Correct. Is that your only question?
Consider the ring R = {0,2,4,6,8,10,12} with the operations being addition and multiplication mod 14.
(a) Show that R is a ring with identity by ﬁnding the multiplicative identity (and prove that it’s the identity by computation).
(b) Find the multiplicative inverse of 10 in R.
Would I be doing this correctly:
For part (a):
8 would be the multiplicative identity since every element in R multiplied by 8 equals itself (mod 14)?
I hate multiplication tables. Honestly, I do.
What we want to do here is try to find a ring homomorphism between $R = \{0,2,4,6,8,10,12\}$ and $\Bbb Z_7$. The trick here is to find an IDEMPOTENT element of $R$, because identity elements are idempotent (they satisfy: $e^2 = e$).
We can eliminate 2 right off the bat, and:
4*4 = 2
6*6 = 8
10*10 = (-4)*(-4) = 4*4 = 2
12*12 = (-2)*(-2) = 2*2 = 4
so the only idempotent (besides 0) is 8: 8*8 = 8.
Next we need to show that 8*a = a, for all of our a's. Note that we have a = 2b (b = 0,1,2,3,4,5,6). So, mod 14, we have:
8a = 8(2b) = 16b = 2b = a. That establishes identity.
Since any non-zero element is a generator (because $(R,+)$ is an additive cyclic subgroup of $(\Bbb Z_{14},+)$, and since any ring homomorphism maps identity to identity), this suggests the homomorphism:
$\phi: \Bbb Z_7 \to R$ given by $\phi(k \text{ (mod }7)) = 8k \text{ (mod }14)$.
Of course, we have to verify the multiplicative property:
$\phi(km) = \phi(k)\phi(m)$. This means that:
$\phi(km) = 8km = (8)(8)(km) = (8k)(8m) = \phi(k)\phi(m)$, which is true since 8 is idempotent.
Now to find the inverse of 10 in $R$, we need to do 2 things:
a) find the pre-image of 10 under $\phi$
b) find the inverse of the pre-image in $\Bbb Z_7$
To accomplish (a), let's calculate explicitly what $\phi$ does:
$\phi(0) = 0$
$\phi(1) = 8$
$\phi(2) = 2$ (16 = 2 (mod 14))
$\phi(3) = 10$ <--we can stop, now, the pre-image of 10 is 3.
Now we need to find the inverse of 3 (mod 7). The easiest way to do this is to find integers r,s with 3r + 7s = 1. This means that 3r = 1 (mod 7), so taking r (mod 7) will give us the inverse. To find these integers, use the gcd division algorithm:
7 = 2(3) + 1, so:
1 = 7 - 2(3), that is: (-2)(3) + (1)(7) = 1, so r = -2, and s = 1. We don't need s, so forget about it. We DO need r, and we need to find -2 (mod 7), which is 5. Hence 5 is the multiplicative inverse of 3 (mod 7):
(3)(5) = 1 (mod 7) (because 15 = 1 (mod 7)).
So the inverse of 10 (mod 14) is going to be $\phi(5)$, which is 40 (mod 14), which reduces to 12 (40 = 12 + 2(14)).
The advantage to doing it this way is avoiding explicitly creating a multiplication table with 49 entries (although if you're clever, you can reduce it to just 21 entries, which is still more calculation than I like to do).