Let $R$ be a commutative ring with $1_R$ such that every subring of it is an ideal. Prove that $R=\mathbb{Z}$ or $R=\mathbb{Z}_n$.
Can someone give me a hint to start solving this?
Any help is appreciated.
Thank you in advance.
[Math] Ring such that every subring is an ideal
abstract-algebraidealsring-theory
Best Answer
The hint by Orat is correct .You need the fact that There is a homomorphism from $\mathbb{Z}$ into $R$ uniquely determined (or recursively defined by $(n+1)1 =n1 +1 (1 =1_R )$ the identity of $R$ ) . $S = \{n1 |n \in \mathbb{Z} \}$ is a sub-ring hence an ideal , Why is $S = R$
If $h(n) =n1$ , the set $K= h^{-1}(\{0\})$ is an ideal in $\mathbb{Z}$ (the kernel of $h$) .what are the ideals of $\mathbb{Z}$?
(Answer : $(x) = \{nx |n \in \mathbb{Z}\}$, ($x =0,1,2,3,4 ...$) so $h$ induces an isomorphism of $\mathbb{Z}$/$(x)$ with $S = R$ .
By the way you should for completeness also show that every every sub-ring of $\mathbb{Z}$ is an ideal (and the same for sub-rings of $\mathbb{Z}$/$(x)$ )
Stuart M.N.