Assume $R$ is a commutative local ring, $I$ is a proper ideal in $R$ and $M$ is a finitely generated $R$-module. Is it true that $\operatorname{Ann}(M/IM)=I+\operatorname{Ann}(M)$?    (Note that $\supseteq$ is clear.)  If it's true I would like (a hint for) an elementary proof (preferably avoiding any homological algebra).
Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an module over $A$. Is it true that $\operatorname{Ann}(M/IM) = \operatorname{Ann}(M) + I$?    One inclusion is certainly true, but I don't know about the other one. If the above statement does not hold in general, does it perhaps for finitely generated modules?
counting with max and min restrictions $$x_1 + x_2 + x_3 + x_4 = 10$$ But $x_1 \ge 1$ and $x_2 \ge2$.  Finally, $x_1&lt;6,  x_2&lt;6, x_3&lt;6$ and $x_4&lt;6$.  I started off by finding the total amount that you can get for the original equation. That would be $C (13, 3)$. Then i calculated for the cases of numbers being greater than or equal to 5. I dont know how to incorporate $x_1\ge1$ and $x_2\ge2$. Any help?  $x_1, x_2, x_3, x_4$ are all integers that could be >= 0(But $x_1$ and $x_2$ have their own restriction). What is the recommended wording for a generic error message Homomorphism between free groups Let $F_a$ be the group which is freely generated by $a$ elements. How to show that there is a homomorphism from $F_a$ onto $F_b$ if and only if $b\le a$?  I was thinking one possibility is if $F_a$ generated freely by $\{x_1, \dots , x_a\}$ and $F_b$ generated freely by $\{y_1, \dots , y_b\}$ then consider map which says $\phi(x_i)=y_i$ for $1\le i \le b$ and $y_1$ otherwise. I do not know if this works or even how to proceed. How to show this fact? Can I leave the UK and come back in, on a Tourist visa? $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\ $ [LCM Universal Property] I updated the image in my Gravatar but my old picture still shows up.  Is this a cache issue?  Do I need to log out and then log back in? Is there some trick to burning a cd with the CD/DVD creator?    I'm installing k3b because that's a known entity.  What is this, and where is the "burn" button?  Help just tells me this is Nautilus 3.4.1, but I don't see a .  There's no option to burn in the Files menu, nor does clicking on the whitespace indicate an option to burn anything. How are "Suggested Tags" chosen? After you've typed in a question, you get suggested tags that appear that look like this:        It appears that you need a minimum amount of characters ():     you must have entered a title (any length) and a minimum body length of 220 characters.   But other than that, how are the tags chosen? Is it:   A match of the title/body with other questions (taking the tags from those questions) A match of the title/body with tag excerpts (taking tags with overlap) A match of the title/body with tag wikis (taking tags with overlap) All of the Above None of the Above Other (please explain)   Some additional info from :  The suggested tags seem to only pop up on meta.so (they do not work on regular Stack Overflow, or on other SE sites).  While we're on the topic, why isn't this feature implemented network-wide? How to remove spikes from switching transistor? For all infinite cardinals $\kappa, \ (\kappa \times \kappa, <_{cw}) \cong (\kappa, \in).$ I don't understand the proof to the a/m claim. How we know that $\eta &lt; \kappa$ and $(\alpha,\beta)&lt;_{cw} (0, \eta) $ and hence $h: \mu \to \eta \times \eta$ is injective? Appreciate if anyone could advise.  Here is an excerpt of the proof:  For an infinite ordinal $\kappa,$ the canonical well-ordering of $\kappa \times \kappa,$ denoted by $&lt;_{cw}$ is defined as follows: $(\alpha_1, \beta_1)&lt;_{cw} &lt; (\alpha_2, \beta_2)$ iff either one of the following holds.  $(1): \max\{\alpha_1,\beta_1\} &lt; \max\{\alpha_2,\beta_2\}$  $(2): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1&lt;\alpha_2$   $(3): \max\{\alpha_1,\beta_1\} =\max\{\alpha_2,\beta_2\}$ and $\alpha_1 = \alpha_2$ and $\beta_1 &lt; \beta_2.$  Claim: For all infinite cardinals $\kappa, (\kappa \times \kappa, &lt;_{cw}) \cong (\kappa, \in).$  Proof: (It has been show that it is true for $\kappa = \omega$) Assume there is a counterexample  $\kappa.$ Let $A= \{\lambda \in \kappa + 1| \lambda\geq \omega \ is \ a \ cardinal \ and \ (\lambda \times \lambda, &lt;_{cw}) \not\cong (\lambda, \in)\}.$ Then, $\kappa \in A.$ Let $\mu$ be least element of $A.$ Thus, $\mu &gt; \omega$ and for all $\lambda,$ if $\omega \leq \lambda &lt; \mu$ and $\lambda$ is a cardinal, then $(\lambda \times \lambda, &lt;_{cw}) \cong (\lambda, \in).$  By Comparability theorem, $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, &lt;_{cw})$ since $|\mu \times \mu| \geq \mu$ and $\mu$ is cardinal. Let $(\alpha, \beta) \in \mu \times \mu$ such that $(\mu, \in)$ is isomorphic to initial segment of $(\mu \times \mu, &lt;_{cw})$ given by $(\alpha, \beta).$ Let $h$ be the isomorphism. Let $\eta= \max\{\alpha, \beta\}+ \omega.$ Then $\eta &lt; \kappa$ and $(\alpha,\beta)&lt;_{cw} (0, \eta).$ HENCE, $h: \mu \to \eta \times \eta$ is injective. Why am I constantly running out of space on my Galaxy S2? Every planar graph has a vertex of degree at most 5. How to differ sessions in browser-tabs? Why particle hole symmetry and chiral symmetry are called symmetries? $PHP^{-1}=-H$ (particle-hole symmetry) and   $\Gamma H \Gamma^{-1}=-H$ (chiral symmetry)   I understand why we get the negative signs but im just a bit confused as to why such equalities mean $H$ is particle hole and chiral symmetric.   When we say $H$ is symmetric under some operation, like time reversal,  don't we usually mean that $H$ is invariant under the said transformation?  $THT^{-1}=H$ if $H$  is time reversal symmetric.  On the contrary, we say $H$ is symmetric under particle hole and chiral operations when $H$ picks up a negative sign. This seems to contradict our usual convention of what we mean by $H$ being symmetric(invariant). Cannot add custom launcher to Ubuntu Dock ('Add to Favorites' option does not show up) I have a custom .desktop file: [Desktop Entry] Type=Application Terminal=false Icon[en_US]=/home/alex/bin/idea-IC-173.4127.27/bin/idea.png Name[en_US]=IntelliJ Exec=/home/alex/bin/idea-IC-173.4127.27/bin/idea.sh Name=IntelliJ Icon=/home/alex/bin/idea-IC-173.4127.27/bin/idea.png  Output of ls -lha .local/share/applications/intellij.desktop: -rwx--x--x 1 alex alex 248 Dez 30 10:54 .local/share/applications/intellij.desktop  Answers posted to a previous question () don't work for me. I can find this launcher in Activities and I can double click on that file and IntelliJ will start. Nevertheless, I cannot add this launcher as a favorite to the dock. The option does not show up. How can I install a Nautilus script? I found a nautilus script I would like to use. How do I install it? equivalent norms in Banach spaces of infinite dimension How do I transcode a Javascript string to ISO-8859-1? ActionBarSherlock with ViewPager not calling ViewPager fragments lifecycle methods when going to and returning from backstack Take a look at this example schematic:  The chip in question is the  and as you can see power is provided on both sides (with a 100nF cap to smooth noise I guess). Is this strictly required or could you put power into one side and just ignore the other? I know that I have just put power in one side and everything still seems to work fine but I feel uneasy with doing that. I feel that they would not waste pins that could be doing something useful by having an extra pair of power in/out pins. For what purpose does this PIC MCU (and others like it) have multiple power pins?