Thread: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

1. Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

Can someone help me get started on the following problem.

Determine the degree over $\mathbb{Q}$ of $\ 2 + \sqrt{3}$ and of $1 + \sqrt[3]{2} + \sqrt[3]{2}$

Peter

2. Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

$Q \subset Q\left(2+\sqrt{3}\right)\subset Q\left(\sqrt{3}\right)$

So the degree of $2+\sqrt{3}$ is 2 since it cannot be 1

Besides, $2+\sqrt{3}$ satisfies the irreducible polynomial $x^2-4x+1$

3. Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

$1+\sqrt[3]{2}+\sqrt[3]{2} = 1+2\sqrt[3]{2}$. Are you sure that is what you want? Or did you mean $1+\sqrt[3]{2}+\sqrt[3]{3}$? Assuming it is, $1+\sqrt[3]{2}+\sqrt[3]{3} \in \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$. Then, $x^3-2$ and $x^3-3$ generate that field extension. So, is $x^3-3$ reducible over $\mathbb{Q}(\sqrt[3]{2})$? If not, then $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ has degree 3 over $\mathbb{Q}(\sqrt[3]{2})$. Try to figure out why the degree of $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ over $\mathbb{Q}$ equals the degree of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$ times the degree of $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ over $\mathbb{Q}(\sqrt[3]{2})$.

4. Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

To see that $x^3-3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$, assume that it is reducible. Then $x^3-3 = (x-a)(x^2+bx+c)$ with $a,b,c\in \mathbb{Q}(\sqrt[3]{2})$. Multiply that out and try to solve for $a,b,c$. You find that $b-a=0, c-ab=0, abc=3$ (by equating coefficients). You find that any solution must have $a=\sqrt[3]{3}$, and since $\sqrt[3]{3}\notin \mathbb{Q}(\sqrt[3]{2})$, $x^3-3$ is indeed irreducible over $\mathbb{Q}(\sqrt[3]{2})$.