's flesh and skin are human in strength, but the healing factor makes it superhumanly durable and strong. Still, when Wolverine slashes something, or someone, extremely hard, dense, or thick (as he has to use great amount of force), the claws don't tear through his hand and leave his hand through the opposite side of the palm (because of the action having equal and opposite reaction). "He is loved", is 'loved' an adjective or a verb? He is loved.   This is something that I've always kind of wondered.  In a sentence like this, is loved a verb or an adjective? Can it be considered either? How can I show or hide boot messages when Ubuntu starts? Is there a way to easily turn on/off showing the boot messages (loading the services) when Ubuntu starts? Is it something in Grub2?  I am running 10.04. If $A\subset\mathbb{R^2}$ is countable, is $\mathbb{R^2}\setminus A$ path connected? Possible Duplicate:          As the topic,if $A\subset\mathbb{R^2}$ is countable, does $\mathbb{R^2}\setminus A$ path connected??? I know the answer is it is path connected but not sure how to prove it. What to call the attitude of pursuing something Turn a string into a valid filename? I have a string that I want to use as a filename, so I want to remove all characters that wouldn't be allowed in filenames, using Python.  I'd rather be strict than otherwise, so let's say I want to retain only letters, digits, and a small set of other characters like "_-.() ". What's the most elegant solution?  The filename needs to be valid on multiple operating systems (Windows, Linux and Mac OS) - it's an MP3 file in my library with the song title as the filename, and is shared and backed up between 3 machines. I've already defeated the Big Bug.   I see the hint that the small bugs disguised as cursors are vulnerable to the three words "Charity", "Uganda", and "Bram".  I think I have all keys available, I've tried undoing with the bugs on the three words, I've tried motions which include the three words in their range, I have each of the three words stored in a named register, but I haven't figured out how to defeat the three small bug minions underground in Level 14 of Vim Adventures.  I suspect I should have defeated them before defeating Big Bug, maybe even before finding 'undo'. Is there instead something I should do in the 'lorem' buffer where I found the three cursors (Blue, White, and Pink) at the beginning of the level?  How does one defeat the three small bug minions in Level 14? Is there any way to tell Google to display the updated title instead of displaying the old blog title in search results? As a matter of optimization, a few weeks ago I modified the way the titles of my blog posts are displayed (I made a change also in the title of the blog). Then I used the Google Webmaster Tools to re-index the old pages and update titles. Immediately I could update the titles and posts appear correctly on Google.  The problem is that now the titles of posts began to appear incorrectly in the search, using the old blog title. Some posts continue to display the updated title, but many recent posts show the old title.  I've tried re-index the pages with Google Webmaster Tools, but the titles remain unchanged in the search results.  Is there any way to "force" Google to display the correct titles? How to use GROUP BY to concatenate strings in SQL Server? How do I get:  id       Name       Value 1          A          4 1          B          8 2          C          9   to  id          Column 1          A:4, B:8 2          C:9 How to initialize all members of an array to the same value? 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