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Why must one verify that (xy+yx+I)v=0(xy+yx+I)v=0 before calling CnC^n an R-module?

One must verify this to ensure the scalar multiplication is well-defined. Since xy+yxxy+yx is in the ideal II, its coset is the zero element in RR. If the action of this zero element does not yield the zero vector, the action would depend on the chosen representative rather than the coset itself.

Conditions

  • R=C⟨x,y⟩/⟨xy+yx⟩R = \mathbb{C}\langle x,y\rangle / \langle xy+yx\rangle.
  • The proposed action is (x+I)v=Xv(x+I)v=Xv and (y+I)v=Yv(y+I)v=Yv.
  • X,YX,Y satisfy XY+YX=0XY+YX=0.

Reasoning, step by step

  1. Recognize that xy+yx∈Ixy+yx \in I, so xy+yx+I=0Rxy+yx+I = 0_R in the quotient ring.
  2. Check that the proposed action of 0R0_R on any vector vv results in the zero vector 0M0_M.
  3. Compute (xy+yx+I)v=(x+I)(y+I)v+(y+I)(x+I)v(xy+yx+I)v = (x+I)(y+I)v + (y+I)(x+I)v.
  4. Substitute the matrix actions to get XYv+YXv=(XY+YX)vXYv + YXv = (XY+YX)v.
  5. Conclude that this equals 00 because XY+YX=0XY+YX=0, proving the action is well-defined.

Example

The lecturer performs the 'Check OK' computation: `(xy+yx+I)v=(x+I)(y+I)v+(y+I)(x+I)vv = (x+I)(y+I)v + (y+I)(x+I)v = XYv + YXv = (XY+YX)v=0v = 0`, showing that the relation is respected.

Common misconceptions

  • Thinking that defining the action on generators xx and yy is automatically sufficient without checking relations.
  • Confusing the zero element of the ring RR with the zero vector of the module MM.

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