Exercise 1.7.1 ($F^ imes$ acts on $F$ by $g\cdot a = ga$)

Question

Let F be a field. Show that the multiplicative group of nonzero elements of $F$ (denoted by $F^{	imes}$) acts on the set $F$ by $g\cdot a = ga$, where $g \in F^{	imes}, a \in F$ and $ga$ is the usual product in $F$ of the two field elements (state clearly which axioms in the definition of a field are used).

Richard Ganaye · Accepted Answer

\begin{proof} 
We define the action $\varphi$ by
$$
\varphi 
\left\{
\begin{array}{ccl}
F^	imes 	imes F & 	o & F\
(g,a) & \mapsto & g\cdot a = ga
\end{array}
ight.
$$

\begin{itemize}
\item
Since the multiplication in $F$ is associative, then

$(i) \quad \forall g \in F^	imes,\ \forall h \in F^	imes, \ \forall a \in F,\ (g\cdot h) \cdot a = (gh)a = g(ha) = g \cdot (h \cdot a).$

\item
Since $1$ is the identity of $F^*$,

$(ii) \quad \forall g \in F^	imes, \ 1 \cdot g = 1 g = g.$

\end{itemize}
Therefore $F^{	imes}$ acts on the set $F$ by $g\cdot a = ga$.
\end{proof}

Exercise 1.7.1 ($F^\times$ acts on $F$ by $g\cdot a = ga$)

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Exercise 1.7.1 ($F^\times$ acts on $F$ by $g\cdot a = ga$)

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