a) Find all the subrings of $\mathbb{Z}_{12}$ (give a generator for each subring)
b) For each nonzero proper subring of $\mathbb{Z}_{12}$, determine whether there is a multiplicative identity.
So I found the subrings to be:
$\langle 2 \rangle = \{0,2,4,6,8,10\}$
$\langle 3 \rangle = \{0,3,6,9\} $
$\langle 4 \rangle = \{0,4,8\} $
$\langle 6 \rangle = \{0,6\} $
For b is where I was unsure, I have:
$\langle 6 \rangle$ No multiplicative identity b/c this is a trivial product structure, every multiple of 6 will 6 or 0.
$\langle 4 \rangle$ has a multiplicative identity of 4? (or is the same as for $\langle 6 \rangle$)?
$\langle 3 \rangle$ No multiplicative identity b/c $4 \cdot 3 \equiv 12 \equiv 0 \pmod{12} \implies 4 \cdot (3y) \equiv 0 \pmod{12} \implies 4 \cdot e \equiv 0$ (where e is inverse and this can never happen)
$\langle 2 \rangle$ has a multiplicative identity 7, $7 \cdot 2 = 14 \equiv 2 \pmod{12}$ (so every multiple of 2)
so $7 \cdot (2y) \equiv 2y \pmod{12}$
I'm not sure if I did this right and am looking for clarification/ verification of what I have done, especially for $\langle 4 \rangle$.
Thank you.
Best Answer
When checking whether there is a multiplicative inverse you should check whether there is such $x \in S \subseteq R$, s.t. $x\cdot a = a \cdot x = a$ for all $a \in S$, not for all $a \in R$.
Therefore as you've noted $\langle 6 \rangle$ has no multiplicative identity, as the only possibility is $6$ (multiplicative identity must be non-zero) and $6 \cdot 6 = 0$
For $\langle 4 \rangle$ we have a multiplicative identity, namely $4$. Similarly for $\langle 3 \rangle$ the multiplicative identity is $9$, as $9 \cdot 6 = 6$, $9 \cdot 9 = 9$ and $9 \cdot 3 = 3$.
On the otherside $\langle 2 \rangle$ has no multiplicative identity as $6 \cdot a = 0$ for all even $a$.
Furthermore you have forgotten two more subrings, the zero subring $\langle 0 \rangle$ and $\langle 1 \rangle$, the whole ring.