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On Line 2 of Page 40 of the book introduction to commutative algebra by M. F. Atiyah and I. G. Macdonald, it is said that $m/s =0$ implies $tm=0$ for some $t \in S$. I think that if $m/s=0$, then $m/s = 0/t_1$ for some $t_1 \neq 0, t_1 \in S$. Therefore there is some $u \in S$ such that $(t_1m-0)u=0$. Therefore $ut_1 m=0$. Let $t=t_1 u$, then $tm=0$. In the third line of Page 40, it is written that $$ 1/s \otimes m = t/st \otimes m = \cdots $$ I think that $t$ should be non-zero. But $t=ut_1$ can be zero since $u\in S$ can be zero. I am confused. Thank you very much.

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LJR
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    Recall the definitions. The zero in $S^{-1} M$ is given by $0/1$, and it is not required that the denominators are $\neq 0$. They only have to be in $S$. – Martin Brandenburg Jun 12 '13 at 08:24
  • @MartinBrandenburg, thank you very much. Now I understand, $t$ can be $0$. – LJR Jun 12 '13 at 12:34

1 Answers1

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If $t=0\in S$ then $S^{-1}A=0$ from Example 2 above.

Boris Novikov
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