Prove that the only ring homomorphism is the trivial one

Question

I am stuck in this exercise. it asks me to:

Proof that the only ring homomorphism $\phi:\Bbb Z_{n} \rightarrow \Bbb Z$ is the trivial one $\phi(n)=0$

I believe that I have to show that the kernel $K=\{n \in R|\phi(n)=0_{s} \}$

But I am not sure how to start, any hints will be appreciated

Answer 1

$\phi :\Bbb Z\to \Bbb Z_n$ such that $\phi(x)=\bar{x}$. Then $$\rm{ker}\phi=\{x\in \Bbb Z:\phi(x)=\bar{0}\}$$ $$\rm{ker}\phi=\{x\in \Bbb Z:\phi(x)=\bar{x}=\bar{0}\}$$ $$\rm{ker}\phi=\{x\in \Bbb Z:x\sim 0\}$$ $$\rm{ker}\phi=\{x\in \Bbb Z:x-0=nk, k\in \Bbb Z\}$$ $$\rm{ker}\phi=\{x\in \Bbb Z:x=nk, k\in \Bbb Z\}=\langle n\rangle$$

Hence $\Bbb Z/\langle n\rangle\cong \Bbb Z_n$.

This is only because $\phi(1)=\bar{1}$ determines $\phi$.

$\phi(n)=\phi(1)+\cdots+\phi(1)=\bar{1}+\cdots +\bar{1}$ for $n>0$.

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Actually, There is one and only unital ring homomorphism from $\Bbb Z\to R$

Answer 2

We have $0=\phi(\overline{0})=\phi(\overline{n})=\phi(n\overline{1})=n\phi(\overline{1})$, so $\phi(\overline{1})=0$.

So $\phi(\overline{k})=k\phi(\overline{1})=0$ for all $1\leq k\leq n-1$.

Answer 3

If $x\in\mathbb{Z}_n$, then $nx=0$, so also $\phi(nx)=0$.

For a homomorphism, $\phi(nx)=n\phi(x)$. Now use that $n>0$.