- 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 $d ≥ 1$. - 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 $6 x^{5}+3 x^{2}=x^{2}\left(6 x^{3}+3\right)$.

**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. - over $𝔽_{2}$: irreducible
- 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. - 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.

- 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 $\left[ℚ\left(2^{1/3}, 3^{1/4}\right): ℚ\right]=12$. - 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). - 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.