1
$\begingroup$

I was wondering how one shows that $\phi(m):\mathbb{Z}\rightarrow\mathbb{Z}_n$ is a homomorphism? I know that we define $m:=qn + r$, but I don't really know how to continue from there.

$\endgroup$
  • $\begingroup$ Well, it depends on what $\phi$ is. $\endgroup$ – Randall Oct 27 '17 at 0:53
  • $\begingroup$ I got this from jupiter.math.nctu.edu.tw/~weng/courses/alg_2007/Algebra%202006/… on the 4th page. I don't really know if I'm defining something incorrectly in my question $\endgroup$ – K.M Oct 27 '17 at 1:04
  • $\begingroup$ use $\phi(z)= z \ (mod \ n)$ $\endgroup$ – Philip White Oct 27 '17 at 1:05
  • $\begingroup$ What's your definition of $\mathbf Z_n$? $\endgroup$ – Bernard Oct 27 '17 at 1:05
  • 1
    $\begingroup$ @PhilipWhite: So I just show that $(a+b)mod n = amodn +bmodn$ by definition of modular addition? $\endgroup$ – K.M Oct 27 '17 at 1:11
1
$\begingroup$

$\phi(z) = z \pmod n$

i.e. for $z,m \in\mathbb Z$

so we have

$\phi(z) = z \pmod n$

$\phi(m) = m \pmod n$

adding the above two

$\phi(z)+\phi(m)=z+m \pmod n $

similarly we have

$\phi(z+m) = z+m \pmod n$

and finally we see

$\phi(z+m) =\phi(z) + \phi(m)$

not sure how deep into the division algorithm you needed to go...

$\endgroup$
1
$\begingroup$

You have $m=qn + r,$ and there you should mention that $r\in\{0,1,2,\ldots,n-1\}.$

The homomorphism would be $\varphi(m) = r.$

Showing that that is a homomorphism means showing that $\varphi(m_1+m_2) = \varphi(m_1)+\varphi(m_2).$

That means if $m_1 = q_1 n + r_1$ and $m_2 = q_2 n + r_2$ and $m_1+m_2 = q_3 n + r_3$ then $r_1+r_2\equiv r_3\pmod n.$ That means $(r_1+r_2)-r_3$ is a multiple of $n.$ So observe that $$ r_1+r_2-r_3 = (q_1+q_2-q_3) n. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.