# Ring such that every subring is an ideal

Let $R$ be a commutative ring with $1_R$ such that every subring of it is an ideal.Prove that $R=\mathbb{Z}$ or $R=\mathbb{Z}_n$ or $R=\{0\}$.

Can someone give me a hint to start solving this?

Any help is appreciated.

• Hint: consider a subring $S := \{\, n1_R \mid n \in \mathbb{Z} \,\} \subseteq R$. – Orat May 2 '17 at 21:21

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.

The formulation seems to imply that subrings need to share the identity with the ring.

Let $S$ be a subring of $R$ (the intersection of all subrings of $R$); then $S$ is also an ideal, by assumption. Since $1\in S$, for every $r\in R$ we have $r1=r\in S$.

Thus $S=R$: hence $R$ is the only subring of itself.

Consider the (unique) ring homomorphism $\chi\colon\mathbb{Z}\to R$ (see below). Since the image of $\chi$ is a subring of $R$, by the argument above, the image of $\chi$ is $R$, that is, $\chi$ is surjective. The statement follows (look at the spoiler, if not able to finish).

Hence$$R\cong\mathbb{Z}/\ker\chi$$by the isomorphism theorem.

If subrings need not share the identity, the situation is more complicated. For instance the property is obviously true for “zero rings”, that is, additive abelian groups with the multiplication defined by $ab=0$, for every $a,b$: here every subgroup is both a subring and an ideal.

The unique ring homomorphism $\chi$ is defined by $n\mapsto n1_R$, where $n1_R=\underbrace{1_R+\dots+1_R}_{\text{$n$times}}$ for $n>0$, as usual, with the standard extension for $n\le0$.