Final answer:
The function f from Z₆ to Z₂xZ₃ is a ring isomorphism as it is bijective and preserves both addition and multiplication, verified by using the addition and multiplication tables for both rings.
Step-by-step explanation:
To show that a function f from Z₆ to Z₂xZ₃ is a ring isomorphism, we need to verify two main properties: that f is a ring homomorphism and that f is bijective. A ring homomorphism between two rings preserves the addition and multiplication operations, and a function is bijective if it is both injective (one-to-one) and surjective (onto).
From the given specification of values function f maps:
- 0↦(0,0)
- 1↦(1,1)
- 2↦(0,2)
- 3↦(1,0)
- 4↦(0,1)
- 5↦(1,2)
we can deduce that every element in
Z₆
has a unique image in
Z₂xZ₃
, satisfying injectivity.
Additionally, every element of Z₂xZ₃ is the image of some element of Z₆, which ensures surjectivity. Furthermore, we examine the preservation of addition and multiplication by looking at the addition and multiplication tables for Z₆ and Z₂xZ₃. For example, let's verify the preservation of addition for a pair of elements:
- Addition in Z₆: (1 + 2) mod 6 = 3
- Addition in Z₂xZ₃: (1,1) + (0,2) = (1+0 mod 2, 1+2 mod 3) = (1,0)
Since f(1+2) = f(3) = (1,0) which is equal to f(1)+f(2), addition is preserved. The preservation of multiplication is verified in a similar manner. Hence, the function f is a ring isomorphism between Z₆ and Z₂xZ₃.