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."
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.