Suppose $A$ is a domain and $S=A \setminus \{0\}$ is a left denominator set. I have the following argument:
Let $\phi \colon A \to S^{-1}A$ be the natural map to the ring of fractions. Then I know that $S^{-1}A = \{ \phi(s)^{-1}\phi(a) : s\in S, a\in A \}.$ Now suppose $\phi(s)^{-1}\phi(a)$ is a nonzero element in $S^{-1}A.$ Then $a$ is nonzero (if it were, we would have $\phi(a) = 0,$ hence $\phi(s)^{-1}\phi(a) = 0)$, so $a \in S = A \setminus \{0\}.$ Then $\phi(a)^{-1}\phi(s)$ is an inverse of $\phi(s)^{-1}\phi(a).$
Does this mean that $S^{-1}A$ is always a division ring whenever $S = A \setminus \{0\}$ is left localisable? This doesn't seem right to me, but I can't find what I got wrong in my reasoning. Could somebody please help?
EDIT: The reason I ask this is that I have the following theorem in my notes: every Noetherian domain has a division ring of fractions. It is proved as follows:
Let $S = A \setminus \{0\}$, which is a left Ore set. It consists of regular elements because $A$ is a domain. Therefore, the natural map $\phi \colon A \to S^{-1}A$ is injective, and $S^{-1}A = \{s^{-1}a : s\in S,a\in A\}.$ Now if $s^{-1}a \in S^{-1}A$ is nonzero, then $a,s \in S,$ and $a^{-1}s$ is the inverse of $s^{-1}x$.
However, by my reasoning above, I don't understand why we need injectivity in the proof. It feels like we only need to know the existence of the left quotient ring of $A$ with respect to $S$.