Algebraic Extensions

Is $\sqrt{2-\sqrt{i}}$ a transcendental number?

Find $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]$. Hence find $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3},\sqrt[6]{72}):\mathbb{Q}]$.

Suppose $\alpha$ is an element of some extension of $\mathbb{Q}$ with a minimal polynomial of $p(x)=x^{5}+2x+1\in\mathbb{Q}[x]$. Prove that $\sqrt{2}\notin\mathbb{Q}(\alpha)$.

Consider $\mathbb{Q}(\sqrt{3},\sqrt{5})$.

1. Prove that $\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{3}+\sqrt{5})$.

2. Deduce the degree of the minimal polynomial of $\sqrt{3}+\sqrt{5}$.

Prove 1.26.

Prove that $E=\mathbb{Q}(2^{\frac{1}{2}},2^{\frac{1}{3}},2^{\frac{1}{4}},\dots,2^{\frac{1}{n}},\dots)$ is an algebraic extension of $\mathbb{Q}$ but not a finite extension of $\mathbb{Q}$.

Find $[\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}(\sqrt{15})]$, and hence find

1. the minimal polynomial of $\sqrt{3}+\sqrt{5}$ over $\mathbb{Q}(\sqrt{15})$; and

2. a basis for $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$.

(Hint: consider an exercise given elsewhere in the book.)

Prove that $\mathbb{Q}(\sqrt{2},\sqrt[3]{2})=\mathbb{Q}(\sqrt[6]{2})$.

Let $a,b\in\mathbb{Q}$. Show that $\mathbb{Q}(\sqrt{a},\sqrt{b})=\mathbb{Q}(\sqrt{a}+\sqrt{b})$.

Suppose $F$ is a field and $E$ is a finite extension where $[E:F]=p$ where $p$ is a prime. Prove that for every $\alpha\in E$ we have $F(\alpha)=F$ or $F(\alpha)=E$.

Let $F$ be a field and $E/F$ be a field extension. Prove that if $\alpha\in E$ is transcendental over $F$, then $\alpha^{n}$ is transcendental over $F$ for all positive integers $n$.

Let $F$ be a field and $E$ be an extension field of $F$.

1. Let $a_{0},a_{1},\dots,a_{n}\in E$ be algebraic over $F$. Prove that the zero(es) of the polynomial

   are also algebraic over $F$.

2. Deduce that if $\alpha$ and $\beta$ are both transcendental over $F$, then either $\alpha+\beta$ or $\alpha\beta$ is also transcendental.

Suppose $\alpha\in\mathbb{C}$ is a zero of $x^{2}+x+1$ over $\mathbb{Q}$. Prove that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{\alpha})$.

Prove that $\sqrt{2}\notin\mathbb{Q}(\pi)$. You may assume that $\pi$ is a transcendental number and that $\sqrt{2}$ is irrational.(Hint: refer to the proof of 1.7 on the form of an element in $\mathbb{Q}(\pi)$.)

Let $F$ be a field and $f(x)$ is a non-constant polynomial in $F[x]$ with degree $n$. Let $E$ be the splitting field of $f(x)$ over $F$.

1. By considering the binomial coefficient ${a+b\choose a}$, explain why $a!b!$ divides $(a+b)!$ for all non-negative integers $a$ and $b$.

2. Prove that $[E:F]$ divides $n!$.(Hint: use strong induction on $n$; break into two cases, where $f(x)$ is irreducible or $f(x)$ is reducible.)