I have a question about how I have to do this exercise for my math study:

Let d, n $\in$ $\mathbb{Z}$>0 with d|n.

a) Prove that there is a homomorphism f: $\mathbb{Z}$/n$\mathbb{Z}$ $\rightarrow$ $\mathbb{Z}$/d$\mathbb{Z}$ and that f(a mod n) = (a mod d) for every a $\in$ $\mathbb{Z}$.

b) Is f surjective?

To prove homomorphism, I have to prove that the general rule: f(r + s) = f(r) + f(s) holds, and that the function f is well-defined, but I haven't worked a lot with residue classes, so I don't actually know how to do this exercise. I have a lot more of this exercises to do, so I thought that maybe you could show me this one, and then I can do the rest on myself. Your help would be very much appreciated, because I'm stucked here for a few days now.

Thanks in advance!!!

  • $\begingroup$ I recommend that you choose some explicit numbers and write them down. See if that gets you started. Perhaps $15$ and $5$, for instance. $\endgroup$ – davidlowryduda Dec 12 '14 at 22:08

Just take $f: Z_{n} \to Z_{d}$ by $f([a]_{n}) = [a]_{d}$. Check it is well defined.

If $n\mid a-b$, then $[a]_{n} = [b]_{n}$. But then $d\mid a-b$ since $d\mid n$. So $[a]_{d} = [b]_{d}$.

b). Let $p = \dfrac{n}{d} \geq 2$, then for any $y \in Z_{d}$, i.e $0 \leq y \leq d-1$, then take $x = y +(p-1)d$, then $x \in Z_{n}$, and $0 \leq x \leq n-1$. So $f([x]_{n}) = [y]_{d}$, hence $f$ is surjective.

  • $\begingroup$ Thanks. That answers my question about how to prove that the function is well-defined. Now I think I have the answer about how to prove that it is a homomorphism. Prove: Let $[a]_{n}$ , $[b]_{n}$ $\in$ $\mathbb{Z}$/n$\mathbb{Z}$. Then f($[a]_{n} + [b]_{n}$) = f($[a + b]_{n}$) = $[a + b]_{d}$ = $[a]_{d}$ + $[b]_{d}$. Is that correct? And how to explain that f is surjective? $\endgroup$ – Peter Dec 13 '14 at 15:17
  • $\begingroup$ That is correct. $\endgroup$ – DeepSea Dec 14 '14 at 2:23

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