$p\mathbb{Z}$ is not a direct summand of $\mathbb{Z}$

Question

Let $p$ be a prime number and let $p\mathbb{Z}\leq\mathbb{Z}$ be a submodule.

Prove that $p\mathbb{Z}$ is not a direct summand of $\mathbb{Z}$.

What about this proof? Thank you so much.

Answer 1

Suppose that $p \mathbb{Z}$ has a complement, say $q \mathbb{Z}$ where $q \neq 0$.

$$\mathbb{Z}= p\mathbb{Z} \oplus q \mathbb{Z}$$

Then $0 \neq pq \in p\mathbb{Z} \cap q \mathbb{Z}$, which is impossible.

Answer 2

Any direct summand of a ring $R$ considered as a left (resp. right) $R$-module must be the principal left (resp. right) ideal $Re$ (resp. $eR$) for some idempotent $e \in R$. In particular, since the only idempotents in $\mathbb{Z}$ are $0$ and $1$, it follows that the only direct summands of $\mathbb{Z}$ are $0$ and $\mathbb{Z}$, so $p\mathbb{Z}$ is not a direct summand.