Confusion on the tensor product $I\otimes_A A/I$

Question

Sorry for this trivial question. But I am confusing myself with it.
Given an ideal $I$ of a ring $A$.Now $I$ can be regarded as an $A$-module with map $$A\times I\to A\\ (a,i)\mapsto a\cdot i$$ I belive that by module theory, one should have an $A$-module isomorphism $$I\otimes_A A/I\cong I/I^2$$ However, since the tensor $I\otimes_A A/I$ is generated by elements $i\otimes \bar{1}$ for all $i\in I$. Don't we have the following equalities? $$i\otimes\bar{1}=(i\cdot 1)\otimes \bar{1}=i(1\otimes\bar{1})=1\otimes (i\cdot \bar{1})=1\otimes \bar{i}=0$$ which means the tensor should be zero.
Any help is appreciated! Thank you.

Answer 1

The problem is in the second equality. $1$ is not an element of $I$ (unless $I = A$), so $1 \otimes \bar{1}$ is not an element of $I \otimes_A A/I$.