Let $R$ be a commutative ring with unity and $M$ be a free $R$-module. Let $S$ be a finite basis of $M$. If $T\subset M$ has cardinality greater than $S$, I am wondering whether $T$ is linear dependent over $R$.   I know that this is true when $R$ is field. But I am not sure if $R$ is not field.  If it is not true when $R$ is not field, could you suggest me an example such that $T$ is linear independent over $R$?
I'm trying to prove that if $A\neq 0$ is a commutative ring and there is an injective $A$-module homomorphism $A^m\hookrightarrow A^n$ then $m\leq n$ must necessarily hold.    This is exercise 2.11 from Atiyah and MacDonald's Introduction to Commutative Algebra, but unfortunately all the solutions available online are either very sparse with regard to this question or seem to use (to my surprise) a generalized version of Cramer's rule. Is there perhaps some other "cleaner" approach to solving this problem?
How to disable autorefresh in snap Is it possible to disable autorefresh in snap? This  says that Snaps are updated automatically in the background to the latest version, every day. Some snaps (e.g. vlc) are built daily, and it makes no sense to install them every day. Prove by induction. How do I prove $\sum\limits_{i=0}^n2^i=2^{n+1}-1$ with induction? How to remove lines included in one file from another file? I have two lists, a complete list and a partial list. I want a bash script that will search the complete list for any lines that are in the partial, if there are, they will be removed from the complete list. Can anyone help? Why didn't the Hobbits get frostbite? When is the sum of divisors a perfect square? Why is the [I'm Done] button unavailable after only flagging comments in the First Posts review queue? I just flagged two comments during a review of a question in the First Posts queue (on another SE site). Then I wanted to select the button I'm Done, but it's unavailable (I can only select No action needed). Even if I refresh my page (while still on that review page), it remains the same.  If however I also downvote the question (only temporary, as a test, since I don't want to downvote the question being reviewed), then it seems to work as I'd expect it to: the button I'm Done becomes available, and No action needed becomes unavailable.  Why is the I'm Done button unavailable after only flagging comments in the First Posts review queue?  Notes:   I'm not convinced that you're supposed to review the actual post, and not (only) the comments underneath it (as mentioned in 's comment). Because "if" I would post a new comment, or just upvote an existing comment, I can actually select I'm Done. While in this case I flagged existing comments which IMO had become obsolete. Flagging comments is "acting on comments". And if my flag gets accepted (in this case: "no longer needed"), and the comment is actually deleted (by the mod), then it makes the question a bit more readable (= don't waste your time digesting a comment which is obsolete). Therefor "IMO", in the end I also "acted on the post". Think about all those posts with a zillion of comments, of which many are obsolete (e.g because the post got updated already). I am not sure (yet) if a similar issue exist in any other review queue (i.e. the Late Answers queue). I have Ubuntu 13.1 installed with the default window manager and DE.   I am trying to enable single click so I can open files and folders with only one click. I tried googling and all results said to open files, then go to preferences from top bar but my top bar has no options.  picture of desktop: I installed kubuntu-desktop just to experiment with its interface.  Then I decided to remove and did the traditional apt-get purge and autoremove but it seems elements of kubuntu are still there. For example interface elements, scrollbars, bootscreen etc are still there.  Is there any method I can completely remove without installing a fresh copy of ubuntu? Does the Sun produce audible sound? I have noticed that if you look directly at tubelight (I have only white ones in my home) and close your eyes little, you see little random shaped colourless things sometimes with slow speed and sometimes with falling down.   They are shaped like circles with something in between and in chains of random length.   My father has also seen them. So, I am sure I am not imagining things. What the hell are they? When I get a notification on android that someone had posted an answer or comment, if I click the notification, it opens the Stack Exchange app but shows the main view with the inbox opened, not the comment or answer that was in the notification.  If there is only one notification pending, when you tap it it should show you the actual contents rather than just the inbox. What is the "correct" way of embedding text into math mode? Lets say I have something like:   \[ Let x = number of cats \]   (This is a contrived example, but pretend we are actually inside a long align environment block...) "Let" and "number of cats" need to be written like normal text, while x= needs to be math type text.  I've seen this:  \[ \mbox{Let } x = \mbox{ number of cats} \]   and this:  \[ \textrm{Let } x = \textrm{ number of cats} \]   They seem to produce equivalent output, but I'm not sure which is the "correct" one. Also, they are not ideal in that the extra space needs to be placed inside the text mode block, otherwise it is ignored as math mode and butts the text right up against the math.  Ideally there would be some command which allowed embedding text into an equation, which correctly inserted leading and trailing space around the block as required without adding space inside the text block.  How does one do that "correctly"? String concatenation: concat() vs "+" operator $\DeclareMathOperator{\card}{card}$From Problem-Solvers Topology  Prove the following: CANTOR'S THEOREM If $A$ is a set, then $$\card A &lt; \card \mathcal{P}(A)$$ where $\card A$ stands for the cardinality of set $A$.  My Answer If $A = \emptyset$, then $\card A =0$ and $\card \mathcal{P}(A)=1$. If $A = \{a\}$, then $\card A=1$ and $\card \mathcal{P}(A)=2$. So suppose that $A$ has at least two elements. Define a function $f: A \rightarrow \mathcal{P}(A)$ such that $f(x) = \{x\}$ for all $x \in A$. Then $f$ is injective. But it cannot be surjective, because for any two distinct elements $a,b \in A$, there is no element in $A$ that is sent to the set $\{a,b\}$ in $\mathcal{P}(A)$. Therefore, there is no bijection, and $\card A &lt; \card\mathcal{P}(A)$. Do you think my answer is correct? Thanks in advance How can I use wget to get files recursively if the username contains an @? Firstly my main goal is to try and get FTP files recursively in Centos.  However the actual FTP login contains an @ which ruins the syntax.  Is there some way to place the username in "double quotes" or some such equivalent such that I can use @in the username? Holomorphic self map on the open unit disc Does there exist a holomorphic function $f:\mathbb{D}\longrightarrow\mathbb{D}$ such that $f(1/2)=-1/2$ and $f'(1/4)=1$, where $\mathbb{D}=\{z:|z|&lt;1\}$? Cause of Alpha or Black Artefacts in Render Show that $A_n$ is the kernel of a group homomorphism of $S_n \rightarrow \{−1,1\}$. Possible Duplicate:          I have seen  in SQL. I wonder how would you approach the opposite problem: splitting coma delimited string into rows of data:  Lets say I have tables:  userTypedTags(userID,commaSeparatedTags) 'one entry per user tags(tagID,name)   And want to insert data into table  userTag(userID,tagID) 'multiple entries per user   Inspired by  question  EDIT  Thanks for the answers, actually more then one deserves to be accepted but I can only pick one, and the  with recursions seems pretty clean to me. It works on SQL Server 2005 and above.   For earlier version of SQL Server the solution  can be used.  For working with text data type  will be helpful. Thanks again. Prove divisibility law $\,b\mid a,c\,\Rightarrow\, b\mid ka + lc$ for all $k,l\in\Bbb Z$