$R\cong\mathbb{Z}$ or $R\cong\mathbb{Z}/p\mathbb{Z}$ ($p$ prime)

Question

Let $R$ be an integral domain and each subgroup of additive subgroup of $R$ forms an ideal of $R$. Prove that either $R\cong\mathbb{Z}$ or $R\cong \mathbb{Z}/p\mathbb{Z}$ ($p$ prime).

Help me.

Answer 1

Let $n$ be the characteristic of $R$. If $n =0$, then we know that a copy of $\mathbb{Z}$ is sitting in $R$, namely the one generated by 1, which is an additive subgroup...

If $n > 0$, we know that $n$ must be a prime p. (If you don't know this, think about how you can get a contradiction if n has two distinct prime factors.) Now, same as above: we have a copy of $\mathbb{Z}/p \mathbb{Z}$ sitting in $R$, which is an additive subgroup...