most recent 30 from http://mathoverflow.net 2013-05-18T23:55:20Z http://mathoverflow.net/feeds/question/109962 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109962/generators-of-the-kernel-of-multiplication-of-a-ring Tom 2012-10-18T00:30:26Z 2013-02-15T21:32:28Z

<p>Is the following true?</p> <p>Let $R$ be a commutative ring and $S$ be an $R$-algebra. Let $\mu: S\otimes_R S\to S$ be the multiplication. Let $I$ be the kernel of $\mu$.</p> <p>Let $J\subset I/I^2$ be the smallest ideal containing $1\otimes s-s\otimes 1$ for all $s\in S.$</p> <p>Then $J=I.$</p>

http://mathoverflow.net/questions/109962/generators-of-the-kernel-of-multiplication-of-a-ring/109965#109965 Will Sawin 2012-10-18T01:27:11Z 2012-10-18T01:27:11Z

<p>Assuming by $J \subset I$ you mean $J\subset S\otimes_R S$, yes. Let $ab=cd$. Then we must show that $a\otimes b-c\otimes d\in J$. But </p> <p>$a\otimes b-c\otimes d=a\otimes b+ab\otimes 1 - c\otimes d -cd\otimes 1$</p> <p>$=a \otimes 1(1\otimes b-b \otimes 1) -c \otimes 1 (1 \otimes d -d \otimes 1)\in J$</p> <p>An identical argument works for a longer formula like $a_1b_1+a_2b_2+\dots a_nb_n=c_1d_1+\dots+c_md_m$.</p>

http://mathoverflow.net/questions/109962/generators-of-the-kernel-of-multiplication-of-a-ring/109969#109969 Mariano Suárez-Alvarez 2012-10-18T02:47:44Z 2012-10-18T02:47:44Z

<p>The map $\mu$ has a right $S$-linear section $\sigma:s\in S\mapsto 1\otimes s\in S\otimes_RS$. It follows that the kernel of $\mu$ is the set of elements of the form $x-\sigma\mu(x)$ with $x\in S\otimes S$. Since $\mu$ and $s$ are right $S$-linear and $S\otimes_RS$ is generated as a right $S$-module by the elements of the form $s\otimes 1$, that kernel is in fact generated as a right $S$-module by the elements of the form $s\otimes 1-\sigma\mu(s\otimes 1)=s\otimes 1-1\otimes s$.</p>

http://mathoverflow.net/questions/109962/generators-of-the-kernel-of-multiplication-of-a-ring/121945#121945 Martin Brandenburg 2013-02-15T21:32:28Z 2013-02-15T21:32:28Z

<p>If $M,N$ are $S$-modules, then $M \otimes_S N$ is the quotient of $M \otimes_R N$ by the $R$-submodule generated by $ms \otimes n - m \otimes sn$ with $m \in M, n \in N, s \in S$; simply because both sides satisfy the same universal property. For $M=N=S$ the result follows. Actually the same holds when $\mathrm{Mod}(R)$ is replaced by an arbitrary cocomplete symmetric monoidal category, and $S$ by a commutative algebra object.</p>