Exercise on Rings 4
We want to determine whether the function f(x) = 3x defines a ring homomorphism between the rings (Z6, +, *) and (Z6', +, *), where both addition and multiplication are taken modulo 6.
As a first step, let’s construct the addition table modulo 6.
| a +6 b | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
Now let’s build the multiplication table modulo 6.
| a ·6 b | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 0 | 2 | 4 |
| 3 | 0 | 3 | 0 | 3 | 0 | 3 |
| 4 | 0 | 4 | 2 | 0 | 4 | 2 |
| 5 | 0 | 5 | 4 | 3 | 2 | 1 |
We now verify whether the function satisfies the first condition for being a ring homomorphism:
$$ f(a + b) = f(a) + f(b) $$
Since f(x) = 3x, we have:
$$ 3 \cdot (a + b) = 3a + 3b $$
This identity holds true modulo 6, so the function preserves addition.
Next, we test whether f(x) also preserves multiplication:
$$ f(a \cdot b) = f(a) \cdot f(b) $$
Substituting again:
$$ 3 \cdot (a \cdot b) = 3a \cdot 3b $$
By associativity and modular arithmetic, the equality is verified.
Therefore, the function f(x) = 3x is a ring homomorphism from (Z6, +, *) to itself.
Verification
Since we are working with a small, finite ring, it’s useful to explicitly display the tables for the homomorphism to confirm the result.
For addition, both 3(a + b) and 3a + 3b yield the same table:
