2
$\begingroup$

so I was given the problem: find the characteristic of $Z_3\times Z_4$ and I got $\operatorname{char}(Z_3\times Z_4)=12$, is it true that for any $Z_n \times Z_m$, $\operatorname{char}(Z_n \times Z_m)=n*m$???

so for instance, does $\operatorname{char}(Z_2\times Z_3)=6$?

For $\operatorname{char}(Z_{10}\times Z_{20})$ would you take 20*10, or 10 because thats the GCF, or 2 because thats the lowest common factor????

Thanks!

$\endgroup$
2
  • 2
    $\begingroup$ A tip: When you hover your mouse on top of a tag, you get a brief description of what it means. Doing that here and reading it would reveal that [tag: characteristic-function] is not appropriate for your question. $\endgroup$ Dec 15, 2014 at 6:14
  • $\begingroup$ But have you tried applying the definition of characteristic? What problems did you encounter while doing that? $\endgroup$ Dec 15, 2014 at 6:24

1 Answer 1

2
$\begingroup$

Hint: I'd say $char(\mathbb{Z}_m\times \mathbb{Z}_n)=lcm(m,n)$. Can you see why?

Note $\mathbb{Z}_m\times \mathbb{Z}_n$ is a ring with unity $(1,1)$, so if we can find the characteristic of $(1,1)$, then we are done.

Let $k$ be its characteristic.

Then, $k(1,1)=(0\pmod m,0\pmod n)$

$(k,k)=(0\pmod m,0\pmod n)$

$m|k$ and $n|k\Rightarrow lcm(m,n)|k$

It is fairly simple to show that $k|lcm(m,n)$

$\endgroup$
3
  • $\begingroup$ So for Z10*Z20 it would be 10? $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:22
  • $\begingroup$ $lcm(10,20)=20$ $\endgroup$ Dec 15, 2014 at 6:23
  • $\begingroup$ Sorry, I misread the beginning of the post... that makes sense! $\endgroup$
    – Kaitlyn
    Dec 15, 2014 at 6:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .