# Subrings of $\mathbb Z_{18}$

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.

• 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? Oct 23, 2017 at 15:30
• @Arthur I would assume 0 because anything times 0 gives us 0, however in modulo 18, anything times 9 could give 0 or 9. Oct 23, 2017 at 15:32
• 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 $B$? Oct 23, 2017 at 16:29

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.
• 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? Oct 23, 2017 at 15:57
• @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.