Exercise 1: Let $K$ be a finite field and $p$ its characteristic. Show that there exists $n \in \mathbb{N}$ such that $|K|= p^n$.

Solution: Let $p$ be the characteristic of $K$. Since $K$ is finite $p>0$ and so $p$ is a prime. Let $F$ be the unique subfield of $K$ with $F \cong \mathbb{Z} /p\mathbb{Z}$. Then $K$ is a vector space over $F$. $K$ must be finite dimensional, because $|K|$ is finite. Let $n$ be the ($F$) dimension of $K$. Then $K \cong F^n$ as $F$ vector spaces and so \begin{equation} |K| = |F^n| = |F|^n= p^n. \end{equation}

Exercise 2: Let $p$ a prime, $n \in \mathbb{N}$ and $K$ a field with $|K|=p^n$. Let $m \in \mathbb{N}$. Show that $F:= \{ a \in K : a^{p^m} =a \}$ is a subfield of $K$ and that for every every $ b \in F$ there exists $a \in K$ such that $a^p= b$.

Solution: By exercise $1$ and the uniqueness of the prime factor decomposition $p$ is the characteristic of $K$. Let $\varphi : K \to K$, $\varphi (a) := a^p$. Then $\varphi$ is a field homomorphism (called the Frobenius homomorphism), because $p$ is the characteristic of $K$. Since $K$ is finite and $\varphi$ injective, $\varphi$ is surjective and therefore a field automorphism. Now $F$ is the fixed field of $\varphi^m$. This shows that $F$ is a field. Let $b \in F$. Then $b = \varphi^m (b)= (\varphi^{m-1} (b))^p$. And so $b = a^p$ with $a:= \varphi^{m-1} (b)$.

Exercise 3: Let $p$ a prime, $n \in \mathbb{N}$ and $K \subset \overline{\mathbb{Z}/ p \mathbb{Z}}$ with $|K|=p^n$. Let $\alpha \in K$ such that $\alpha^{p^n-1} \in K \setminus \{0,1\}$. Show that $K(\alpha)/K$ is a Galois extension. Let $G$ be the Galois group of $K(\alpha)/K$. Show that $G$ is cyclic and that $\forall \sigma \in G$: $\sigma(\alpha)/\alpha \in K^*$.

Solution: Let $f:= X^{p^n-1} - \alpha^{p^n-1} \in K[x]$. Then $f(\alpha)=0$ and \begin{equation} f = \prod_{ b \in K^*} (X-b\alpha). \end{equation} To show this: Let $ b \in K^*$. Since $|K|=p^n$, $K^*$ is cyclic with $K^* \cong \mathbb{Z}/ (p^n-1) \mathbb{Z}$ and so $b^{p^n-1} =1$. Therefore $f(b \alpha)=0$. Therefore the $p^{n}-1$ distinct roots of $f$ are $b \alpha$, $b \in K^*$ and the identity follows. This shows that $K(\alpha)$ is the splitting field of $f$ and therefore $K(\alpha)/K$ is normal. Since $K$ is finite it is perfect and therefore $K(\alpha)/K$ is separable. This shows that $K(\alpha)/K$ is Galois. Furthermore $f$ is irreducible, because if $g,h \in K[x]$ with $\deg f, \deg g \geq 1$ and $f = gh$, then there exists $S, S^\prime \subset K^*$ both non-empty with $S \cup S^\prime = K^*$ and $S \cap S^\prime = \varnothing$ and \begin{equation} g =c \prod_{b \in S } (X-b\alpha), \quad h =c^\prime \prod_{b \in S^\prime} (X-b\alpha), \end{equation} where $c , c^\prime \in K^*$. This is because the decomposition of $f$ into irreducible factors is unique (up to reordering and associates). Then $ \alpha^{|S|} \in K$ and $\alpha^{|S|^\prime} \in K$. But this is not possible, because: Let $j \in \{ 1, \dots, p^{n}-2 \}$ with $\alpha^j \in K$. Then ($\alpha \neq 0$ since fields are zero divisor free) \begin{equation} 1 = (\alpha^j)^{p^n-1} = (\alpha^{p^n-1} )^j \end{equation} from which it follows that $ p^n-1 | j$ (because $\alpha^{p^n-1} \in K \setminus \{ 0,1 \}$). This contradicts $0<j < p^{n}-1$. Therefore $f$ is irreducible, and $f = M_{\alpha, K}$ and so $p^n-1 =[K(\alpha): K ]= |G|$. Define $\psi : K^* \to G$ by $\psi(b) :=$ the unique extension of $K \to K(\alpha)$ with $\psi (b) (\alpha) = \alpha b$. The uniqueness and existence of such an extension follows from a well known extension theorem. Then $\psi$ is an injective group homomorphism (easily verified) and hence surjective (since $|K^*| = p^n-1 = |G|$). Evidently $G$ is cyclic since $K^*$ ist and the second assertion is also easy to see.

Exercise 4: Let \begin{equation} f:= 5 X^4 +6X^3 +7X+4581 \in \mathbb{Q} [X]. \end{equation} Is $f$ irreducible? Is $f$ separable?

Solution: Reduction Criterion (Eisenstein does not work). Let $\pi : \mathbb{Z} \to \mathbb{Z}/2 \mathbb{Z}$ the projection and $\varphi : \mathbb{Z}[X] \to \mathbb{Z}/ 2\mathbb{Z}[X]$ the unique ring homomorphism with $\varphi|_{\mathbb{Z}} = \pi$ and $\varphi(x) =x$. Now \begin{equation} \bar f := \varphi (f) = X^4 + X + 1. \end{equation} Then $\deg \bar f= \deg f$ and so if $ \bar f$ is irreducible then so is $f$ by the reduction criterion. $\bar f (0) =1$ and $\bar f(1) = 1$. Therefore $\bar f$ does not have any zeros. Assume $\bar f = gh $ with $g, h \in \mathbb{Z}/2\mathbb{Z}[X]$ and $\deg g , \deg h \geq 1$. Then we must have $\deg g =\deg h =2$, because $\bar f $ does not have any zeros. The polynomials of degree 2 are (the number of them is $2^2 =4$): \begin{equation} X^2, \ X^2 + X + 1, \ X^2 +1 , \ X^2 +X. \end{equation} All of these have a zero (and therefore can not be equal to $g$ or $h$) except $X^2 + X +1$. Therefore $g = h = X^2 + X +1$. \begin{equation} gh = (X^2 + X +1)(X^2 + X +1) = X^4 + X^3 + X^2 + X^3 + X^2 +X + X^2 + X +1 \end{equation} And so \begin{equation} gh = X^4 + 2 X^3 + 3 X^2 + 2 X +1 \neq \bar f. \end{equation} Therefore $\bar f$ is irreducible and so $f$ is too.

Exercise 5: Let $K:= \mathbb{Q}$ and $E:= K(\sqrt{2},\sqrt{3})$.

Show that $E/K$ is Galois
Show that the Galois group $G$ of $E/K$ is isomorphic to $(\{-1,1\},\cdot)^2$
What are the subextensions of $E/K$?
Find $\alpha \in E$ so that $E = K(\alpha)$
What is the minimal polynomial of $\alpha$ over $K$/ $K(\sqrt{2})$
Is $K (\sqrt{2}, \sqrt{3})/ K (\sqrt{2})$ normal?

Solution: To 1.: $E$ is the splitting field of $f:= (X^2-2)(X^2-3)$. Therefore $E/K$ is normal. Since $K$ is perfect and $E/K$ algebraic, $E/K$ is separable. And so in total $E/K$ is Galois.

To 2.: $X^2 -2$ is irreducible over $K$ by Eisensteins criterion (with $p=2$). Therefore $[K(\sqrt{2}):K]=2$. Now $X^2 -3 = (X+\sqrt{3})(X-\sqrt{3})$ is also irreducible over $K(\sqrt{2})$, because otherwise $\sqrt{3} \in K(\sqrt{2})$. However if $\sqrt{3} \in K(\sqrt{2})$: Then there exist $a,b \in \mathbb{Q}$ such that \begin{equation} \sqrt{3} = a + b \sqrt{2} \end{equation} and then (assuming $a,b \neq 0$) \begin{equation} 3 = a^2 +2ab \sqrt{2} +2 \Rightarrow \sqrt{2} \in \mathbb{Q} \end{equation} Which is a contradiction. Therefore the minimal polynomial of $\sqrt{3}$ over $K(\sqrt{2})$ is $X^2-3$ and so in total \begin{equation} [K(\sqrt{2},\sqrt{3}) :K] = [K(\sqrt{2})(\sqrt{3}) :K(\sqrt{2})] [K(\sqrt{2}):K] = 4. \end{equation} Therefore $|G|=4$. For $s \in \{-1,1\}^2$ let $\sigma \in G$ be the unique extension of the injection $K \to E$ with $\sigma ( \sqrt{2} ) := s_1 \sqrt{2}$ and $\sigma(\sqrt{3}):= s_2 \sqrt{3}$. This extension exists and is unique, because of the just found minimal polynomials and a general result on extension of field homomorphisms. The map $\{-1,1\}^2 \ni s \mapsto \sigma_s \in G$ is an injective group homomorphism. This is easily verified on the basis of $E/K$ given by $(1,\sqrt{2}, \sqrt{3}, \sqrt{6})$. The group homomorphism is also surjective, because $|G| =4 = |\{-1,1\}^2|$. Therefore $G$ is isomorphic to $\{-1,1\}^2$.

To 3.: Using the fundamental theorem of Galois theory we only need to find all subgroups of $\{-1,1\}^2$. The order of any subgroup must divide $4 =2 2$. Therefore the order must be one of $1,2,4$. The subgroups are given by \begin{equation} H_1 := \{(1,1) \}, \ H_2 :=\{-1,1\}^2, \ H_3 := \{(1,1) , (1,-1) \}, \end{equation} \begin{equation} H_4 := \{(1,1), (-1,1)\}, \ H_5 := \{ (1,1), (-1,-1) \}. \end{equation} Therefore (the non trivial) subextensions of $E/K$ are (easily verified on the above basis): \begin{equation} E^{H_3} = K(\sqrt{2}), \ E^{H_4} = K ( \sqrt{3}), \ E^{H_5} = K (\sqrt{6}). \end{equation}

To 4.: The roots of the minimal polynomial of $\sqrt{2}$ are $\sqrt{2}$ and $-\sqrt{2}$. The roots of the minimal polynomial of $\sqrt{3}$ are $ \sqrt{3}$ and $- \sqrt{3}$. Let $\mu :=1 $. Then \begin{equation} \mu \neq 0 = \frac{\sqrt{2}- \sqrt{2}}{-\sqrt{3}- \sqrt{3}} \end{equation} and \begin{equation} \mu \neq \frac{ \sqrt{2} -(-\sqrt{2})}{- \sqrt{3}- \sqrt{3}} = -\frac{\sqrt{2}}{\sqrt{3}}. \end{equation} Let $ \alpha = \sqrt{2}+ \mu \sqrt{3}$. Then $E= K(\alpha)$ by the primitive element theorem.

To 5.: The minimal polynomial of $\alpha$ over $K$ has degree 4. And \begin{equation} 3 = (\alpha - \sqrt{2})^2= \alpha^2 - 2 \sqrt{2} \alpha + 2 \end{equation} and so \begin{equation} \alpha^2 -1 = 2 \sqrt{2} \alpha \Rightarrow (\alpha^2 - 1)^2 = 8 \alpha^2 \end{equation} and therefore $\alpha$ is a root of \begin{equation} (X^2-1)^2 - 8 X^2 \in K[X], \end{equation} which is the minimal polynomial, because it has degree $4$ and is normalized. The minimal polynomial of $\alpha$ over $K(\sqrt{2})$ has degree 2. It is given by (see above equation) \begin{equation} (X- \sqrt{2})^2-3 \in K(\sqrt{2})[X]. \end{equation} Since this polynomial has a root at $\alpha$ and is normalized.

To 6.: $K(\sqrt{2}, \sqrt{3})$ is the splitting field of $X^2-3$ over $K(\sqrt{2})$ and therefore the extension is normal.