The exact question is:

Is it possible that a ring with unity may simultaneously contain subrings isomorphic to $\mathbb{Z}_n$ and $\mathbb{Z}_m$, where $n \ne m$?

The answer says it's possible and says $\mathbb{Z}_2 \times \mathbb{Z}_3$ has a subring isomorphic to $\mathbb{Z}_2$ and one isomorphic to $\mathbb{Z}_3$. I'm just trying to understand the problem better, but I can't see which two subrings are isomorphic. Could someone explain this solution to me? Thank you.

  • $\begingroup$ HINT: $\Bbb Z_2 \times \Bbb Z_3 \cong \Bbb Z_6$. $\endgroup$ – Marc Bogaerts Nov 15 '16 at 14:37

If you are working in the category of rings, yes, as shown in another answer.

But if you are working in the category of rings with unity, that is, if you insist on subrings having the same unity as the ring in which they are contained, then you have a problem.

In fact, in $\mathbb{Z}_n$ you have $n \cdot 1 = 0$ (where $1$ is the unity, and $n \cdot 1$ denotes the $n$-th multiple of $1$), and $k \cdot 1 \ne 0$ for $0 < k < n$. In $\mathbb{Z}_m$ you have $m \cdot 1 = 0$, and $k \cdot 1 \ne 0$ for $0 < k < m$.

If $m < n$ or $n < m$ we obtain a contradiction. So $m = n$.

  • $\begingroup$ So are you saying it isn't possible? This problem is from Fraleigh Abstract Algebra by the way, section 27 #25, if you need additional reference. The solutions seem to say it is possible... now I'm getting confused. $\endgroup$ – Max Nov 15 '16 at 9:02
  • 2
    $\begingroup$ It depends very much on the exact definitions. Do you insist on subrings to have the same identity as the ring? Then my argument applies. You allow subrings to have a different identity? Then the argument in the other answer applies. $\endgroup$ – Andreas Caranti Nov 15 '16 at 9:55

The two are $\mathbb{Z}_2\times \{0\}$ and $\{0\}\times \mathbb{Z}_3$ with the most natural isomorphism $$\mathbb{Z}_2\times \{0\}\cong \mathbb{Z}_2,$$ and analogously for the other subring.

  • $\begingroup$ The problem is that these aren't sub(rings with unity). They are subrngs, they both have a unit, but the unit doesn't correspond to the unit of the full ring. $\endgroup$ – Najib Idrissi Nov 15 '16 at 10:08
  • $\begingroup$ Sure. That's why in that case the other answer applies. $\endgroup$ – b00n heT Nov 15 '16 at 10:16

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