1. Let $K$ be a finite field. Show that there exists a positive integer $d$ and a prime number $p$ such that ${|K|}=p^{d}$. Hint: what is the prime subfield of $K$ ?
   Solution. The field $K$ contains the image of ℤ by the unique map of rings $ϕ: ℤ → K$ sending $1_{ℤ}$ to $1_{K}$. The image of $ϕ$ is a domain (since it is contained in a field) and hence $\ker(ϕ)$ is a prime ideal. Hence the image of $ϕ$ is isomorphic to $𝔽_{p}=ℤ/p ℤ$ for some prime number $p$ (because the prime ideals of $ℤ$ are all of the of the form $(p)$ for some prime number $p$ - see Rings and Modules). Hence we may view $K$ as a $𝔽_{p}$-vector space. Furthermore, $K$ is a finite-dimensional $𝔽_{p}$-vector space since it is a finite set. We conclude that there is an isomorphism of $𝔽_{p}$-vector spaces $K ≃ 𝔽_{p}^{d}$ for some $d ≥ 1$. Hence we have $\# K=p^{d}$ for some $d ≥ 1$.
2. Factorise $f(x)=x^{6}+x^{3}+1$ into irreducible factors over $K$ for each of $K=𝔽_{2}, 𝔽_{3}, 𝔽_{19}, ℚ$.
   Calculate the formal derivative $D f$. Over which of these fields $K$ do the irreducible factors of $f$ have distinct roots in any splitting field for $f$ ?
   Solution.
   • over $𝔽_{2}$: irreducible
     Check first that $f(0), f(1) ≠ 0\pmod2$. We deduce that if $f(x)$ is reducible, it must have a factor of degree$≤3$. Now check that $f(x)$ is not divisible by $a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ when $a_{i}=0,1\pmod2$.
   • over $𝔽_{3}$: $f(x)=(x+2)^{6}$.
     Find the root $-2$ and compute that $D(f)=6 x^{5}+3 x^{2}=0\pmod2$ in $𝔽_{3}[x]$. Using Q4 (e) in sheet 1 we deduce that $(x+2)^{2}$ divides $f(x)$ and we compute $$ f(x)/(x+2)^{2}=x^{4}+2 x^{3}+2 x+1=g(x) . $$ Now $g(x)$ also has the root $-2$ and $D(g)=4 x^{3}+2$, so that $D(g)(-2)=0$. We conclude similarly that $g(x)$ is divisible by $(x+2)^{2}$. Finally, we compute $g(x)/(x+2)^{4}=(x+2)^{2}$ so that $f(x)=(x+2)^{6}$.
   • over $𝔽_{19}$: one checks by hand that $f(x)$ has six distinct roots in $𝔽_{19}$. This gives $$ f(x)=(x+2)(x+3)(x+10)(x+13)(x+14)(x+15) $$
   • over ℚ: $f(x)$ is irreducible because $f(x)$ is monic, has coefficients in $ℤ$ and is irreducible over $𝔽_{2}$ (apply Gauss's lemma - Lemma 2.12 in the notes). The formal derivative of $f(x)$ is $6 x^{5}+3 x^{2}=x^{2}\left(6 x^{3}+3\right)$.
   For the rest of the exercise, we shall need the
   Lemma. Let $𝔽_{p}$ be a finite field and let $P(x) ∈ 𝔽_{p}[x]$ be an irreducible polynomial. Then $P(x)$ is separable. In particular, any finite extension of a finite field is separable.
   Proof. Let $𝔽≔𝔽_{p}[x]/(P(x))$. Let $α≔x\pmod{P(x)}∈ 𝔽$. The map $\text{Frob}: 𝔽 → 𝔽$ such that $\operatorname{Frob}(y)=y^{p}$ is a map of rings (see Sheet 1, solution to Q6). The map Frob is injective because if $y^{p}=0$ then $y=0$ (since 𝔽 is a field, in particular a domain). The map Frob is also surjective, because its image has the same cardinality as 𝔽 (because Frob is injective). Hence Frob defines a permutation of the elements of 𝔽 and in particular Frob has a finite order $c$ in the group of permutations of 𝔽 (because this group is finite). In particular $α^{p^{c}}=α$. Since $P(x)$ is the minimal polynomial of $α$, we thus have $P(x) ∣ x^{p^{c}}-x$. Now $x^{p^{c}}-x$ is a separable polynomial, because $\left(x^{p^{c}}-x, D\left(x^{p^{c}}-x\right)\right)=\left(x^{p^{c}}-x,-1\right)=(1)$. Hence $P(x)$ is also separable.
   From the lemma we conclude that over $𝔽_{2}, 𝔽_{3}, 𝔽_{19}$, the irreducible factors of $f(x)$ have no repeated roots in any splitting field of $f(x)$.
   Any finite extension of ℚ is separable (see before Def. 3.7 in the notes) so the same is true over ℚ.
3. Show that if $f$ is a polynomial of degree $n$ over $K$, then its splitting field has degree less than or equal to $n!$ over $K$.
   Solution. By induction on $n$. The statement clearly holds if $n=1$. Let $P(x)$ be an irreducible factor of $f(x)$ and $L≔K[x]/(P(x))$. Let $α≔x\pmod{P(x)}∈ L$. Let $Q(x)=f(x)/(x-α) ∈ L[x]$. Let $M$ be a splitting field of $Q(x)$ over $L$. Then $M$ is a splitting field of $f(x)$ because $f(x)$ splits over $M$ and is generated over $K$ by the roots of $f(x)$ in $M$. Applying the inductive hypothesis, we see that $[M: L] ≤(n-1)!$. Applying the tower law, we see that $[M: K] ≤ n !$.
   Note that the solution of Q3 follows the construction of the splitting field given in Th. 3.13 (a) in the notes.
4. Find the degrees of the splitting fields of the following polynomials.
   • $x^{3}-1$ over ℚ;
   • $x^{3}-2$ over ℚ;
   • $x^{5}-t$ over $𝔽_{11}(t)$;
   Solution.
   • degree 2;
     we have $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$, so the degree of the splitting field of $x^{3}-1$ is the degree of the splitting field of $x^{2}+x+1$, which is generated by $\sqrt{-3}=i \sqrt{3}$ and thus has degree 2.
     Note. The splitting field of $x^{3}-1$ over ℚ is an example of a cyclotomic field. See section 5.1 in the notes.
   • degree 6;
     the splitting field is $ℚ\left(2^{1/3}, j\right)$ where $j$ is a primitive cube root of 1. Indeed the roots of $x^{3}-2$ in ℂ are $2^{1/3}, j 2^{1/3}, j^{2} 2^{1/3}$ and so the splitting field of $x^{3}-2$ is $$ ℚ\left(2^{1/3}, j 2^{1/3}, j^{2} 2^{1/3}\right)=ℚ\left(2^{1/3}, j\right) . $$ Now note that $\left[ℚ\left(2^{1/3}\right): ℚ\right]$ is 3 because $x^{3}-2$ is irreducible by Eisenstein's criterion. On the other hand, since $j$ is not real and $j^{2}+j+1=0$, the tower law implies that $\left[ℚ\left(2^{1/3}, j\right): ℚ\right]$ is at least $3 ⋅ 2=6$. Since the degree of the splitting field is at most $3 !=6$ (see Q3) we thus have $\left[ℚ\left(2^{1/3}, j\right): ℚ\right]=6$.
   • degree 5;
     let $x_{0}$ be a root of $x^{5}-t$ in the splitting field $L$ of $x^{5}-t$. We have $$ x^{5}-1=(x+2)(x+6)(x+7)(x+8)(x+10) $$ in $𝔽_{11}[x]$ so $x^{5}-1$ also splits in $\left(𝔽_{11}(t)\right)[x]$. The roots of $x^{5}-t$ in $L$ are all of the form $x_{0} ⋅\left(\text{root of }x^{5}-1\right)$ (why?) so $x^{5}-t$ splits in the simple subfield $\left(𝔽_{11}(t)\right)\left(x_{0}\right)$ of $L$. Therefore $\left(𝔽_{11}(t)\right)\left(x_{0}\right)=L$.
     Now $x^{5}-t$ is irreducible over $𝔽_{11}(t)$. To see this suppose for contradiction that $P(x)$ is an irreducible factor of $x^{5}-t$, which is of degree $d=2$ (note that $d=1$ is impossible because $t$ has no 5 th root in $𝔽_{11}(t)$ (why?)). By the above, the roots of $P(x)$ in $L$ are of the form $x_{0} l$ and $x_{0} m$, where $l$ and $m$ are roots of $x^{5}-1$. So $$ P(x)=x^{2}-x_{0}(l+m) x+lmx_{0}^{2} . $$ We conclude that $lmx_{0}^{2} ∈ 𝔽_{11}(t)$ and so $$ \left(lmx_{0}^{2}\right)^{3}=(lm)^{3} x_{0}^{6}=(l m)^{3} x_{0} t ∈ 𝔽_{11}(t) . $$ But then $x_{0}∈ 𝔽_{11}(t)$, a contradiction.
     Finally, since $x^{5}-t$ is irreducible, $$ \left[L: 𝔽_{11}(t)\right]=\left[\left(𝔽_{11}(t)\right)\left(x_{0}\right): 𝔽_{11}(t)\right]=5 $$ Note. This is an example of a Kummer extension. See section 5.2 in the notes.
5. Let $L=ℚ\left(2^{1/3}, 3^{1/4}\right)$. Compute the degree of $L$ over ℚ.
   Solution. Notice that $L$ contains $ℚ\left(2^{1/3}\right)$, which is of degree 3 over ℚ (because $x^{3}-2$ is irreducible by Eisenstein's criterion). The field $L$ also contains $ℚ\left(3^{1/4}\right)$, which is of degree 4 over ℚ by a similar reasoning. Hence, by the tower law, the degree of $ℚ\left(2^{1/3}, 3^{1/4}\right)$ over ℚ is divisible by 3 and by 4, hence by 12. On the hand, the degree of $ℚ\left(2^{1/3}, 3^{1/4}\right)$ over ℚ is smaller than 12, because the degree of $ℚ\left(2^{1/3}, 3^{1/4}\right)$ over $ℚ\left(2^{1/3}\right)$ is smaller or equal to 4. Hence $\left[ℚ\left(2^{1/3}, 3^{1/4}\right): ℚ\right]=12$.
6. Recall that $α ∈ ℂ$ is algebraic over ℚ if $α$ satisfies a (monic) polynomial over ℚ, equivalently if $[ℚ(α): ℚ]<∞$. Let $$ 𝔸=\{α ∈ ℂ: α \text{ is algebraic over } ℚ\} $$
   • Show that 𝔸 is the union of all the subfields $L$ of ℂ which are finite extensions of ℚ.
   • Prove that 𝔸 is a subfield of ℂ. [Hint: if $α, β ∈ 𝔸$, consider the extension $ℚ(α, β)|ℚ$.]
   • Prove that $𝔸|ℚ$ is not a finite extension.
   Solution.
   • We have $[ℚ(α): ℚ]=\deg\left(m_{α}(x)\right)$. Hence $α$ is contained in a finite extension of ℚ (contained in ℂ). On the other hand, let $L ∣ ℚ$ be a finite extension (with $L ⊆ ℂ$) and $α ∈ L$. Consider the map of rings $ϕ: ℚ[x] → L$, such that $$ ϕ\left(\sum_{k=0}^{d} a_{k} x^{k}\right)=\sum_{k=0}^{d} a_{k} α^{k} $$ Then the kernel of $ϕ$ is not $(0)$, for otherwise the finite-dimensional ℚ-vector space $L$ would contain a vector space isomorphic to $ℚ[x]$, which is infinite-dimensional. Now let $P(x) ∈ \ker(ϕ)$ with $P(x) ≠ 0$. Then $P(α)=0$ and so $α$ is algebraic over ℚ.
   • Let $α, β ∈ 𝔸$. By the tower law, the subfield $ℚ(α, β)=ℚ(α)(β)$ is finite over ℚ and thus $ℚ(α, β) ⊆ 𝔸$. Since, $α β, α^{-1}, α+β ∈ ℚ(α, β) ⊆ 𝔸$, we see that 𝔸 is a field.
   • Let $p$ be a prime number and $n ≥ 1$. Let $α=\sqrt[n]{p}$. Then $ℚ[x]/\left(x^{n}-p\right) ≃ ℚ(α)$ is a finite extension of degree $n$ of ℚ because $x^{n}-p$ is irreducible (by Eisenstein's criterion), and $ℚ(α) ⊆ 𝔸$ by (a). Since $n$ is arbitrary, 𝔸 cannot be finite over ℚ.
   Note. One usually writes $\overline{ℚ}$ for 𝔸. One can show that 𝔸 is algebraically closed (ie. every polynomial with coefficients in 𝔸 splits).
7. Which of the following fields are normal extensions of ℚ ?
   • $ℚ(\sqrt{2}, \sqrt{3})$;
   • $ℚ\left(2^{1/4}\right)$;
   • $ℚ(α)$, where $α^{4}-10 α^{2}+1=0$.
   Solution.
   • $ℚ(\sqrt{2}, \sqrt{3})$ is normal over ℚ; indeed, $ℚ(\sqrt{2}, \sqrt{3})$ is the splitting field of $\left(x^{2}-2\right)\left(x^{2}-3\right)$ (apply Th. 3.16 in the notes).
   • $ℚ\left(2^{1/4}\right)$ is not normal over ℚ. Indeed, the minimal polynomial of $2^{1/4}$ is $x^{4}-2$ (because $x^{4}-2$ annihilates $2^{1/4}$ and $x^{4}-2$ is an irreducible polynomial by Eisenstein's criterion). On the other hand $x^{4}-2$ has the complex root $i 2^{1/2}$, which does not lie in $ℚ\left(2^{1/4}\right) ⊆ ℝ$.
   • $ℚ(α)$ is normal over ℚ.
     To prove this, we shall show that $ℚ(α)$ is the splitting field of $P(x)=x^{4}-10 x^{2}+1$. This together with Th. 3.16 in the notes implies that $ℚ(α)$ is normal over ℚ. Now note that $P(-α)=0$ and also that $P( ± 1/α)=0$ (divide $α^{4}-10 α^{2}+1$ by $α^{4}$). The elements $α,-α, 1/α,-1/α$ are all distinct (why?) so that $P(x)=(x-α)(x+α)(x-1/α)(x+1/α)$.
     Note. A polynomial $P(x)$ of degree $d$ such that $P(x)=x^{d} P(1/x)$ is called palindromic. The polynomial $x^{4}-10 x^{2}+1$ is palindromic.