I have this exercise:
Find the direct summands of the $\mathbb Z$-module $M = \mathbb Z/36 \mathbb Z$.
If $\bar T$ and $\bar N$ are direct summands of $M$ then $\bar T \cap \bar N = \{\bar 0\} = \{36 \mathbb Z\}$ (1). And $\bar T = T/36 \mathbb Z$ and $\bar N = N/36\mathbb Z$, where $T$ and $N$ are $\mathbb Z$-modules of $\mathbb Z$ so they are ideals of $\mathbb Z$, so they are $T = a\mathbb Z$ and $N = b\mathbb Z$ and $a,b \in \mathbb Z$ are divisors of $36$.
The condition (1) implies that for any $x \in \mathbb Z$ such that $a|x$ and $b|x$, we have $36 |x$. What does this mean? and how to proceed?
I am stuck and I need some help. Thanks.
Continuing with your notation, let $T = a \mathbb {Z}$, $N = b \mathbb {Z}$, $\bar {T} = T/36 \mathbb Z$ and $\bar{ N} = N/36\mathbb Z$, where $a,b \in \mathbb Z$ are divisors of $36$.
The equality $\bar {T} + \bar{N} = \mathbb{Z} / 36 \mathbb {Z}$ implies that for all $d \in \mathbb {Z} $ there exists $x, y \in \mathbb {Z}$ such that $ ax + by + 36 \mathbb {Z} = d + 36 \mathbb {Z}$. By Bezout's identity $a$ and $b $ are be coprime.
But as you noted, for any $n \in \mathbb Z$ such that $a|n$ and $b|n$, we have $36 |n$. So the union of the divisors of a and b equals the divisors of $36$.
Hence $a=4$ and $b=9$.