# Is the ring $(\mathbb Z/n, +, \times)$ a subring of $(\mathbb Z, +, \times)$?

Is the ring $(\mathbb Z/n, +, \times)$ a subring of $(\mathbb Z, +, \times)$?

I don't think it is, but I can't exactly see why.

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Once you specify an inclusion map ${\mathbb{Z}}_n\to\mathbb Z$, you see that this map does not preserve addition. –  Lubin Nov 7 '12 at 18:18

No. A subring of $\mathbb{Z}$ must contain $1$. Then it must contain all elements $1+1+\cdots+1$ and all their additive inverses. And also zero. So any subring of $\mathbb{Z}$ must be $\mathbb{Z}$ and in particular it is infinite while $\mathbb{Z}/(n)$ is finite.

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No, they have different characteristics: $1+1+\cdots+1$ ($n$ times) is zero in $\mathbb Z/n$ but never zero in $\mathbb Z$.

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$\Bbb Z/n$ isn’t even a subset of $\Bbb Z$; for that reason alone it cannot be a subring, strictly speaking. However, $\Bbb Z/n$ as a set is often identified with the subset $\{0,1,\dots,n-1\}$ of $\Bbb Z$, and that’s probably what you’re expected to do here. In that case just consider $(n-1)+(n-1)$. What is it in $\Bbb Z/n$? What is it in $\Bbb Z$?

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Hint  It fails already at the group level: an infinite cyclic group has no nontrivial finite subgroup.

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