Krystal Villanueva

2022-02-15

Let A be a ring, S a multiplicatively closed subset. Is it true that $\frac{a}{1}\in {S}^{-1}A$ is a non zero-divisor if and only if $a\in A$ is?

uporabah5pn

Set . Then $\stackrel{^}{2}$ is a zerodivisor in A and $\frac{\stackrel{^}{2}}{\stackrel{^}{1}}$ is not zerodivisor in ${S}^{-1}A$. (Actually it is invertible in ${S}^{-1}A$.)