# is the intersection of all non-trivial subrings $\mathbb{Z}_n$?

Let $$R$$ be a ring.

Define $$\phi:\mathbb{Z}\rightarrow R: n\mapsto n\cdot 1_R$$

This is a ring homomorphism and below is a theorem using this homomorphism:

Let $$R$$ be a ring with $$char(R)=n$$.

Then, there exists a subring $$S$$ of $$R$$ which is ring isomorphic to $$\mathbb{Z}_n$$.

Is this $$S$$ the intersection of all non-trivial subrings of $$R$$?

• $S=\phi(\mathbb Z)$, $\ker \phi=n\mathbb Z$. – lhf Nov 14 '14 at 15:01

Since you have mentioned $1_R$, I assume that $R$ has unit. I'll write $1=1_R$, $2=2\cdot 1_R$, etc.

If $S'$ is a subring, then $1\in S'$ so $\{0,1,2,\ldots,n-1\}\subset S'$. That is, $S\subset S'$.

You have defined a homomorphism $\phi:\mathbb{Z}\rightarrow R$ by $\phi(a)=1_R\cdot a$ where $1_R$ is the unity of $R$.

Clearly, $\phi$ is a map from $\mathbb{Z}$ onto $\phi(\mathbb{Z})$.

Now, show that $Ker\,\, \phi=\langle n \rangle$

By, first isomorphism theorem, you have

$\frac{\mathbb{Z}}{\langle n \rangle}\cong \phi(\mathbb{Z})$

So, you've just shown that every ring of characteristic $n$ with unity has a subgroup isomorphic to $\mathbb{Z}_n$. So what happens when you take intersection of all such rings?