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The action of the nonzero elements of a field on a vector space by left multiplication forms a faithful group action


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.7

Let $F$ be a field and $V$ an $F$-vector space. The axioms for vector spaces over a field demand that $F^\times$ act on $V$ by scalar multiplication. Show that this action is faithful.


Solution: Suppose to the contrary that there exist distinct $\alpha_1, \alpha_2 \in F^\times$ such that $$\alpha_1 \cdot v = \alpha_2 \cdot v$$ for all $v \in V$; we may choose a particular $v \neq 0$. Then $\alpha_1 \cdot v = \alpha_2 \cdot v$ means $\alpha_1 v = \alpha_2 v$, so that $\alpha_1 v - \alpha_2 v = 0$, hence $$(\alpha_1 - \alpha_2) v = 0.$$ This is only possible if $\alpha_1 - \alpha_2 = 0$; that is, if $\alpha_1 = \alpha_2$. This is a contradiction, so no such $\alpha_1, \alpha_2$ exist and the action is faithful.

Linearity

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