Two facts about unital modules

Let R be a ring with 1 and let M be a left unital R-module. Prove that for all m \in M, 0 \cdot m = 0 and (-1) \cdot m = -m.


Let m \in M. Now 0 \cdot m + 1 \cdot m = (0+1) \cdot m = 1 \cdot m by the definition of module. Since M is an abelian group, subtracting 1 \cdot m from both sides we have 0 \cdot m = 0.

Now note that 1 \cdot m + (-1) \cdot m = (1 - 1) \cdot m = 0 \cdot m = 0. By the uniqueness of inverses in a group, -m = -(1 \cdot m) = (-1) \cdot m.

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