**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.7 Exercise 1.7.1**

Solution: Let $a \in F$. We have $1 \cdot a = 1a = a$ by the definition of identity. Now let $g_1, g_2 \in F^\times$. Then $$g_1 \cdot (g_2 \cdot a) = g_1 \cdot g_2 a = g_1(g_2a) = (g_1g_2)a = (g_1g_2) \cdot a,$$ by the associativity of multiplication in $F$. So left multiplication by nonzero elements is in fact a group action.

## Joseph Yung

23 Oct 2020Is it necessary to additionally show for the case that "a" equals the identity of additive group of the field (i.e. a = 0)? since if a = 0, then "a" is not in the multiplicative group of the field and we thus cannot use the group axioms the multiplicative group such as associativity and identity.

## Linearity

25 Oct 2020No. There is no need to consider $a=0$ as $a\notin F^\times$.