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Find subrings of $\mathbb Z_{18}$ which illustrate the following:

  1. A is a ring with unity, B is a subring of A, but B is not a ring with unity.

  2. A and B are rings with unity, B is a subring of A, but the unity of B is not the same as the unity of A.

So for both 1 and 2, I would let A=$\mathbb Z_{18}$. The unity is 1. for 1. B={0,3,6,9,12,15}. This is a subring since its closed under addition and multiplication but it does not have the unity=1. However when I look at 2. I would say maybe B= {0,9} but then I get that B doe not have the unity=1, making me think that 0 is the unity for B.

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  • $\begingroup$ In your second $B$, what is $0\cdot 9$? Which of those two elements seems to have the unital property based on that one calculation? $\endgroup$
    – Arthur
    Oct 23, 2017 at 15:30
  • $\begingroup$ @Arthur I would assume 0 because anything times 0 gives us 0, however in modulo 18, anything times 9 could give 0 or 9. $\endgroup$
    – K Math
    Oct 23, 2017 at 15:32
  • $\begingroup$ And what is the unital property? What is the defining feature of unity, and does either of $9$ or $0$ fulfill that as elements of the ring $B$? $\endgroup$
    – Arthur
    Oct 23, 2017 at 16:29

1 Answer 1

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A unit must be idempotent. $B=\{0,9\}$ works because $9^2 \equiv 9 \bmod 18$ and $9$ is the unit of $B$.

The other nontrivial idempotent in $\mathbb Z_{18}$ is $10$ and so the additive subgroup generated by $10$ is another solution for 2.

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  • $\begingroup$ So then for a group without unity, could this be any subgroup of $\mathbb Z_{18}$ since they will not have a different identity or will they have their own unique identity? $\endgroup$
    – K Math
    Oct 23, 2017 at 15:57
  • $\begingroup$ @KellyR, I'm not sure what you're saying but the point is that units in subrings $B$ must be idempotents in the main ring $A$. You cannot choose units arbitrarily. $\endgroup$
    – lhf
    Oct 23, 2017 at 17:14

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