If $A \subseteq B$, then $A_{\mathfrak p} \subseteq B_{\mathfrak q}$? Let $A \subseteq B$ be commutative rings with identity, $\mathfrak q$ a prime ideal of $B$, and $\mathfrak p = A \cap \mathfrak q$.  Is the 'identity' ring homomorphism $A_{\mathfrak p} \rightarrow B_{\mathfrak q}$ always an injection?  The 'obvious' method of proving this seems to lead to a dead end: if $\frac{a}{s} \in A_{\mathfrak p}$ is the zero element in $B_{\mathfrak q}$, then there exists a $t \in B$, but not in $\mathfrak q$, such that $ta = 0$.  But it isn't clear whether $t$ may be chosen to lie in $A$.  So my guess would be that this map doesn't need to be an injection.  The reason I'm asking is that it is an exercise in Hartshorne (II, 2.18) to show that a homomorphism of rings $\phi: A \rightarrow B$ is injective if and only if the corresponding morphism of sheaves $f^{\#}: \mathcal O_X \rightarrow f_{\ast} \mathcal O_Y$ is injective ($X$ and $Y$ are the spectra of $A, B$, and $f: Y \rightarrow X$ is the map $f(\mathfrak q) = \phi^{-1} \mathfrak q$).  Assuming $\phi$ is injective, I want to show that $f^{\#}$ is injective by showing the corresponding homomorphism on the stalks $A_{\phi^{-1}\mathfrak q} \rightarrow B_{\mathfrak q}$ is injective.  But if I am not mistaken, this homomorphism on the stalks is exactly the map $\frac{a}{s} \rightarrow \frac{\phi(a)}{\phi(s)}$ I just described.    EDIT: Someone has asked the same question before, and as I suspected the homomorphism $A_{\mathfrak p} \rightarrow B_{\mathfrak q}$ does not have to be an injection.  So my question now is why this counterexample does not invalidate exercise II.2.18(b) in Hartshorne.  EDIT 2: I just now realized the homomorphism on the stalks isn't the natural map $A_{\phi^{-1}\mathfrak q} \rightarrow B_{\mathfrak q}$; the codomain isn't correct.  Rather, if $\mathfrak p = \phi^{-1}\mathfrak q$, $(f_{\ast} \mathcal O_Y)_{\mathfrak p}$ is the direct limit of the rings $\mathcal O_{Spec B}(f^{-1}U)$, where $U$ runs through the open neighborhoods of $\mathfrak p$.
Injectivity of Homomorphism in Localization Let $\alpha:A\to B$ be a ring homomorphism, $Q\subset B$ a prime ideal, $P=\alpha^{-1}(Q)\subset A$ a prime ideal. Consider the natural map $\alpha_Q:A_P\to B_Q$ defined by $\alpha_Q(a/b)=\alpha(a)/\alpha(b)$. Suppose that $\alpha$ is injective. Then is $\alpha_Q$ always injective?    I think so, but I'm clearly being too dense to prove it! My argument goes as follows.  Let $\alpha(a)/\alpha(b)=0$. Then $\exists c \in B\setminus Q$ s.t. $c\alpha(a)=0$. If $B$ is a domain we are done. If not we must exhibit some $d\in A\setminus P$ s.t. $da=0$. Obviously this is true if $c =\alpha(d)$. But I don't see how I have any information to prove this!  Am I wrong and this is actually false? If so could someone show me the trivial counterexample I must be missing?   Many thanks!
Is there a symbol for "unitless"? I have two laptops (same manufacture), with the same type of power connector.  However, the power supplies/transformers are slightly different.    The output on the first laptop's power supply is 15.6&nbsp;V at 8.0&nbsp;A. The output on the second laptop's power supply is 15.6&nbsp;V at 5&nbsp;A.  Clearly the voltages are the same, but the currents are different. I assume the second laptop's power supply can not be used on the first, because it can't supply enough power to the laptop.   However, can the first laptop's power supply be safely used on the second laptop? Is there any way to call the parent version of an overridden method? (C# .NET) In the code below I tried in two ways to access the parent version of methodTwo, but the result was always 2. Is there any way to get the 1 result from a ChildClass instance without modifying these two classes?  class ParentClass {     public int methodOne()     {         return methodTwo();     }      virtual public int methodTwo()     {         return 1;     } }  class ChildClass : ParentClass {     override public int methodTwo()     {         return 2;     } }  class Program {     static void Main(string[] args)     {         var a = new ChildClass();         Console.WriteLine("a.methodOne(): " + a.methodOne());         Console.WriteLine("a.methodTwo(): " + a.methodTwo());         Console.WriteLine("((ParentClass)a).methodTwo(): "          + ((ParentClass)a).methodTwo());         Console.ReadLine();     } }   Update ChrisW posted this:     From outside the class, I don't know   any easy way; but, perhaps, I don't   know what happens if you try   reflection: use the Type.GetMethod   method to find the MethodInfo   associated with the method in the   ParentClass, and then call   MethodInfo.Invoke   That answer was deleted. I'm wondering if that hack could work, just for curiosity. How to add lists with unequal length? Search for text files where two different words exist (any order, any line) I'm looking for a way to search files where two word instances exist in the same file.  I've been using the following to perform my searches up to this point:  find . -exec grep -l "FIND ME" {} \;   The problem I'm running into is that if there isn't exactly one space that between "FIND" and "ME", the search result does not yield the file.  How do I adapt the former search string where both words "FIND" and "ME exist in a file as opposed to "FIND ME"?  I'm using AIX. How is the authenticity of Debian packages guaranteed? I'm new to regular expressions and would appreciate your help.  I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.  Example input:   This is a string that "will be" highlighted when your 'regular expression' matches something.   Desired output:  This is a string that will be highlighted when your regular expression matches something.   Note that "will be" and 'regular expression' retain the space between the words. Even though JRE 8 is installed on my MAC OS Yosemite, the error - " No Java Runtime present,requesting to install " I would like AUCTeX (that is, Emacs with auctex loaded) to suggest more LaTeX macros when using completion (M-Tab or C-Ret) than it does by default. I would like to have macros from packages I have downloaded on CTAN and macros I have written. I would also like it if it would complete \section or other macros that can be accessed via C-c C-s, C-c C-e etc.  Solution: I could not make the code quoted by  work. i get an error on loading emacs but did not looked much further in because I ca bear the nuisance of appending a few lines to my files.  The link given by Ben ( contains a working solution, thus I accept his answer. I had come across this page in my researchs but somehow messed up things. What eventually worked for me was using TeX-auto-generate TEX AUTO to create a bunch of .el files in the ~/.emacs.d/auctex/auto direcoty. Note that this directory must be created beforhand. For some reasons TeX-macro-private did not achieve the wanted result. Converting from shapefile or raster formats to LAS? What purpose does downvoting questions serve? Rotate Geographic map How fast is Earth moving through the universe? As the galaxy is moving and the solar system orbiting the galaxy and the Earth orbiting the sun. So how fast is each object moving and what is the fastest we move at?  Do we even know how fast the galaxy is moving that is not relative to another galaxy (although I guess velocity has to be measured relative to something). Weapon damage bonuses and critical hits I would like to draw a squiggly line with an arrow somewhere in the middle of it. TikZ has a pathmorphing decoration that draws a squiggly line and a markings decoration to place an arrow tip. So a solution would be to use them together:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{decorations,decorations.pathmorphing,decorations.markings} \begin{document}  \begin{tikzpicture}  \draw[          decoration={snake},          decorate,          postaction={decoration={markings,mark=at position 0.6 with {\arrow{>}}},decorate}      ]         (0,0) -- (2,0); \end{tikzpicture}  \end{document}   Unfortunately this does not work and results in  ! Dimension too large. &lt;to be read again&gt;                     \relax  l.11         (0,0) -- (2,0);   I suspect there is an infinite recursion going on somewhere.  Is there a workaround? If not, is there a different simple solution to the problem?  (Note: This is probably related to my previous question ) Why General Leibniz rule and Newton's Binomial are so similar? The binomial expansion:  $$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$$  The General Leibniz rule (used as a generalization of the product rule for derivatives):  $$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$  Both formulas can be obtained simply by induction; Newton's binomial also has a combinatorial proof . It's striking how these formulas are similar; is there a possible connection between them? I was thinking that maybe the General Leibniz rule could be obtained using a combinatorial argument as well (hence the binomial coefficients)... Python dictionary: are keys() and values() always the same order? I would like to typeset the top part of Pascal's triangle. To get the triangle with the names of the binomial coefficients, i.e., {n \choose k}, I used the following code  \begin{tikzpicture} \foreach \n in {0,...,4} {   \foreach \k in {0,...,\n} {     \node at (\k-\n/2,-\n) {${\n \choose \k}$};   } } \end{tikzpicture}   The result is this   Now I want to be equally lazy and do something like this for the values of the binomial coefficients, i.e., replace {\n \choose \k} in the node label with \CalculateBinomialCoefficient{\n}{\k} where \CalculateBinomialCoefficient is a hypothetical macro that calculates the binomial coefficient. Has anyone done something like that?  The result should look like this: If $\gcd (a,b) = 1$, what can be said about $\gcd (a+b,a-b)$? $n$-th digit in the sequence of natural numbers