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Question:

Find the number of non trivial ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$. ($f$ is not necessarily unitary, i.e., $f(1)$ need not be $1$.)

Suppose $f$ is a ring homomorphism from $\mathbb Z_{12}$ to $\mathbb Z_{28}$.

Consider $f$ as a additive group homomorphism. Let $k= |\ker f|$ and $ t = |\operatorname{im}(f)|$. Then $k\mid 12$ and $t\mid 24$ and $kt=12$, by first isomorphism theorem of groups.

There are two possibilities $k=3$, $t=4$ and $k=6$, $t=2$.

For the first case $f$ should map $1$ to an element of the subgroup generated by $7$ as there is a unique subgroup of $\mathbb Z_{28}$ of order $4$ generated by $7$. For the second case $1$ has to map to $14$, for the same reasoning.

So there are at most two ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$. Question is how to check the possible maps which are ring homomorphisms.

Thanks.

  • 0
    What about $k=12$?2012-10-23
  • 0
    We need non trivial homomorphisms.2012-10-23
  • 0
    For counting the group homomorphisms, it's true there's only one subgroup of order 4, but you can map 1 to *any* generator of that subgroup.2012-10-23
  • 0
    @GerryMyerson you are right. Thanks2012-10-23
  • 0
    So why didn't you change your incorrect statement that $f$ should map $1$ to $7$ when you edited the question?2012-10-23
  • 0
    Related: http://math.stackexchange.com/questions/263063/the-number-of-ring-homomorphisms-from-mathbbz-m-to-mathbbz-n/1564242#15642422015-12-07

4 Answers 4