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One of the ways we characterize prime ideals is as follows: For any elements $a$ and $b$ of $R$, if $aRb \subseteq P$, then $a \in P$ or $b \in P$. Suppose $P$ is a prime ideal and $R/P$ has no nonzero nilpotent elements. Suppose $$\overline a \overline b = (a + P)(b + P) = ab + P = 0 + P = \overline 0.$$ Let $r \in R$ be arbitrary and consider ...


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Yes. By the way, both of these are also what mathematicians call an exterior algebra.


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We have the short exact sequence $$ 0 \to N \xrightarrow[]{i} M \xrightarrow[]{\pi} M/N \to 0$$ Tensor on the right by $A$ and get a right exact sequence $$N \otimes A \xrightarrow[]{i\otimes 1_A} M \otimes A \xrightarrow[]{\pi\otimes 1_A} (M/N)\otimes A \to 0$$ Therefore, the kernel is the submodule of $M\otimes A$ generated by $i(n)\otimes a$ with ...


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$\DeclareMathOperator{\Tor}{Tor}$ Let $N\subseteq M$ and $A$ be as in the question. In general, there is not a simple description of the kernel of the map $M\otimes A\to M/N\otimes A$. Given modules $X$ and $Y$ (on the right and on the left, respectively) there is a standard way to construct an abelian group $\Tor(X,Y)$; this can be done in several ways, ...


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Let $K\stackrel f\to M\stackrel g\to L\to 0$ and $K'\stackrel {f'}\to M'\stackrel {g'}\to L'\to 0$ be two short exact sequences. Then $$\ker (g\otimes g')=\operatorname{im}(f\otimes1_{M'})+\operatorname{im}(1_M\otimes f').$$ In your case $\ker(\pi\otimes 1_A)=\operatorname{im}(\iota\otimes1_{A})$, where $\iota:N\to M$ is the inclusion. One can also ...


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I guess you are wondering about the map $ M \otimes A \rightarrow M/N \otimes A$. $A$ is free and so flat as $A$-module which means $\ker \pi \otimes A$ is a submodule of $M \otimes A$. It is clearly in the kernel of $\pi \otimes 1$. Now think of a an element of $\pi \otimes 1$. You can see it's of the form $u \otimes 1$ for $u \in M$. Then again by ...


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It seems to me that if $[A,B]=C$ with $[C,A]=[C,B]=0$, it will be a simple case. We have $$e^{A+B}=e^A e^B e^{-\frac{1}{2}[A,B]}$$ and $$\cos(A+B)=\frac{e^{i(A+B)}+e^{-i(A+B)}}{2}=\frac{e^{iA}e^{iB}e^{\frac{1}{2}[A,B]}+e^{-iA}e^{-iB}e^{\frac{1}{2}[A,B]}}{2}$$ ...


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If $R=M_2(k)$ is the ring of $2\times 2$ matrices over a field, then the $2$-dimensional left module of column vectors is a generator, but doesn't have $R$ (i.e., $Re$ for $e=1$) as a direct summand. More generally, if $R$ contains any indecomposable projective module $P$ as a direct summand with multiplicity greater than one, then you can choose a ...



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