Question

(max new score 9/12) For each pair of rings below, name and explain two ring invariants that the rings differ on. Z₁₀ and Z₁₁ ,Z ₁₂ and Z₂⊕Z₆R and M₂(R)

Answer 1

The invariants that differentiate Z₁₀ and Z₁₁ include characteristic and units. Z₁₂ and Z₂⊕Z₆ differ in direct sum decomposition and nilpotent elements. R and M₂(R) differ in the presence of non-identity elements and commutativity.

Step-by-step explanation:

The student's question asks for two ring invariants that differentiate the given pairs of rings: Z₁₀ and Z₁₁, and Z₁₂ and Z₂⊕Z₆, as well as R and M₂(R). For the first pair, Z₁₀ and Z₁₁, the two ring invariants that differ are:

1. Characteristic: Z₁₀ has a characteristic of 10, meaning the smallest positive number of times the multiplicative identity (1) must be added to itself to yield the additive identity (0) is 10. Z₁₁ has a characteristic of 11 as the process requires 11 repetitions to reach 0.
2. Units: Z₁₀ has 4 units (1, 3, 7, 9), as these are the elements that have multiplicative inverses. Z₁₁ is a field and thus every non-zero element is a unit, so it has 10 units (1, 2, ..., 10).

For the second pair, Z₁₂ and Z₂⊕Z₆, the two ring invariants are:

1. Direct Sum Decomposition: Z₁₂ is a cyclic ring and cannot be decomposed as a direct sum of two non-trivial rings, whereas Z₂⊕Z₆ is the direct sum of Z₂ and Z₆.
2. Nilpotent Elements: Z₁₂ has nilpotent elements, such as 6 (since 6² = 36 ≡ 0 mod 12), while in Z₂⊕Z₆, there are no nilpotent elements as the direct sum of two rings does not allow non-zero nilpotent elements.

Finally, for R and M₂(R), R represents a commutative ring, and M₂(R) typically refers to the ring of 2x2 matrices over R, and their invariants differ on:

1. None identity elements: In R, all non-zero elements are not necessarily units, while in M₂(R), not all non-zero matrices are invertible.
2. Commutativity: R is commutative by definition, whereas M₂(R) is non-commutative since matrix multiplication is not commutative in general.