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

  • $\begingroup$ What is $Z_n$ and $Z$? Also what is $R$ in your $K$, and $f$? $\endgroup$ – Pink Panther Jan 19 '18 at 8:11
  • $\begingroup$ R is the ring and Z is the integers Zn is the n congruence class $\endgroup$ – user420309 Jan 19 '18 at 8:16
  • $\begingroup$ and what about $f$? $\endgroup$ – Pink Panther Jan 19 '18 at 8:18
  • $\begingroup$ ahh sorry, that should be $\phi$ $\endgroup$ – user420309 Jan 19 '18 at 8:20

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 $1\leq k\leq n-1$.

  • $\begingroup$ Thank you, but what is the bar above n,0,1? $\endgroup$ – user420309 Jan 19 '18 at 8:24
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    $\begingroup$ The equivalent classes of $k$ in ${\bf{Z}}_{n}$. $\endgroup$ – user284331 Jan 19 '18 at 8:25

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 $n>0$.

  • $\begingroup$ If $x \in Zn$ then nx=0 (can I ask you why)? $\endgroup$ – user420309 Jan 19 '18 at 8:45
  • $\begingroup$ @Djhoe $x=a+n\mathbb{Z}$, for some $a$; then $nx=na+n\mathbb{Z}$ (by definition) and $na\equiv0\pmod{n}$. At a higher level: $\mathbb{Z}_n,+$ is a torsion group and $\mathbb{Z},+$ is a torsionfree group; any homomorphism from a torsion group to a torsionfree group is trivial. $\endgroup$ – egreg Jan 19 '18 at 8:49
  • $\begingroup$ Okay thank you very much. unfortunately I only know about rings so far. $\endgroup$ – user420309 Jan 19 '18 at 8:57

$\phi :\Bbb Z\to \Bbb Z_n$ 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 $n>0$.

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


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