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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.

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  • $\begingroup$ What about $k=12$? $\endgroup$
    – Phira
    Oct 23 '12 at 5:56
  • $\begingroup$ We need non trivial homomorphisms. $\endgroup$
    – user38764
    Oct 23 '12 at 6:08
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    $\begingroup$ 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. $\endgroup$ Oct 23 '12 at 6:22
  • $\begingroup$ @GerryMyerson you are right. Thanks $\endgroup$
    – user38764
    Oct 23 '12 at 6:31
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    $\begingroup$ So why didn't you change your incorrect statement that $f$ should map $1$ to $7$ when you edited the question? $\endgroup$ Oct 23 '12 at 11:51
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Number of ring homomorphism from $\mathbb Z_m$ into $\mathbb Z_n$ is $2^{[w(n)-w(n/\gcd(m,n))]}$ , where $w(n)$ denotes the numbers of prime divisors of positive integer n. From this formula we get number of ring homomorphism from $\mathbb Z_{12}$ to $\mathbb Z_{28}$ is $2$.

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  • $\begingroup$ Will above formula work for Zn to Zn. Please reply $\endgroup$
    – sabeelmsk
    Dec 7 '19 at 3:59
  • $\begingroup$ What is $w(1)$ defined $\endgroup$
    – sabeelmsk
    Dec 14 '19 at 15:41
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Even if not necessarily $f(1)=1$, we still need $f(1)\cdot f(1)=f(1\cdot 1)=f(1)$, hence $f(1)$ must be a solution of $x^2-x=0$ in $\mathbb Z_{28}$. The solutions for this are the residue classes of: $0, 1, 8, 21$. Therefore, $f\colon \mathbb Z_{12}\to\mathbb Z_{28}$ is necessarily induced by $\mathbb Z\to\mathbb Z$, $x\mapsto cx$ with $c\in \{0,1,8,21\}$. Also, we need $28|12c$, i.e. $c$ must be a multiple of $7$. This leaves only the cases $c=0$ and $c=21$.

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A ring homomorphism $f:Z_{12}\to Z_{28}$ must map $1$ to $1$, because ring homomorphisms do that (if you are not assuming your ring homomorphisms to be unitary, you should always be explicit about it!)

Now $\underbrace{1+1+\dots+1}_{\text{$12$ summands}}=0$ in $Z_{12}$, so applying $f$ we must have $$\underbrace{1+1+\dots+1}_{\text{$12$ summands}}=\underbrace{f(1)+f(1)+\dots+f(1)}_{\text{$12$ summands}}=0$$ in $Z_{28}$. As this is not true, there is no morphism!

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    $\begingroup$ We assume that f is not unitary. $\endgroup$
    – user38764
    Oct 23 '12 at 6:33
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    $\begingroup$ Then please do add that information to the question. $\endgroup$ Oct 23 '12 at 6:52
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    $\begingroup$ You should not have accepted this, as this does not do what you wanted" $\endgroup$ Oct 23 '12 at 13:37
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Hint $\ $ Applying the hom $\rm\,f\,$ to $\rm\:12\cdot 1 = 0\:$ and $\:1^2\!= 1,\:$ denoting $\rm\:e = f(1),\:$ we infer that $\rm\: 12\,e = 0\:\Rightarrow\:28\:|\:12\,e\:\Rightarrow\: 7\:|\:e,\:$ hence $\rm\:mod\ 28\!:\ e \equiv 0,7,14,21;\:$ also $\rm\:e^2 \equiv e\:$ hence $\rm\:e\equiv\,\ldots$

To understand better the role played by idempotents $\rm\:e^2 = e\:$ look up the Peirce decomposition.

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