1. ## Fields

1) Find all
c 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.

2. 1) Find all
c ∈ Z3 such that Z3 [x] /(x3 + x2 + c)is a field.
$\mathbb{Z}_3[x]/(x^3+x^2+c)$ is a field if and only if $x^3+x^2+c$ is irreducible if and only if $x^3+x^2+c$ has no zeros in $\mathbb{Z}_3$.

2) Find the degree of and a basis for E = Q
$(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})$
Let $\alpha = \sqrt[12]{2}$ then since $x^{12}-2$ is an irreducible polynomial which has $\alpha$ as a root it means $\{1,\alpha,\alpha^2,...,\alpha^{11}\}$ is a basis. This means $\alpha^6,\alpha^4,\alpha^3$ are linearly independent. Thus, $\{ \sqrt{2}. \sqrt[3]{2}, \sqrt[4]{2} \}$ are linearly independent and so they spam a vector space of dimension three.

3. Originally Posted by ThePerfectHacker
[tex]
Let $\alpha = \sqrt[12]{2}$ then since $x^{12}-2$ is an irreducible polynomial which has $\alpha$ as a root it means $\{1,\alpha,\alpha^2,...,\alpha^{11}\}$ is a basis. This means $\alpha^6,\alpha^4,\alpha^3$ are linearly independent.
Thus, $\{ \sqrt{2}. \sqrt[3]{2}, \sqrt[4]{2} \}$ are linearly independent and so they spam a vector space of dimension three.

that vaecor space is not $\mathbb{Q}(\sqrt{2},\sqrt[4]{2},\sqrt[3]{2}).$
you see my comment in red! the degree is 12. let $a=\sqrt{2}, \ b=\sqrt[4]{2}, \ c=\sqrt[3]{2}.$ then:

$[\mathbb{Q}(a,b,c):\mathbb{Q}]=[\mathbb{Q}(a,b)(c):\mathbb{Q}(a,b)][\mathbb{Q}(a)(b):\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}]=3 \times 2 \times 2 = 12.$

a more constructive (and easier) way is to prove that $\mathbb{Q}(\sqrt[12]{2})=\mathbb{Q}(a,b,c)$:

clearly $\mathbb{Q}(\sqrt[12]{2}) \supseteq \mathbb{Q}(a,b,c).$ the reverse inclusion is also clear because $\sqrt[12]{2}=\frac{ab}{c^2}.$