Your unfair advantage to profitable dropshipping
Your unfair advantage to profitable dropshipping
Your unfair advantage to profitable dropshippingExercise 4.1.2: Let $K$ be a field and $G$ a subgroup of $\operatorname{Aut}(K)$. Show that $K^G = {a \in K \mid \sigma(a) = a \text{ for all } \sigma \in G}$ is a subfield of $K$.
Exercise 4.1.1: Let $K$ be a field and $\sigma$ an automorphism of $K$. Show that $\sigma$ is determined by its values on $K^{\times}$.
Solution: Let $a \in K$. If $a = 0$, then $\sigma(a) = 0$. If $a \neq 0$, then $a \in K^{\times}$, and $\sigma(a)$ is determined by its values on $K^{\times}$.
($\Leftarrow$) Suppose every root of $f(x)$ is in $K$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$, showing that $f(x)$ splits in $K$.
Solution: Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $L = K(\alpha_1, \ldots, \alpha_n)$, and $[L:K] \leq [K(\alpha_1):K] \cdots [K(\alpha_1, \ldots, \alpha_n):K(\alpha_1, \ldots, \alpha_{n-1})]$.
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Exercise 4.1.2: Let $K$ be a field and $G$ a subgroup of $\operatorname{Aut}(K)$. Show that $K^G = {a \in K \mid \sigma(a) = a \text{ for all } \sigma \in G}$ is a subfield of $K$.
Exercise 4.1.1: Let $K$ be a field and $\sigma$ an automorphism of $K$. Show that $\sigma$ is determined by its values on $K^{\times}$. abstract algebra dummit and foote solutions chapter 4
Solution: Let $a \in K$. If $a = 0$, then $\sigma(a) = 0$. If $a \neq 0$, then $a \in K^{\times}$, and $\sigma(a)$ is determined by its values on $K^{\times}$. Exercise 4
($\Leftarrow$) Suppose every root of $f(x)$ is in $K$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$, showing that $f(x)$ splits in $K$. Show that $\sigma$ is determined by its values
Solution: Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $L = K(\alpha_1, \ldots, \alpha_n)$, and $[L:K] \leq [K(\alpha_1):K] \cdots [K(\alpha_1, \ldots, \alpha_n):K(\alpha_1, \ldots, \alpha_{n-1})]$.