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Let $C(l)$ be a cyclic ring with elements $0,e,2e,...,(n-1)e$. Addition is defined by adding coefficients. Multiplication is defined such that $e^2=le$. So if we multiply two elements together, say $xe$ and $ye$, we get $xe*ye=xye^2=xyle$. Let $C(f)$ be the same ring except with $e^2=fe$.

If $C(l)$ is infinite, can $C(l)$ ever be isomorphic to $C(f)$?

If $C(l)$ is finite, what is the condition on $l$ and $f$ in order that there be an isomorphism?

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What are the multiplicative identities of both rings ? –  Amr Nov 12 '12 at 16:40
    
These rings are not with unit. Essentially they are $l\mathbb Z/nl\mathbb Z$. –  Hagen von Eitzen Nov 12 '12 at 16:53
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If $C(l)$ is infinite then it is isomorphic (via $e\mapsto l$) to $l\mathbb Z$. If $C(l)\cong C(f)$ then the only isomorphisms as additive gorups are given by $l\mapsto f$ or $l\mapsto-f$. If we assume $l,f>0$ then this is an isomorphism of rings only if $f=l$ because $l\cdot l=\underbrace{l+\cdots+l}_l$ and $f\cdot f=\underbrace{f+\cdots+f}_f$.

In the finite case, for $l\mathbb Z/ln\mathbb Z\cong f\mathbb Z/fn\mathbb Z$, there are $\phi(n)$ candidates for additive group homomorphisms $\psi\colon l\mathbb Z/ln\mathbb Z\to f\mathbb Z/fn\mathbb Z$, from sending the generator $l+ln\mathbb Z$ of the first group to $kf+fn\mathbb Z$ with $(k,n)=1$ and $1\le k\le n$. For a ring isomorphsim we additionally need that $\psi(l\cdot l)=\psi(l)\cdot\psi(l)$, i.e. $klf\equiv k^2f^2\pmod {fn}$. This implies $kl\equiv k^2f\pmod n$ and finally $l\equiv kf\pmod n$. This may have no or one or several solutions $k$, depending on $l,f,n$.

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