I am stuck on the first one but there are 5 questions on this so I really need help with the process. If anyone can help with any of the following.
i) Find a Basis for the field K = $\mathbb{Q}(\alpha)$ as a vector space over $\mathbb{Q}$ where $\alpha$ is the root of the polynomial 3x + 1.
ii) K = $\mathbb{Q}(\alpha)$ where $\alpha$ is the root of the polynomial $x^3$ + 1
iii) K = $\mathbb{Q}(\alpha)$ where $\alpha$ is the root of the polynomial $x^3$ - 1
iv) K = $\mathbb{Q}(\alpha)$ where $\alpha$ is the root of the polynomial $x^4$ + 1
v) K = $\mathbb{Q}(A)$ where A is the set of all roots over $\mathbb{C}$ of the polynomial $x^3$ - 2.
Please help! I have read the proof for the Tower law over and over. I know I need to make sure the polynomial is irreducible first.
$\newcommand{\QQ}{\mathbb{Q}}$
i) $\alpha$ is a root of the polynomial $3x+1$. This polynomial only has one root, namely... $-1/3$. Hence $\QQ(\alpha) = \QQ$, and a basis is just $\{1\}$.
ii) $\alpha$ is a root of $x^3+1$. This polynomial is irreducible over $\QQ$, so adjoining a root of it, namely $\alpha$, will produce a degree 3 extension of $\QQ$. Hence, $\QQ(\alpha)$ has dimension 3 over $\QQ$. Here's a basis $1,\alpha,\alpha^2$. You can see they're linearly independent by noting that $\QQ(\alpha)$ is just $\QQ[x]/(x^3+1)$, and clearly $1,x,x^2$ are linearly independent in that space. Since $\QQ(\alpha)$ has dimension 3 over $\QQ$, $\{1,\alpha,\alpha^2\}$ must be a basis.
iii) Since $x^3-1$ is not irreducible over $\QQ$, you have to be more careful. $x^3-1$ factors as $(x-1)(x^2+x+1)$. If $\alpha = 1$, then $\QQ(\alpha) = \QQ$, and $\{1\}$ is a basis. If $\alpha$ is a root of $(x^2+x+1)$, then $\QQ(\alpha)$ has degree 2 over $\QQ$ (since $x^2+x+1$ is irreducible over $\QQ$), and again you can see that $\{1,\alpha\}$ is a basis.
iv) same as ii.
v) Now you're adjoining all the roots of $x^3-2$. If you adjoined just one root, since $x^3-2$ is irred over $\QQ$, you'd get a degree 3 extension. However, there are two other roots. Say you adjoin $\sqrt[3]{2}$. The other roots of $x^3-2$ are $\omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}$, where $\omega$ is a primitive 3rd root of 1. It's clear that adjoining $\sqrt[3]{2}$ does not get you the other two roots. On the other hand, if you have both $\sqrt[3]{2}$ and $\omega\sqrt[3]{2}$ in a field, then their quotient $\omega$ is in the field, and hence $\omega^2\sqrt[3]{2}$ is in your field. Conversely, if you have both $\omega$ and $\sqrt[3]{2}$, then you have all 3 roots of the polynomial $x^3-2$ as well. Hence, adjoining all the roots is the same as adjoining just $\sqrt[3]{2}$ and $\omega$. It's clear that $\sqrt[3]{2}$ has degree 3 over $\QQ$. Furthermore, $\omega$ is a root of $x^2+x+1$. You know $\omega\notin\QQ(\sqrt[3]{2})$, hence $x^2+x+1$ must be irreducible over $\QQ(\sqrt[3]{2})$, and $\QQ(\sqrt[3]{2},\omega)$ must have degree 2 over $\QQ(\sqrt[3]{2})$. This shows that $[\QQ(\sqrt[3]{2},\omega):\QQ] = [\QQ(\sqrt[3]{2},\omega):\QQ(\sqrt[3]{2}][\QQ(\sqrt[3]{2}):\QQ] = 2*3 = 6$. Finding a basis in this situation is a bit more tricky. One way is to prove that $\{1,\omega,\sqrt[3]{2},\omega\sqrt[3]{2},\sqrt[3]{4},\omega\sqrt[3]{4}\}$ is linearly independent, hence a basis. Another way is to show that $\QQ(A) = \QQ(\beta)$ for some single element $\beta$, in which case a basis will just be $\{1,\beta,\beta^2\,\ldots,\beta^5\}$. I'll leave this to you :-)