solutions.tex

\section*{Some solutions}

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% \begin{sol}{xca:algebraic_bijective}
%     Let $\sigma\colon C\to C$ be a homomorphism. As $\sigma$ is
%     injective, we need to prove that $\sigma$ is surjective. Let $y\in C$. Note that $y$ is algebraic over $K$. Let 
%     $R$ be the set of roots of the minimal polynomial 
%     $f(y,K)$ of $y$ over $K$. 
%     The map 
%     $\sigma|_R\colon R\to R$ is injective. Since 
%     $R$ is finite, $\sigma|_R$ is then bijective. In particular, 
%     there exists $x\in R$ such that $y=\sigma(x)$.
% \end{sol}


\begin{sol}{xca:Q(i)}
Assume that $\Q[i]$ and $\Q[\sqrt{2}]$ were isomorphic 
and let $\varphi:\Q[i]\to \Q[\sqrt{2}]$
be a field isomorphism.
Then 
    \[
    \varphi(i)^2=
    \varphi(i^2)=
    \varphi(-1)=
    -\varphi(1)=-1.
    \]
But $\varphi(i)\in\Q[\sqrt{2}]$ and
$\Q[\sqrt{2}]\subseteq \R$ where every square is 
positive, a contradiction.
\end{sol}

\begin{sol}{xca: field characteristic}
    Let $t>0$ be the characteristic of a field $K$ and 
    let $\varphi:\Z\to K,$ $x\mapsto x1$.
    Then, by definition, $\ker\varphi$ is 
    the ideal generated by $t$.
    On the other hand, the image $\varphi(\Z)$ is 
    a domain, being a subring of a field
    and is isomorphic to $\Z/\ker\varphi$.
    Therefore, $\ker\varphi$ is a prime ideal of $\Z$,
    i.e. $t$ is a prime number.
\end{sol}

\begin{sol}{xca: char 0}
Let $\varphi:\Z\to K,$ $x\mapsto x1$.

We first prove that \textbf{1)} implies all the other 
properties. So suppose that the characteristic of $K$ is zero, i.e. $\ker\varphi=\{0\}$. 

Then $m1=0$ if and only if $m=0$, i.e.
the order of $1$ is infinite.

Let $0\neq x\in K$. If $mx=0$, then $0=mx=(m1)x$.
But $K$ is a field and $x\neq 0$, hence $m1=0$,
so $m\in\ker\varphi=\{0\}$.
Hence the order of $x$ is infinite.

By definition, the prime subring of $K$ is
the image of $\varphi$, which 
so it is isomorphic to $\Z/\ker\varphi=\Z$.

Finally, we prove that \textbf{4)} implies \textbf{1)}.
Take $m\in\ker\varphi$. Then $m1=0$,
but 1 has infinite order, hence $m=0$.
Therefore $\ker\varphi=\{0\}$, 
i.e. $K$ has characteristic 0.
\end{sol}

\begin{sol}{xca: char p}
Let $\varphi:\Z\to K,$ $x\mapsto x1$.

We first prove that \textbf{1)} implies all the other 
properties. So suppose that the characteristic of $K$ is $p>0$ i.e. $\ker\varphi$ is the ideal generated by $p$. 

Then $m1=0$ if and only if $p$ divides $m$, i.e.
the order of $1$ is $p$.

Let $0\neq x\in K$. If $mx=0$, then $0=mx=(m1)x$.
But $K$ is a field and $x\neq 0$, hence $m1=0$,
so $p$ divides $m$.
Hence $x$ has order $p$.

By definition, the prime subring of $K$ is
the image of $\varphi$, which 
so it is isomorphic to $\Z/\ker\varphi\cong \Z/p$.

Finally, we prove that \textbf{4)} implies \textbf{1)}.
Take $m\in\ker\varphi$. Then $m1=0$,
but 1 has order $p$, hence $p$ divides $m$.
Therefore $\ker\varphi$ is generated by $p$, 
i.e. $K$ has characteristic 0.
\end{sol}

\begin{sol}{xca: Frobenius hom}
Let $\Phi: K\to K$ be the map 
$x\mapsto x^{p}$.
Since the map $x\mapsto x^{p^n}$
is exactly $\Phi^n$,
it is enough to prove that $\Phi$ is a field homomorphism.

As $K$ is commutative under multiplication, 
for all $x,y \in K$ 
        \[
        \Phi(xy) = (xy)^p = x^py^p = \Phi(x) \Phi(y).
        \]
Moreover, for all $x,y \in K$ 
\[
\Phi(x+y) =
(x+y)^p \sum_{k=0}^p \binom{p}{k}x^py^{p-k}=
x^p + y^p + \sum_{k=1}^{p-1} \binom{p}{k}x^ky^{p-k} ,
\]
where $\binom{p}{k} = \frac{p!}{k!(p-k)!}$,
which can also be written as
        \[
        p! = \binom{p}{k} \cdot k! \cdot (p-k)!
        \]
But $p$ divides $p!$, so $p$
has to divide at least one 
factor on the right side. 
But $p$ doesn't divide $i$
 for $1 \leq i \leq p-1$, therefore if $k \leq p-1$, $p$ doesn't divide
$k!$ and if $1 \leq  k$, $p$ doesn't divide $(p-k)!$. 
Hence, if $1 \leq k \leq p-1$, $p$ has to divide $\binom{p}{k}$ and 
\[
\sum_{k=1}^{p-1} \binom{p}{k}x^ky^{p-k}=0.
\]
Therefore, $\Phi$ is a field homomorphism.
\end{sol}

\begin{sol}{xca: prime field is fixed}
    By definition $K_0=\{m1\colon m\in \Z\}$ and 
    $\sigma:K\to K$ is a field homomorphism,
    so $\sigma(1)=1$.
    Hence, for every $m\in\Z$
    \[
    \sigma(m1)=m\sigma(1)=m1,
    \]
    i.e. $\sigma|_{K_0}$ is the identity.
\end{sol}

\begin{sol}{xca: X^3-2}
    If $X^3-2$ were reducible, since it has degree 3,
    it would have a linear
    factor in the decomposition in irreducibles.
    Therefore it would have a rational root.
    But the roots of $X^3-2$ are $\sqrt[3]{2},\sqrt[3]{2}\xi,\sqrt[3]{2}\xi^2$,
    where $\xi$ is a primitive third root of unity,
    So all the roots are not in $\Q$, a contradiction.
\end{sol}

\begin{sol}{bxca:Eisenstein's criterion}
Recall first the following:
\begin{lemma}[Gauss' Lemma]
    Let $A$ be a unique factorization domain and $K$ be its fraction field.
    A non-constant polynomial $f\in A[X]$ is irreducible if and only if it is primitive and irreducible in $K[X]$.
\end{lemma}

    Suppose that $f$ is reducible in $K[X]$.
    Then $g=c^{-1}f$, where $c$ is the content of $f$,
    would be reducible and primitive.
    Hence, by Gauss' Lemma, $g$ is also 
    reducible in $A[X]$.
    So $c^{-1}f=g=hl$, for some non-constant polynomials $h,l\in A[X]$.
    Now consider $\pi:A\to A/(p)$, $a\mapsto \overline{a}$ the natural surjection.
    We know that $\overline{a_i}=0$ for all 
    $i\in\{0,1,\dots, n-1\}$ and $\overline{a_n}\neq 0$.
    Therefore 
    \[
    \overline{\pi}(ch)\overline{\pi}(l)=\overline{c}\:\overline{\pi}(h)\overline{\pi}(l)=\overline{\pi}(f)=\overline{a_n}X^n\in  A/(p)[X].
    \]
    But $A/(p)[X]$ is a UFD so the only possibility
    is that $\overline{\pi}(ch)=\overline{d}X^t$ and $\overline{\pi}(l)=\overline{f}X^s$, for
    some $f,d\in A/(p)\setminus\{\overline{0}\}$
    and $t,s\in\{1,\dots, n-1\}$.
    In particular, $\overline{\pi}(ch)$ and $\overline{\pi}(l)$ have both constant term 
    equal to 0.
    Hence $p$ divides $ch(0)$ and $l(0)$ in $A$.
    Therefore $p^2$ divides $ch(0)l(0)=f(0)$,
    a contradiction.
\end{sol}

\begin{sol}{xca: using Eisenstein}
It is easy to see that $f$ satisfies the Eisenstein criterion for $p=2$
and $g$ satisfies it for $p=5$. 
\end{sol}

\begin{sol}{xca: counterexample os Eisenstein in Z}
    $f=3(X^{10}+5X^2-15)$ is a product of $3$ and $(X^{10}+5X^2-15)$,
    which are both non-invertible elements of $\Z[X]$.
    Hence $f$ is reducible.
\end{sol}

\begin{sol}{xca: Q[sqrt2]=Q(sqrt2)}
Clearly for every field extension $L/K$ and 
    every $\alpha\in L$ we have that
    $K[\alpha]\subseteq K(\alpha)$.

    Vice versa take $\frac{a+\sqrt{2}b}{c+\sqrt{2}d}\in \Q(\sqrt{2})$,
    then we can write:
    \[
    \frac{a+\sqrt{2}b}{c+\sqrt{2}d}=\frac{(a+\sqrt{2}b)(c-\sqrt{2}d)}{(c+\sqrt{2}d)(c-\sqrt{2}d)}=
    \frac{ac-2bd+(bc-ad)\sqrt{2}}{c^2-2d^2}.
    \]
    Hence
    \[
    \frac{a+\sqrt{2}b}{c+\sqrt{2}d}=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt{2}\in \Q[\sqrt{2}].
    \]
\end{sol}


\begin{sol}{xca:degree_of_x}
    Let $f=f(x,K)$ be the minimal polynomial of $x$ over $K$ of degree $\deg(f)=n$.
    We claim that $\{1,x,\dots, x^{n-1}\}$ is a basis of $K(x)$ as a $K$-vector space. 

    To prove that $\{1,x,\dots, x^{n-1}\}$ is a generating set, recall that $K(x)=K[x]$, since $x$ is algebraic over $K$. 
    Let $z\in K(x)=K[x]$, say $z=h(x)$ for some $h\in K[X]$. 
    Divide $h$ by $f$ to obtain polynomials $q,r\in K[X]$ 
    such that $h=fq+r$, where either $r=0$ or $\deg r<\deg f=n$. Then 
    \[
		z=h(x)=f(x)q(x)+r(x)=r(x).
	\]
	Write $r=\sum_{i=0}^{n-1}c_iX^i$ for some $c_0,\dots,c_{n-1}\in K$. 
    Thus $z=\sum_{i=0}^{n-1}a_ix^i\in \langle 1,x,\dots,x^{n-1}\rangle$.
        
    We now prove that $\{1,x,\dots, x^{n-1}\}$ is linearly independent. If not, 
    there exists a linear combination
    \[
    0=\sum_{i=0}^{n-1}a_ix^i
    \]
    with $a_0,\dots,a_{n-1}\in K$ not all zero. 
    Then $h(X)=\sum_{i=0}^{n-1}a_iX^i\in K[X]\setminus\{0\}$
    has $x$ as a root and 
        \[
        n=\deg(f)\leq \deg(h)\leq n-1,
        \]
       a contradiction. 
\end{sol}

\begin{sol}{xca:algebraic}
$a$ is algebraic over $K$, 
so, by Theorem \ref{thm:simple extesnions},
it has finite degree over $K$
and $K[a]=K(a)$.
$b$ is algebraic over $K$,
so it is also algebraic over $K(a)$, hence it has finite degree over $K(a)$ and 
$K(a)[b]=K(a,b)$.
This implies that the extension
$K(a,b)/K$ is a finite extension
since it is a tower of finite extensions.
Hence, by Corollary \ref{cor:finite=>algebraic}, $K(a,b)/K$ is an algebraic extension.
Therefore, since $a+b,ab\in K(a,b)$,
this implies that $a+b$ and $ab$
are algebraic over $K$.
\end{sol}

\begin{sol}{xca:finite type}
    Assume that $K(S)/K$ is algebraic, then, by Corollary \ref{cor:finite type algebraic}, 
    $x$ is algebraic over $K$ for all $x\in S$.
    By Corollary \ref{cor:finite type finite}, we conclude that $K(S)/K$ is finite.

    On the other hand, if $K(S)/K$ is finite, then
    $K\subseteq K(x)\subseteq K(S)$ for all $x\in S$, so
    $K(x)/K$ is finite for all $x\in S$.
    Then, by Theorem \ref{thm:simple extesnions},
    $x$ is algebraic over $K$ for all $x\in S$.
    Hence, by Corollary \ref{cor:finite type algebraic}, $K(S)/K$ is algebraic.
 \end{sol}

\begin{sol}{xca:degree of sqrt[3]2}
    $\sqrt[3]{2}$ is a root of the monic polynomial $f=X^3-2\in \Q[X]$.
    Therefore $\sqrt[3]{2}$ is algebraic over $\Q$ 
    and $\Q[\sqrt[3]{2}]=\Q(\sqrt[3]{2})$.
    In Exercise \ref{xca: X^3-2}, we proved that
    $f$ is irreducible in $\Q[X]$.
    Hence $f$ is the minimal polynomial of $\sqrt[3]{2}$
    over $\Q$ and, by Theorem \ref{thm:simple extesnions},
    $[\Q(\sqrt[3]{2}):\Q]=\deg f=3$.
\end{sol}

\begin{sol}{xca:Q(i,sqrt2)}
    $i$ is a root of the monic polynomial $X^2+1\in\Q[X]$
    and $\sqrt{2}$ is a root of the monic polynomial of 
    $X^2-2\in\Q[X]$.
    So, by Corollary \ref{cor:finite type finite},
     $\Q[i,\sqrt{2}]=\Q(i,\sqrt{2})$ and it is algebraic over $\Q$.
    By Eisenstein's criterion with $p=2$,
    $X^2-2$ is irreducible in $\Q[X]$, so 
    $[\Q(\sqrt{2}):\Q]=2$.
    Since $i$ is a root of $X^2+1\in\Q[X]$,
    then $[E:\Q(\sqrt{2})]\leq 2$.
    Moreover, $i\notin \R\supseteq \Q(\sqrt{2})$.
    Therefore $[E:\Q(\sqrt{2})]=2$ and,
    by Proposition \ref{pro:multiplicativity of degree},
    \[
    [E:\Q]=[E:\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]=4.
    \]
\end{sol}

\begin{sol}{xca:Q(sqrt2,sqrt[3]5)}\
    \begin{enumerate}
        \item      
        We know that $[\Q(\sqrt[3]{5}):\Q]$ is the same as
    the degree of the minimal polynomial of $\sqrt[3]{5}$
    over $\Q$.
    Clearly $\sqrt[3]{5}$ is a root of $X^3-5\in\Q[X]$.
    Moreover, by Eisenstein's criterion with
    $p=5$, we get that $X^3-5$ is irreducible in $\Q[X]$.
    Hence $f(\sqrt[3]{5},\Q)=X^3-5$ and $[\Q(\sqrt[3]{5}:\Q]=3$.
    We also know that $X^2-2$ is the minimal polynomial of $\sqrt{2}$ over $\Q$. 
    So $[\Q(\sqrt{2}):\Q]=2$.
    Therefore we are in the following situation:
    \
    \begin{center}
        \begin{tikzcd}
	& {E=\Q(\sqrt[3]{5},\sqrt{2})}\arrow[rd,no head]\\
	{\Q(\sqrt[3]{5})}\arrow[ru,no head,"\leq 2"]&& {\Q(\sqrt{2})}\arrow[ld,no head,"2"]  \\
	& {\Q}\arrow[lu,no head,"3"]
 \end{tikzcd}
    \end{center}
    so on the one hand
    \[
    [E:\Q]=[E:\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]=[E:\Q(\sqrt{2})]2
    \]
    and on the other hand
    \[
    [E:\Q]=[E:\Q(\sqrt[3]{5})][\Q(\sqrt[3]{5}):\Q]=[E:\Q(\sqrt[3]{5})]3\leq 2\cdot 3=6.
    \]
    Therefore $2$ and $3$ divide $[EF:\Q]\leq 6$.
    Hence the only possibility is that $[E:\Q]=6$.
    \item Clearly $\Q(\sqrt{2}+\sqrt[3]{5})\subseteq E$.
    On the other hand, let $\alpha=\sqrt{2}+\sqrt[3]{5}$.
    Then 
    \[
    5=(\alpha-\sqrt{2})^3=\alpha^3-3\sqrt{2}\alpha^2+6\alpha-2\sqrt{2},
    \]
    which implies that 
    \[
    \sqrt{2}=\frac{6\alpha-5}{3\alpha^2+2}\in\Q(\alpha).
    \]
    Moreover, $\sqrt[3]{5}=\alpha-\sqrt{2}\in\Q(\alpha)$, hence $E=\Q(\alpha)$.
    \item From the previous part of this exercise
    we get that 
    \[
    \sqrt{2}(3\alpha^2+2)=\alpha^3+6\alpha-5.
    \]
    Hence, squaring both sides of the previous equality,
    we obtain
    \[
    18\alpha^4+24\alpha^2+8=\alpha^6+36\alpha^2+25+12\alpha^4-10\alpha^3-60\alpha.
    \]
    Therefore $\alpha$ is a root of the 
    polynomial 
    \[
    f(X)=X^6-6X^4-10X^3+12X^2-60X+17.
    \]
    Moreover, from the first part, we know that
    \[
    [E:\Q]=6=\deg f.
    \]
    Hence $f(\alpha,\Q)=f(X)$.
    \end{enumerate}
\end{sol}

\begin{sol}{xca:isqrt[4]3}
Let $\alpha=\sqrt[4]{3}i$. Observe that $\alpha^2 = -\sqrt{3}$,
so $\Q(\sqrt{3})\subseteq \Q(\alpha)$ and 
\[
2\geq \deg f(\alpha,\Q(\sqrt{3}))=[\Q(\alpha):\Q(\sqrt{3})].
\]
Moreover $\alpha\notin\R$,
while $\Q(\sqrt{3})\subseteq \R$.
So $[\Q(\alpha):\Q(\sqrt{3})]>1$, 
hence $[\Q(\alpha):\Q(\sqrt{3})]=2$ and 
$$f(\alpha,\Q(\sqrt{3}))=X^2+\sqrt{3}.$$

Note that the minimal polynomial of $\alpha$
over $\Q(i)$ has degree $[\Q(\sqrt[4]{3}i,i):\Q(i)]$.
Moreover, $\Q(\sqrt[4]{3}i,i)=\Q(\sqrt[4]{3},i)$ and
$$f(\alpha,\Q)=f(\sqrt[4]{3},\Q)=X^4+3,$$
since $X^4+3$ is an irreducible
(due to Eisenstein with $p=3$)
monic polynomial that has $\alpha$ and $\sqrt[4]{3}$
as roots.
Therefore $[\Q(\sqrt[4]{3}):\Q]=[\Q(\sqrt[4]{3}i):\Q]=4$.
Since $\Q(\sqrt[4]{3})\subseteq\R$, while $\sqrt[4]{3}i\notin\R$,
we obtain that $1<[\Q(\sqrt[4]{3}i,i):\Q(\sqrt[4]{3})]\leq [\Q(i):\Q]=2$.
\begin{center}
        \begin{tikzcd}
	& {\Q(\sqrt[4]{3}i,i)=\Q(\sqrt[4]{3},i)}\arrow[rd,no head]\\
	{\Q(\sqrt[4]{3})}\arrow[ru,no head,"2"]& & {\Q(i)}\arrow[ld,no head,"2"]  \\
	& {\Q}\arrow[lu,no head,"4"]
 \end{tikzcd}
    \end{center}
Hence $[\Q(\sqrt[4]{3}i,i):\Q]=8$ and $[\Q(\sqrt[4]{3}i,i):\Q(i)]=4$, which means that $\deg f(\sqrt[4]{3}i,\Q(i))=4$.
But $f(\sqrt[4]{3}i,\Q(i))$ divides $f(\alpha,\Q)=X^4+3,$ so
$$f(\sqrt[4]{3}i,\Q(i))=X^4+3.$$
\end{sol}

\begin{sol}{xca:sqrt{2}+sqrt[3]{5}i}
    Let $\alpha=\sqrt{2}+i\sqrt[3]{5}$ and $E=\Q(\alpha,i)$. 
    The minimal polynomial of $\alpha$ over $\Q(i)$ has
    degree $[E:\Q(i)]$.
    Observe that, since $\alpha-\sqrt{2}=i\sqrt[3]{5}$,
    \[
    \alpha^3-3\sqrt{2}\alpha^2+6\alpha-2\sqrt{2}=-i5.
    \]
    Hence 
    \[
    \sqrt{2}=\frac{\alpha^3\alpha^2+6\alpha+i5}{3\alpha^2+2}\in E
    \]
    and so also $\sqrt[3]{5}=\frac{\alpha-\sqrt{2}}{i}\in E$. 
    Therefore $E=\Q(\sqrt{2},\sqrt[3]{5},i)$.
    To compute $[E:\Q(i)]$ we first compute $[\Q(\sqrt{2},i):\Q(i)]$
    and  $[\Q(\sqrt[3]{5},i):\Q(i)]$.
    
    We know that $i$ has degree 2 over $\Q$,
so $[\Q(\sqrt[3]{5},i):\Q(\sqrt[3]{5})]$ and 
$[\Q(\sqrt{2},i):\Q(\sqrt{2})]$ are both at most 2.
Moreover $\Q(\sqrt[3]{5})$ and $Q(\sqrt{2})$ are contained in $\R$,
while $i\notin\R$.
Hence
\[
[\Q(\sqrt[3]{5},i):\Q(\sqrt[3]{5})]=2=[\Q(\sqrt{2},i):\Q(\sqrt{2})].
\]
\begin{center}
        \begin{tikzcd}
	& {\Q(\sqrt{2},i)}\arrow[rd,no head]\\
	{\R\supseteq\Q(\sqrt{2})}\arrow[ru,no head,"2"]& & {\Q(i)}\arrow[ld,no head,"2"]  \\
	& {\Q}\arrow[lu,no head,"2"]
 \end{tikzcd}
 \begin{tikzcd}
	& {\Q(\sqrt[3]{5},i)}\arrow[rd,no head,"2"]&\\
	{\Q(i)}\arrow[ru,no head,""]& & {\Q(\sqrt[3]{5})\subseteq \R}\arrow[ld,no head,"3"]  \\
	& {\Q}\arrow[lu,no head,"2"]
 \end{tikzcd}
    \end{center}
Therefore $[\Q(\sqrt{2},i):\Q(i)]=2$
    and  $[\Q(\sqrt[3]{5},i):\Q(i)]=3$.
    \begin{center}
        \begin{tikzcd}
	& {\Q(\alpha,i)=\Q(\sqrt{2},\sqrt[3]{5},i)}\arrow[rd,no head]\\
	{\Q(\sqrt{2},i)}\arrow[ru,no head,""]& & {\Q(\sqrt[3]{5},i)}\arrow[ld,no head,"3"]  \\
	& {\Q(i)}\arrow[lu,no head,"2"]
 \end{tikzcd}
    \end{center}
So $[\Q(\alpha, i):\Q(i)]$ is divisible by 2 and 3 and it is also at most 6.
Therefore the degree of the minimal polynomial
of $\alpha$ over $\Q(i)$ is $[\Q(\alpha,i):\Q(i)]=6$.

We already got $\alpha^3-3\sqrt{2}\alpha^2+6\alpha-2\sqrt{2}=-i5$, 
so $2(3\alpha^2+2)^2=(\alpha^3\alpha^2+6\alpha+i5)^2,
$
i.e. 
\[
\alpha^6-6\alpha^4+10i\alpha^3+12\alpha^2+60i\alpha-33=0.
\]
This means that $\alpha$ is a root of the polynomial
\[
f(X)=X^6-6X^4+10iX^3+12X^2+60iX-33\in\Q(i)[X].
\]
Since $f$ is also monic and of degree 6, we can deduce that 
$f(\alpha,\Q(i))=f$.
\end{sol}

\begin{sol}{xca:tower of finite extensions}
By Proposition \ref{pro:multiplicativity of degree},
we know that $[E:K]=[E:F][F:K]$, so 
$[E:K]$ is finite if and only if $[E:F]$ and $[F:K]$ are finite.
\end{sol}

\begin{sol}{xca:composite generated by products}
Let $P$ be the set 
$\left\{\sum_{i=1}^me_if_i:m\in\Z_{>0},e_i\in E,f_i\in F\text{ for all $i\in\{1,\dots,m\}$}\right\}$.
If $\sum_{i=1}^me_if_i\in P$, it is a $E$-linear combination
of elements in $F$, so in particular it is an element
in $E(F)=EF$. Hence $P\subseteq EF$.

Moreover, since $E/K$ and $F/K$ are algebraic extensions,
every element in $E\cup F$ is algebraic over $K$.
So $EF=K(E\cup F)=K[E\cup F]$.
Let $x\in EF$, then $x=f(\alpha_1,\dots,\alpha_k)$,
for some polynomial $f\in K[X_1,\dots X_k]$ 
and $\alpha_1,\dots,\alpha_k\in E\cup F$.
We can then split the polynomial in $f=p+q$ so that 
$x=p(e'_1,\dots,e'_n)+q(f'_1,\dots,f'_m)$,
where $e'_i\in E$ and $f'_j\in F$ and $p,q\in K[X_1,\dots X_k]$.
Since $E$ and $F$ are fields, so closed under multiplication,
we can write $x$ as $x=\sum_{i=1}^Nk_ie_i+\sum_{j=1}^Mh_jf_j$,
for some $k_i,h_j\in K$, $e_i\in E$ and $f_j\in F$.
Then in particular $k_i\in K\subseteq F$ and $h_j\in K\subseteq E$, hence $x\in P$ and $EF\subseteq P$.
\end{sol}

\begin{sol}{xca:sqrt(2),sqrt(3)}
    We know that the minimal polynomials over $\Q$ are
    $f(\sqrt{2},\Q)=X^2-2$ and $f(\sqrt{3},\Q)=X^2-3$.
    So $[\Q(\sqrt{2}):\Q]=[\Q(\sqrt{3}):\Q]=2$.
    Moreover,
    \[
    [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{3})]\leq [\Q(\sqrt{2}):\Q]=2.
    \]
    It remains to check whether $\sqrt{2}\in \Q(\sqrt{3})$ or not.
    A $\Q$ basis of $\Q(\sqrt{3})$ is $\{1,\sqrt{3}\}$.
    If $\sqrt{2}\in \Q(\sqrt{3})$, then $\sqrt{2}=a+b\sqrt{3}$ for 
    some $a,b\in\Q$, so $2=a^2+2ab\sqrt{3}+3b^2$.
    Using the $\Q$-linear independence of $\{1,\sqrt{3}\}$,
    we get that $2ab=0$ and $2=a^2+3b^2$.
    Therefore either $a=0$ and $2/3=b^2$,
    or $b=0$ and $2=a^2$.
    But both cases are not possible because neither $2$ nor $2/3$ are 
    squares in $\Q$.

    We conclude that $\sqrt{2}\notin\Q(\sqrt{3})$.
    So $[\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{3})]=2$ and
    $[\Q(\sqrt{2},\sqrt{3}):\Q]=4$.
    \begin{center}
        \begin{tikzcd}
	& {\Q(\sqrt{2},\sqrt{3})}\arrow[rd,no head]\\
	{\Q(\sqrt{2})}\arrow[ru,no head,""]& & {\Q(\sqrt{3})}\arrow[ldd,no head,"2"]\arrow[ld,no head,""]  \\
	& {\Q(\sqrt{2})\cap\Q(\sqrt{3})}\arrow[lu,no head,""]\\
    & {\Q}\arrow[luu,no head,"2"]\arrow[u,no head]
 \end{tikzcd}
    \end{center}
    Similarly, 
    \[
    [\Q(\sqrt{2}):\Q(\sqrt{2})\cap\Q(\sqrt{3})]\leq [\Q(\sqrt{2}):\Q]=2,
    \]
    but $\sqrt{2}\notin\Q(\sqrt{2})\cap\Q(\sqrt{3})$. Thus 
    \[
    [\Q(\sqrt{2}):\Q(\sqrt{2})\cap\Q(\sqrt{3})]=2
    \] 
    and hence $[\Q(\sqrt{2})\cap\Q(\sqrt{3}):\Q]=1$. Therefore  $\Q(\sqrt{2})\cap\Q(\sqrt{3})=\Q$.
\end{sol}

\begin{sol}{xca:sqrt[3]2,3rd root of 1}
The minimal polynomial of $\sqrt[3]{2}$ is $X^3-2$
since it is monic, irreducible (by Eisenstein) and has $\sqrt[3]{2}$
as a root. Hence $[\Q(\sqrt[3]{2}):\Q]=3$.
$\xi\neq 1$ and it is a root of the polynomial 
\[
X^3-1=(X-1)(X^2+X+1),
\]
so it is a root of $X^2+X+1$, which is monic and
irreducible (it is of degree 2 and the roots are not in $\Q)$.
Hence $f(\xi,\Q)=X^2+X+1$ and $[\Q(\xi):\Q]=2$.
We also have that $[\Q(\sqrt[3]{2},\xi):\Q]\leq 6$.
    \begin{center}
        \begin{tikzcd}
	& {\Q(\sqrt[3]{2},\xi)}\arrow[rd,no head]\\
	{\Q(\sqrt[3]{2})}\arrow[ru,no head,""]& & {\Q(\xi)}\arrow[ldd,no head,"2"]\arrow[ld,no head,""]  \\
	& {\Q(\sqrt[3]{2})\cap\Q(\xi)}\arrow[lu,no head,""]\\
    & {\Q}\arrow[luu,no head,"3"]\arrow[u,no head]
 \end{tikzcd}
    \end{center}
By multiplicity of the degree of extensions, we obtain that
6 has to divide $[\Q(\sqrt[3]{2},\xi):\Q]\leq 6$ and
$[\Q(\sqrt[3]{2})\cap\Q(\xi):\Q]$ has to divide 2 and 3.
Therefore 
\[
[\Q(\sqrt[3]{2},\xi):\Q]=6 \text{ and }
[\Q(\sqrt[3]{2})\cap\Q(\xi):\Q]=1,
\]
which means that
$\Q(\sqrt[3]{2})\cap\Q(\xi)=\Q$.
\end{sol}

\begin{sol}{xca: shift for algebraic}
By definition $EF=E(F)$. If $F/K$ is algebraic, then $EF$ is
generated by algebraic elements over $K$, so also over $E$.
Hence $EF/E$ is an algebraic extension.
\end{sol}

\begin{sol}{xca: shift for finite}
If $F/K$ is finite,
then $F$ is generated by a finite number of algebraic elements over $K$.
The same elements are algebraic over $E$ and generate $EF=E(F)$ over $E$.
Hence $EF/E$ is a finite extension and $[EF:E]\leq [F:K]$.
\end{sol}

\begin{sol}{xca:C/K_bijective}
$\varphi$ is a field homomorphism,
so it is injective.
Take $\sigma:\varphi(C)\to C$ a left inverse
of $\varphi$, i.e such that $\sigma\circ\varphi=\id$.
Then $\sigma$ is a surjective field homomorphism (it is actually an isomorphism).
Moreover $C/K$ is algebraic and $\varphi(C)\subseteq C$,
so, by Proposition \ref{pro:Artin}, there exists 
a field homomorphism $\psi: C\to C$
such that $\psi|_{\varphi(C)}=\sigma$.
But then $\psi$ is injective and 
$\psi(x)=\sigma(\varphi(\psi(x)))=\psi(\varphi(\psi(x)))$, for every $x\in C$.
So $x=\varphi(\psi(x))$ for every $x\in C$.
Hence $\varphi(C)=C$, $\psi=\sigma$ and $\varphi$ is an isomorphism.
\end{sol}

\begin{sol}{xca:dec field of X^3-X-1 over Z3}
We can easily check that $f$ has no roots 
in $\Z/3$, so, having degree 3, it is
irreducible over $\Z/3$.
Note also that $f = X^3-X-1 = X(X-1)(X+1) -1$
So if $\alpha\in E$ is a root of $f$,
then 
\[
f(\alpha+1)=
(\alpha+1)\alpha(\alpha+2)-1=
(\alpha+1)\alpha(\alpha-1)-1=
f(\alpha)=0,
\]
\[
f(\alpha-1)=
(\alpha-1)(\alpha-2)\alpha-1=
(\alpha-1)(\alpha+1)\alpha-1
f(\alpha)=0.
\]
This means that $\alpha+1$ and $\alpha-1$ are also roots of $f$.
Hence $E=\Z/3(\alpha)$ and 
$f(\alpha,\Z/3)=f$,
so $[E:\Z/3]=3$.
\end{sol}

\begin{sol}{xca:dec field X^4-5X^2+5}
Note that, since $f=X^4-5X^2+5$ is an even polynomial
if $\alpha\in \C$ is a root of $f$,
then also $-\alpha$ is a root of $f$.
 Hence, given two roots $\alpha,\beta\in \C$
 such that $\beta\neq-\alpha$,
we have that the decomposition field of $f$ over $\Q$ is
$E=\Q(\alpha,-\alpha,\beta,-\beta)$.
 But $-\alpha,-\beta\in \Q(\alpha,\beta)\subseteq E$
 and so 
\[
 E=\Q(\alpha,-\alpha,\beta,-\beta)\subseteq \Q(\alpha,\beta)\subseteq\Q(\alpha,-\alpha,\beta,-\beta)=E,
 \]
 which means that $E=\Q(\alpha,\beta)$.
 Moreover we can decompose $f$ in $\C[X]$ as 
 \[
 (X-\alpha)(X+\alpha)(X-\beta)(X+\beta)=(X^2-\alpha^2)(X^2-\beta^2)=X^4-(\alpha^2+\beta^2)X^2+\alpha^2\beta^2.
 \]
This implies in particular that $\alpha^2\beta^2=5$, hence $\beta=\pm \frac{\sqrt{5}}{\alpha}\in \Q(\alpha,\sqrt{5})$.

Therefore $E=\Q(\alpha,\beta)\subseteq \Q(\alpha,\sqrt{5})$.
On the other hand $\sqrt{5}=\pm\alpha\beta\in\Q(\alpha,\beta)$,
 hence $\Q(\alpha,\sqrt{5})\subseteq\Q(\alpha,\beta)=E$.
 So we can conclude that $E=\Q(\alpha,\sqrt{5})$.
Using the multiplicative of the degree of finite extension we get that
 \[
 [E:\Q]=[E:\Q(\alpha)][\Q(\alpha):\Q].
\]
But $[\Q(\alpha):\Q]=\deg(f(\alpha,\Q))$.
 Using Eisenstein criterion (Exercise \ref{xca:Eisenstein's criterion}) with $p=5$,
 we have that $f$ is irreducible (and monic), so $f=f(\alpha,\Q)$.
Thus $[\Q(\alpha):\Q]=\deg f=4$.
It remains to compute $[E:\Q(\alpha)]=[\Q(\alpha,\sqrt{5}):\Q(\alpha)]$.
We have the following situation:
\[
\begin{tikzcd}
	& {E=\Q(\alpha,\sqrt{5})}\arrow[rd,no head]\\
	{\Q(\alpha)}\arrow[ru,no head,"\leq 2"]&& {\Q(\sqrt{5})}\arrow[ld,no head,"2"]  \\
	& {\Q}\arrow[lu,no head,"4"] 
 \end{tikzcd}
\]
Observe that $\Q(\alpha,\sqrt{5})$ is equal to the composite of $\Q(\alpha)$ and $\Q(\sqrt{5})$.
 We can use the property of composite extension,
$[LF:L]\leq[F:K]$, to deduce that 
 \[
 [\Q(\alpha,\sqrt{5}):\Q(\alpha)]\leq [\Q(\sqrt{5}):\Q]=2.
 \]
 The last equality is because $f(\sqrt{5},\Q)=X^2-5$,
 as it is monic has $\sqrt{5}$ as a root 
 and it's irreducible 
 (due to Eisenstein's criterion or
 because it is of degree 2 with 2 non-rational roots).
 Finally, we want to understand whether $[\Q(\alpha,\sqrt{5}):\Q(\alpha)]$ is 1 or 2.
 Note that $\alpha^4-5\alpha^2+5=0$, so we can solve the equation for
$\alpha^2$ as it is a root of $X^2-5X+5$, i.e.
 \[
\alpha^2=\frac{5\pm \sqrt{25-20}}{2}=\frac{5\pm \sqrt{5}}{2},
 \]
 hence $\sqrt{5}=\pm (2\alpha^2-5)\in\Q(\alpha)$.
 So $\Q(\alpha,\sqrt{5})\subseteq\Q(\alpha)\subseteq \Q(\alpha,\sqrt{5})$, 
which means that $E=\Q(\alpha)$ and
$[E:\Q]=[\Q(\alpha):\Q]=4$.
\end{sol}


\begin{sol}{xca:Q(sqrt[3]{2},xi) normal}
First of all, note that $\sqrt[3]{2}$ is a root of
the polynomial $f(X)=X^3-2$.
To prove that $\Q(\sqrt[3]{2},\xi)$ is a normal extension 
we use Proposition \ref{pro:normal<=>dec},
so it is enough to prove
that $\Q(\sqrt[3]{2},\xi)$ is the decomposition field of $f$.
We know that the decomposition field $E$ of $f$ over $\Q$ is
$\Q$ extended with the roots of $f$, i.e.
$E=\Q(\sqrt[3]{2},\sqrt[3]{2}\xi,\sqrt[3]{2}\xi^2)$.
But it's easy to see that actually 
\[
\Q(\sqrt[3]{2},\xi)=\Q(\sqrt[3]{2},\sqrt[3]{2}\xi,\sqrt[3]{2}\xi^2)=E.
\]
The inclusion $\subseteq$ is because $\sqrt[3]{2},  \xi=\frac{\sqrt[3]{2}\xi}{\sqrt[3]{2}}\in E$. 
Vice versa $\supseteq$ is due to the fact that 
the roots of $f$ are products of $\sqrt[3]{2}$ and $\xi$, elements in $\Q(\sqrt[3]{2},\xi)$.
\end{sol}


\begin{sol}{xca:Q[sqrt[4]{7}+sqrt{2}]}
Let $\alpha=\sqrt[4]{7}+\sqrt{2}$. Then $(\alpha - \sqrt{2})^4 -7 = 0$.
By expanding the left side, we get
\[
0  =\alpha^4 - 4\sqrt{2}\alpha^3 + 12 \alpha^2 - 8\sqrt{2}\alpha - 3 
 = (\alpha^4 + 12\alpha^2 - 3) -  (4 \alpha^3 + 8 \alpha )\sqrt{2}.
\]
But $4 \alpha^3 + 8 \alpha = 4\alpha (\alpha^2+2) \neq 0$, otherwise $\alpha\in\{0,\pm i\sqrt{2})$.
Therefore $\sqrt{2}=\frac{\alpha^4 + 12\alpha^2 - 3}{4 \alpha^3 + 8 \alpha}\in \Q(\alpha) $.

This allows us to prove that $\Q(\sqrt{2} , \sqrt[4]{7}) = \Q(\alpha)$.
From the definition of $\alpha$ it is clear that 
\[
\Q(\alpha) \subseteq \Q(\sqrt{2} , \sqrt[4]{7}).
\]
On the other hand, we just proved that $\sqrt{2} \in \Q(\alpha)$.
As $\sqrt[4]{7} = \alpha - \sqrt{2} \in \Q(\alpha)$,
we also see that $\sqrt[4]{7} \in \Q(\alpha)$.
It follows that $\Q(\sqrt{2} , \sqrt[4]{7}) \subseteq \Q(\alpha)$.

Moreover, $\sqrt{2}\not\in \Q(\sqrt[4]{7})$. 
Otherwise, as $[\Q(\sqrt[4]{7}):\Q] = 4$ and $[\Q(\sqrt{2}):\Q] = 2$
we would get that $[\Q(\sqrt[4]{7}):\Q(\sqrt{2})] = 2$.
Let $f(\sqrt[4]{7},\Q(\sqrt{2})) = X^2 + \beta X + \gamma$,
with $\beta, \gamma \in \Q(\sqrt{2})$. 
So 
$$0=f(\sqrt[4]{7})=\sqrt{7} + \beta \sqrt[4]{7} + \gamma.$$
Therefore 
$$\beta^2 \sqrt{7}  = (- \sqrt{7} - \gamma)^2 =7 + 2\gamma \sqrt{7}  + \gamma^2.$$
Thus 
\[
(\beta^2 - 2 \gamma)\sqrt{7}= \gamma^2 + 7.
\]
But $\beta^2 - 2 \gamma\neq 0$ because 
$\gamma^2 + \beta \gamma + \frac{\beta^2}{2} = 0$ 
holds only for $\gamma =\frac{\beta}{2} (-1 \pm i)\in \C\setminus\R$,
which is clearly not in $\Q(\sqrt{2})$.
Thus 
$\sqrt{7} = \frac{\gamma^2 + 7}{\beta^2 - 2 \gamma} \in \Q(\sqrt{2})$, a contradiction.

To sum up we have that $\sqrt{2}\not\in \Q(\sqrt[4]{7})$ and
\[
\begin{tikzcd}
	& {E=\Q(\alpha)=\Q(\sqrt{2},\sqrt[4]{7})}\arrow[rd,no head]\\
	{\Q(\sqrt{2})}\arrow[ru,no head,]&& {\Q(\sqrt[4]{7})}\arrow[ld,no head,"4"]  \\
	& {\Q}\arrow[lu,no head,"2"] 
 \end{tikzcd}
\]
\begin{enumerate}
    \item We know that $\sqrt[4]{7} \in \Q(\alpha)$ 
    which has minimal polynomial $f(\sqrt[4]{7},\Q) = x^4-7$.
    One root of this polynomial is $i\sqrt[4]{7}$.
    This root is not in $\Q(\alpha)\subseteq \R$ as it is in $\C\setminus\R$. 
    Therefore $\Q(\alpha)/\Q$ is not normal by Proposition \ref{pro:linear_factorization}.
    \item  As $\sqrt{2} \not\in \Q(\sqrt[4]{7})$,
    we see that $[\Q(\alpha):\Q(\sqrt[4]{7})] > 1$.
    On the other hand, 
    $$[\Q(\alpha):\Q(\sqrt[4]{7})] \leq [\Q(\sqrt{2}:\Q]=2,$$
    which proves that $[\Q(\alpha):\Q(\sqrt[4]{7})] = 2$.
    Therefore,
    \[
     [E:\Q]=[\Q(\alpha):\Q] = [\Q(\alpha):\Q(\sqrt[4]{7})] \cdot [\Q(\sqrt[4]{7}): \Q] = 2 \cdot 4 = 8.
    \]
    \item Let $\sigma\in G=\Gal(E/\Q)$.
    Since $E=\Q(\sqrt{2},\sqrt[4]{7})$ and 
    $\sqrt{2}$ and $\sqrt[4]{7}$ are independent 
    because $\sqrt{2}\not\in \Q(\sqrt[4]{7})$,
    we know that $\sigma$ is completely determined
    by $\sigma(\sqrt{2})$ and $\sigma(\sqrt[4]{7})$.
    By Proposition \ref{pro:conjugate},
    $\sigma(\sqrt{2})\in E$ has to be a root of $f(\sqrt{2},\Q)=X^2-2$ 
    and $\sigma(\sqrt[4]{7})\in E$ has to be a root of $f(\sqrt[4]{7},\Q)=X^4-7$.
    So $\sigma(\sqrt{2})=\pm\sqrt{2}$ and, since $E\subseteq \R$,
    $$\sigma(\sqrt[4]{7})\in E\cap\{\sqrt[4]{7}i^j\mid j\in\{0,1,2,3\}\}=\{\pm\sqrt[4]{7}\}.$$
    Therefore $G$ contains 4 elements $\sigma_{k,l}$ for $k,l\in \Z/2$
    such that $\sigma_{k,l}(\sqrt{2})=(-1)^k\sqrt{2}$ 
    and $\sigma_{k,l}(\sqrt[4]{7})=(-1)^l\sqrt[4]{7}$.
    This gives directly the isomorphism between $G$ and $\Z/2\times\Z/2$. 
\end{enumerate}
\end{sol}



\begin{sol}{xca:dim}
 Let $\{v_i:i\in I\}$ be a basis of $V$ over $K$. For each $i\in I$
 let $f_i\colon V\to F$, $f_i(v_j)=\delta_{ij}$. Then $\{f_i:i\in I\}$ is linearly 
 independent over $F$. In fact, let 
 $\sum a_if_i=0$, where each $a_i\in F$. Then 
 $a_i=0$ for almost all $i$. If $j\in I$, then 
 \[
 0=\left(\sum a_if_i\right)(v_j)=\sum a_if_i(v_j)=a_j.
 \]
 Now assume that $\dim_KV=n$. Let $\{v_1,\dots,v_n\}$ be a basis of $V$ over $K$.
 We claim that $\{f_1,\dots,f_n\}$ is a basis of $\Hom_K(V,F)$ over $F$. If 
 $g\in\Hom_K(V,F)$, then $g=\sum g(v_i)f_i$. If $1\leq k\leq n$, then
 \[
 \left(\sum g(v_i)f_i\right)(v_k)=\sum g(v_i)f_i(v_k)=g(v_k).
 \]
\end{sol}



\begin{sol}{xca:gamma_C}
We need to find a bijective map 
\[
\Hom(E/K,C/K)\to\Hom(E/K,C_1/K).
\]
If $\sigma\in\Hom(E/K,C/K)$, then $\theta^{-1}\sigma\in\Hom(E/K,C_1/K)$. 
If $\varphi\in\Hom(E/K,C_1/K)$, then $\theta\varphi\in\Hom(E/K,C/K)$. The 
maps $\sigma\mapsto\theta^{-1}\sigma$ and 
$\varphi\mapsto\theta\varphi$ are inverse to each other. 
\end{sol}

\begin{sol}{xca:order_reversing}
    We first prove that for every order
    reversing function $\varphi$
    and every element $s,t$ in its domain,
    \[
    \varphi(s\vee t)\leq \varphi(s)\wedge \varphi(t) \text{ and }
    \varphi(s)\vee \varphi(t)\leq \varphi(s\wedge t).
    \]
    Since that $s\leq s\vee t$
    and $t\leq s\vee t$ and $\varphi$
    is order reversing,
    we have that $\varphi(s\vee t)\leq \varphi(s)$ 
    and $\varphi(s\vee t)\leq \varphi(t)$.
    Hence $\varphi(s\vee t)\leq \varphi(s)\wedge \varphi(t).$
    Moreover $s\wedge t\leq s$
    and $s\wedge t\leq t$.
    So $\varphi(s)\leq \varphi(s\wedge t)$ and $\varphi(t)\leq \varphi(s\wedge t)$.
    Hence 
    $\varphi(s)\vee \varphi(t)\leq \varphi(s\wedge t).$

    We can now apply this result
    for $\varphi=f$, $s=x$ and $t=y$
    obtaining that
    \[
    f(x\vee y)\leq f(x)\wedge f(y), \quad
    f(x)\vee f(y)\leq f(x\wedge y).
    \]
    On the other hand, for $\varphi=f^{-1}$, 
    $s=f(x)$ and $t=f(y)$,
    we obtain that 
    \[
    f^{-1}(f(x)\vee f(y))\leq
    f^{-1}(f(x))\wedge f^{-1}(f(y))=
    x\wedge y,
    \]
    and
    \[
    x\vee y=f^{-1}(f(x))\vee f^{-1}(f(y))\leq f^{-1}\big(f(x)\wedge f(y)\big).
    \]
    Thus, applying $f$, which is order reversing,
    it implies that 
    \[
    f(x\wedge y)\leq f\big(f^{-1}(f(x)\vee f(y))\big)=
    f(x)\vee f(y)
    \]
    and
    \[
    f(x)\wedge f(y)=f\big(f^{-1}\big(f(x)\wedge f(y)\big)\big)\leq f(x\vee y).
    \]
 \end{sol}

 \begin{sol}{xca:Cauchy+Galois}
     Since $E/K$ is a Galois extension,
     the order of $\Gal(E/K)$ is precisely $[E:K]=n$.
     So, by Cauchy's Theorem, there exists a subgroup $S$ 
     of $\Gal(E/K)$
     of order $p$.
     Then, by Galois' Theorem, the subextension
     ${}^SE/K$
     has degree equal to the index of $S$,
     which is $n/p$.
 \end{sol}

 \begin{sol}{xca:Sylow+Galois}
     Since $E/K$ is a Galois extension,
     $|\Gal(E/K)|=[E:K]=p^\alpha m$.
     So, by Sylow's Theorem, there exists a subgroup $P$ 
     of $\Gal(E/K)$
     of order $p^{\alpha}$.
     Then, by Galois' Theorem, the subextension
     ${}^PE/K$
     has degree $(\Gal(E/K):P)=m$.
 \end{sol}

\begin{sol}{xca:separable_charp}
    Write $f=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$. Then 
    \[ 
    f'=nX^{n-1}+(n-1)a_{n-1}X^{n-2}+\cdots+2a_2X+a_1.
    \]
    Since $f$ is not separable, $f'=0$. Thus $n=ka_k=0$ in $K$ for all $k\in\{0,\dots,n-1\}$. This implies
    that $p$ divides $k$ whenever $a_k\ne 0$. This means that the only terms in $f$ occur in degree 
    that are multiples of $p$. In particular, $n=pm$ for some $m$. Hence 
    \[
    f=X^{pm}+a_{p(n-1)}X^{p(m-1)}+\cdots+a_pX^p+a_0=g(X^p)
    \]
    for some $g\in K[X]$.  
\end{sol}

% \begin{sol}\
%     \begin{enumerate}
%         \item If $E/K$ is not separable, 
%         then $[E:K]_{\operatorname{ins}}=p^s$ for some $s$. 
%         Then the trace is zero because $K$ is of characteristic $p$. 
%         \item 
%     \end{enumerate}
% \end{sol}

\begin{sol}{xca:solvable+simple}
 If $G$ is solvable, then $[G,G]$ is a proper normal subgroup of $G$. 
 Since $G$ is simple, $[G,G]=\{1\}$ and $G$ is abelian. Thus $G$ is cyclic of prime order.
\end{sol}

\begin{sol}{xca:diagonal}
 Assume that $G$ is simple. Let $A=G\times\{1\}$, $B=\{1\}\times G$ and
 $D=\{(x,x):x\in G\}$ the diagonal subgroup of $G\times G$. 
 Since 
 \[
 (g,h)=(g,1)(1,h)=(gh^{-1},1)(h,h)
 \]
 it follows that $G=AB=AD$. Let $M$ be a subgroup of $G\times G$ that contains $D$. 
 Note that
 \[
 M=M\cap (G\times G)=M\cap AD=(M\cap A)D. 
 \]
 Similarly, $M=(M\cap B)D$. Since $A$ is normal in $G\times G$, $M\cap A$ is normal in $G\times G$ 
 and $(M\cap A)B$ is normal in $MB=G\times G$. Using the second isomorphism theorem, we see that
 \[
 M\cap A\simeq \frac{(M\cap A)B}{B}
 \]
 is a normal subgroup of $(G\times G)/B\simeq A$. Since $A\simeq G$ is simple, either 
 $M\cap A=\{1\}$ or $M\cap A=A$. Thus either $M=D$ or $BD=G\times G$. Therefore $D$ is maximal.
 \end{sol}