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This is definitely not true in general unless $\gcd(n, b) = 1$. For instance $2 \cdot 3 = 4 \cdot 3 \pmod 6$ but $2 \not \equiv 4 \pmod 6$. The problem is that here $3$ is not invertible since it is not relatively prime to $6$.


With conclusion $\,a_1\equiv a_2,\,$ it is true iff $\,b\,$ is invertible mod $n\,$ (iff $\,\gcd(b,n)=1)$ Else $b,n$ share a divisor $c>1$ so $\,\color{#c00}{(n/c)}b = n(b/c)\equiv 0\equiv \color{#c00} 0\cdot b,\,$ but $\,\color{#c00}{n/c\not\equiv 0}\pmod n$ But your hypothesis does not imply that $b$ is invertible. Indeed it is true for all ...


No. To show it's injective, you have to show that, if $\;\sum r_i\otimes m_i\mapsto\sum r_im_i=0$, then $\;\sum r_i\otimes m_i=0$. But that is because $\;\sum r_i\otimes m_i=\sum 1\otimes r_im_i=1\otimes0=0$.


If $\;r\otimes m\to rm=0\;$ then $$r\times m=r(1\otimes m)=1\otimes rm=1\otimes0=0$$ and we've finished. I am assuming, of course, the tensor product is over $\;R\;$ without any further conditions.

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