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Martin Sleziak
Martin Sleziak
4:14 AM
search comments I found two comments using \NN - both in cases where the macro is not defined anywhere on the page. data.stackexchange.com/math/query/556789/… data.stackexchange.com/math/revision/1066862/1318474/…
$\NN$ does have zero. You don't need to add it. — ogerard Feb 10 '15 at 16:46
Since $A$ and $B$ are both subsets of $\NN$, $A \cup B$ is also a subset of $\NN$, therefore the number of equivalence classes is equal to the total number of subsets of $\NN$. Hence, $$\#S=\#P(\NN).$$ I know this is wrong but I can't understand how to do it correctly — 112358 May 3 '16 at 17:14
It seems that these comments use macro \ZZ defined in the answer:
@user437309 the important thing is that it isn't in the $\Bbb{Q}$ image of $A$, since if it had finite order, then for some $n$, $nv_4$ would be in the $\ZZ$ image of A, but then $v_4$ would be in the $\Bbb{Q}$ image. — jgon Aug 5 '18 at 16:10
$A$ is a $4\times 3$ matrix with entries in $\ZZ$. Therefore it represents a map from $\ZZ^3\to\ZZ^4$. However, $\ZZ\subseteq \Bbb{Q}$. Therefore $A$ is also a $4\times 3$ matrix with entries in $\newcommand\QQ{\Bbb{Q}}\QQ$. Hence it also represents a linear map $\QQ^3\to\QQ^4$. Then if $v$ is in $\ZZ^4 \setminus A\QQ^3$, then if $nv\in A\ZZ^3$, we would have that $nv = Aw$ for some $w\in\ZZ^3$, or $v=A(\frac{1}{n}w)$, contradicting that $v$ is not in the image of $\QQ^3$ under $A$. — jgon Aug 7 '18 at 17:07
However, \QQ is defined directly in the comment and works fine.
search posts I did not find posts with unrendered \ZZ. (The single posts returned by the first query uses \renewcommand.) data.stackexchange.com/math/revision/972953/1205843/… data.stackexchange.com/math/revision/1066861/1318473/…
search comments It seems that there are several comments where \ZZ is not rendered. I posted two of them above. And there are two more. data.stackexchange.com/math/query/556789/… data.stackexchange.com/math/revision/1066862/1318474/…
@DerekHolt ah of course: every element of $\ZZ/3\ZZ \oplus \ZZ/3\ZZ$ has order dividing $3$, so $L$ must contain $3\ZZ^3$, and the third isomorphism theorem applies. Thanks! I figured there was probably a simple solution but I missed it. — arkeet Oct 6 '16 at 17:01
More generally, pick $n>1$ and $1\leq m<n$, let $M=N=P=\ZZ/n\ZZ$, let $f:M\to N$ be multiplication by $m$ and $g:M\to N$ be multiplication by $n-m$, and let $k:P\to M$ be any function. — Mariano Suárez-Álvarez Sep 28 '14 at 5:18
In this case, the macro is defined in the question:
It just seems like you could have said $b_1|a_1b_1$ and thus $\frac1{b_1}+\ZZ=\frac{a_1b_1}{b_1}+\ZZ=\ZZ$ a contradiction...obviously, it's the same, but it just seems nicer. — Steven-Owen May 1 '12 at 3:55
Here it is defined in another comment:
The middle step doesn't work. Let's take nonfinitely generated modules, say $M=\Bbb{Z}^{\oplus \Bbb{N}}$, observe that $\newcommand{\ZZ}{\Bbb{Z}}\ZZ\oplus M \cong 0\oplus M$, but you cannot quotient both out by $M$ and preserve the isomorphism, since the isomorphism does not send the factor of $M$ in one to the factor of $M$ in the other. — jgon Aug 20 '18 at 0:54
$\ZZ/p^n\ZZ$ isn't a domain. — jgon Aug 20 '18 at 1:04
And in this case, as far as I can tell, it isn't defined anywhere:
@fretty: the correct statement is that for $p \nmid n$, there is an ideal of $\ZZ[\zeta_n]$ of norm $p$ if and only if $p$ is totally split in the field, which happens if and only if $p = 1 \bmod n$. In particular this never happens when $p = 2$ (unless you count $n = 1$ as an example!). — David Loeffler Dec 12 '12 at 13:10
 

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