Abstract Algebra Dummit And Foote Solutions Chapter 4 Apr 2026

Solution: Clearly, $0, 1 \in K^G$. Let $a, b \in K^G$. Then for all $\sigma \in G$, we have $\sigma(a) = a$ and $\sigma(b) = b$. Hence, $\sigma(a + b) = \sigma(a) + \sigma(b) = a + b$, $\sigma(ab) = \sigma(a)\sigma(b) = ab$, and $\sigma(a^{-1}) = \sigma(a)^{-1} = a^{-1}$, showing that $a + b, ab, a^{-1} \in K^G$.

Solution: The minimal polynomial of $\zeta_5$ over $\mathbb{Q}$ is the $5$th cyclotomic polynomial $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$. Since $\Phi_5(x)$ is irreducible over $\mathbb{Q}$ (by Eisenstein's criterion with $p = 5$), we have $[\mathbb{Q}(\zeta_5):\mathbb{Q}] = 4$. The roots of $\Phi_5(x)$ are $\zeta_5, \zeta_5^2, \zeta_5^3, \zeta_5^4$, and $\mathbb{Q}(\zeta_5)$ contains all these roots. Hence, $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is a splitting field of $\Phi_5(x)$ and therefore a Galois extension. abstract algebra dummit and foote solutions chapter 4

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 $K$. Solution: Clearly, $0, 1 \in K^G$

Solution: ($\Rightarrow$) Suppose $f(x)$ splits in $K$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$ for some $\alpha_1, \ldots, \alpha_n \in K$. Hence, every root of $f(x)$ is in $K$. Hence, $\sigma(a + b) = \sigma(a) + \sigma(b)

Exercise 4.1.1: Let $K$ be a field and $\sigma$ an automorphism of $K$. Show that $\sigma$ is determined by its values on $K^{\times}$.

Exercise 4.3.1: Show that $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is a Galois extension, where $\zeta_5$ is a primitive $5$th root of unity.

Exercise 4.2.1: Let $K$ be a field and $f(x) \in K[x]$. Show that $f(x)$ splits in $K$ if and only if every root of $f(x)$ is in $K$.