From what I understand, the localization of a ring $R$ with a multiplicative set $S$ is a construction of a bigger ring $R^*$ such that $R$ is a subring in this bigger ring and every element in $S$ is a unit.
Furthermore, we want that if we have any ring homomorphism $\phi$ from $R$ to another ring $T$ such that $\phi(d)$ is a unit ($d \in R$) and $\phi(1) = 1$, we also want that the construction of $R^*$ satisfies the universal property. So there is a unique homomorphism $\pi$ from $R^*$ to $T$ such that $\phi = \pi \circ \rho$ where $\rho$ is the homomorphism from $R$ to $R^*$.
Now my question is (sorry it takes so long to get to the point) what if I had that my multiplicative set contains two zero divisors $d$ and $d'$ where $dd'=0$? How can any ring homomorphism map any zero divisors to a unit element in $T$?
Since $\phi(dd') = \phi(0) = 0 = \phi(d)\phi(d')$.
So $\phi(d)$ and $\phi(d')$ are zero divisors, hence they cannot be units.
Where did my reasoning go wrong?
Best Answer
Your reasoning is not wrong. You have shown localization is not injective in general.
Suppose you're localizing at a subset $S\subset A$. You have shown that if $s_1,s_2\in S$ satisfy $s_1s_2=0$ then $0=1\in A[S^{-1}]$. If your $S$ is multiplicatively closed, then your assumptions means $0\in S$, so it may be unsurprising that inverting zero kills the ring.
It can also happen that $s\in S$ is a zero divisor in $A$, but that $s$ does not kill any elements in $S$. In this case, the localization can fail to be injective without being zero. A simple example was given in the comments by user26857: if you localize the quotient ring $\mathbb Z/6\mathbb Z$ at the zero divisor $\hat 2$ you get the field $\mathbb Z/3\mathbb Z$.