1) Find allc ∈ Z3 such that Z3 [x] /(x3 + x2 + c)is a field.
2) Find the degree of and a basis for E = Q
(√2, √3 2, √4 2)
over Q.
$\displaystyle \mathbb{Z}_3[x]/(x^3+x^2+c)$ is a field if and only if $\displaystyle x^3+x^2+c$ is irreducible if and only if $\displaystyle x^3+x^2+c$ has no zeros in $\displaystyle \mathbb{Z}_3$.1) Find all
c ∈ Z3 such that Z3 [x] /(x3 + x2 + c)is a field.
Let $\displaystyle \alpha = \sqrt[12]{2}$ then since $\displaystyle x^{12}-2$ is an irreducible polynomial which has $\displaystyle \alpha$ as a root it means $\displaystyle \{1,\alpha,\alpha^2,...,\alpha^{11}\}$ is a basis. This means $\displaystyle \alpha^6,\alpha^4,\alpha^3$ are linearly independent. Thus, $\displaystyle \{ \sqrt{2}. \sqrt[3]{2}, \sqrt[4]{2} \}$ are linearly independent and so they spam a vector space of dimension three.2) Find the degree of and a basis for E = Q
$\displaystyle (\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})$
you see my comment in red! the degree is 12. let $\displaystyle a=\sqrt{2}, \ b=\sqrt[4]{2}, \ c=\sqrt[3]{2}.$ then:
$\displaystyle [\mathbb{Q}(a,b,c):\mathbb{Q}]=[\mathbb{Q}(a,b)(c):\mathbb{Q}(a,b)][\mathbb{Q}(a)(b):\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}]=3 \times 2 \times 2 = 12.$
a more constructive (and easier) way is to prove that $\displaystyle \mathbb{Q}(\sqrt[12]{2})=\mathbb{Q}(a,b,c)$:
clearly $\displaystyle \mathbb{Q}(\sqrt[12]{2}) \supseteq \mathbb{Q}(a,b,c).$ the reverse inclusion is also clear because $\displaystyle \sqrt[12]{2}=\frac{ab}{c^2}.$