Can we localize rings with zero divisors? Can those zero divisors be in the denominator?
I thought defining $$\frac{a}{b}=\frac{c}{d} \text{ iff }t(ad-bc)=0 \text{ where $b,d,t$ belong to the same multiplicative system}$$
accommodated for that little detail. But my professor thinks not. I am confused.
Any help would be greatly appreciated.
On
Yes, you definitely can localize by sets which contain zero divisors, and the inclusion of the factor $t$ into the definition is there to ensure precisely that. That is, if we don't include $t$ into the definition, then we can still localize by sets without zeros divisors, but then we cannot localize by sets with zero divisors, because then the relation between fractions in $S^{-1}A$ is no longer an equivalence relation,as Hagen Knaf has pointed out. So localization is in fact defined the way it is because we want to localize by sets which may contain zero divisors.
Your definition is correct and even necessary: if you don't include the factor $t$ into the definition, then you will in general not get an equivalence relation between pairs (a,b) of ring elements. This however is necessary to define the notion of a fraction a/b.