I would appreciate help on what should be an easy concept in the proof of a corollary leading up to Nakayama's Lemma.
This link to mathoverflow.com (in the green highlighted section) gives the development as presented in "Atiyah and Macdonald" as well as "Reid."
https://mathoverflow.net/questions/41836/elementary-proof-of-nakayamas-lemma
My question pertains to the second corollary (as in "Reid"):
If $M$ is a finite $A$-module and $M = IM$ then there exists an $x \in A$ such that $x \equiv 1$ mod $I$ and $xM = 0$.
I understand the use of $\phi = id_M$ in the relation of maps to get: $1 + a_1 + \dots + a_n = 0$ with $a_i \in I$. And $x$ is set equal to this, giving $xM = 0$.
Also this satisfies $x \equiv 1$ mod ($I$).
Here is my question:
How can $x \in A$ be = $0$ and be $\equiv 1$ (mod $I$)?
Thanks for straightening out what must be an error in my math understanding.