Solution

The correct answer is neither left ideal nor right ideal.

Key Points

• Given set S = { [ a 0 ; 0 b ] | a,b ∈ ℤ } ⊂ M₂(ℤ).
• To be a left ideal, for any A ∈ M₂(ℤ) and X ∈ S, AX ∈ S must hold.
• Take A = [0 1; 1 0] ∈ M₂(ℤ), X = [a 0; 0 b] ∈ S.
• Compute AX: AX = [0 1;1 0][a 0;0 b] = [0·a+1·0 , 0·0+1·b ; 1·a+0·0 , 1·0+0·b] = [0 b ; a 0].
• This matrix is NOT diagonal ⇒ AX ∉ S. Hence S is not a left ideal.
• To be a right ideal, XA ∈ S must hold.
• Compute XA: XA = [a 0;0 b][0 1;1 0] = [0 a ; b 0].
• This is also NOT diagonal ⇒ XA ∉ S. Hence S is not a right ideal.

Additional Information

• Conclusion:
  • S is closed under addition and multiplication internally ⇒ subring.
  • But multiplication with arbitrary matrices destroys diagonal form.
  • Therefore S is neither left ideal nor right ideal.