Exercise on Rings 3
Is the set of integers Z6 = {0, 1, 2, 3, 4, 5} a ring, with addition defined as modulo 6 addition and multiplication defined as modulo 6 multiplication?
$$ (Z_6 ,+_6, \cdot_6) $$
Since Z6 is a finite set, we begin by constructing the addition table:
| 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 |
And now the multiplication table:
| 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 |
Both addition and multiplication are binary operations that are closed on Z6, so the first requirement for a ring is satisfied.
Let’s now verify whether the remaining ring axioms hold.
The First Operation (+6)
Addition modulo 6 is commutative on Z6:
$$ a +_6 b = b +_6 a \quad \forall \ a, b \in Z_6 $$
It is also associative:
$$ (a +_6 b) +_6 c = a +_6 (b +_6 c) \quad \forall \ a, b, c \in Z_5 $$
The additive identity is 0:
$$ a +_5 0 = 0 +_5 a = a \quad \forall \ a \in Z_5 $$
Moreover, every element in Z6 has an additive inverse.
For example: 0 + 0 = 0, 1 + 5 = 0, 2 + 4 = 0, 3 + 3 = 0, etc.
| 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 |
Therefore, the addition operation satisfies all the axioms required of a ring.
Note. The structure (Z6, +) forms a group, as Z6 is non-empty, addition modulo 6 is closed and associative, there exists an identity element (0), and each element has an inverse. Since addition is also commutative, (Z6, +) is in fact an abelian group.
The Second Operation (·6)
Multiplication modulo 6 is associative over Z6:
$$ (a \cdot_6 b) \cdot_6 c = a \cdot_6 (b \cdot_6 c) \quad \forall \ a, b, c \in Z_6 $$
It also satisfies the distributive laws with respect to addition:
$$ (a +_6 b) \cdot_6 c = a \cdot_6 c +_6 b \cdot_6 c \quad \forall \ a, b, c \in Z_6 $$
$$ a \cdot_6 (b +_6 c) = a \cdot_6 b +_6 a \cdot_6 c \quad \forall \ a, b, c \in Z_6 $$
Hence, the multiplication operation meets all the ring requirements.
Conclusion
To conclude, both binary operations satisfy the axioms of a ring.
Therefore, we can affirm that the algebraic structure (Z6, +, ·) is a ring.
And so forth.
