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The basis for a degree 2 extension $\mathbb{F}(c)$ is ${c, 1}$ with $c^2 \in \mathbb{F}$. This includes $-c$ and so $\mathbb{F}(c)$ is the root field for $(x + c)(x - c) = x^2 - c^2$.
The question I assume is asking to prove every degree 2 field extension is normal. So since $c \in F(c)$, then we can divide the polynomial by $(x - c)$, leaving a degree 1 polynomial $(x - \alpha) \in F(c)[x]$ or that $\alpha \in F(c)$.
Q2
$a(x)$ as a polynomial of degree $n$ over a finite field will have $n$ distinct roots. All roots $c_1, \dots, c_n \in K$ and form a root field over $F$. Assume some roots are in $I$, then the roots of $a(x)$ are still the same, and so $K$ forms a root field over $I$ as well.
Q3
Any polynomial can be factored into a linear combination of complex roots. This means $\mathbb{C}$ is the root field of every polynomial. Since polynomials can also have root fields contained in $\mathbb{R}$, and $\mathbb{R} \subset \mathbb{C}$, this means the root field is either $\mathbb{R}$ or $\mathbb{C}$ of every polynomial.
Roots of $p(x)$ are $\pm \sqrt{\frac{a}{2} \pm d_1}$. We know that $\frac{a}{2} \pm d_1 \in F(d_1)$.
$[F(d_1): F] = 2$. Every extension of degree 2 is a root field from 31C1 above.
Q6
$$\sigma_c(a(x)) = a(c)$$$$\ker \sigma_c = { a(x): a(c) = 0 }$$
Every ideal of a field is principal, so $J = \langle p(x) \rangle$.
For some $a(x) \in J$, $a(x)$ is a multiple of $p(x)$.
$$\sigma_c : F[x] \rightarrow F$$
so $F(c)$ contains $p(x)$ since the ideal $J$ contains all polynomials with $c$ as their root, and all polys are a multiple of $p(x)$.
Q7
From the previous argument $p(x)$ is the lowest degree polynomial in $J$ and hence irreducible.
Otherwise $p(c) = f(c)g(c) = 0 \implies f(c) = 0 \textrm{ or } g(c) = 0$
which would be a contradiction since it implies there would be a lower degree polynomial in $J$.
Every root field is a simple extension $F(c)$, we can convert a polynomial with roots $a, b$ to one with $c$ from theorem 2 in this chapter.
Q8
Give an isomorphism $h(x)$ which fixes F, and a polynomial
$$a(x) = a_0 + a_1 x + \cdots + a_n x^n$$
We can see that where $c$ is an algebraic root in $K$, then
\begin{align*}
a(h(c)) &= a_0 + a_1 h(c) + \cdots + a_n h(c^n) \
&= h(a(c)) = h(0) = 0
\end{align*}
So also $h(c)$ is a root of $a(x)$ and the isomorphism simply permutes roots since it sends unique elements of $K$ to $K'$.
We also observe that the mapping is one to one, fixing $F$, permuting roots, and that $h: K \rightarrow K$.
$$h({c_1, \dots, c_n}) = { c_1, \dots, c_n }$$
Since the polynomial $a(x)$ is irreducible over $F$, and $K$ being a finite extension can be reduced to a simple extension $F(c) = K$, where $c$ is a root of $a(x)$, which means that
$$F(c) \cong F / \langle a(x) \rangle$$
which forms a vector space of $\deg a(x) = n \implies [K:F] = n$.
Given another polynomial $b(x)$ where $b(c) = 0 : c \in K \implies b(h(c)) = 0 \implies$ K is the splitting field for $b(x)$.
This means $b(x)$ is split completely by $K$ into linear factors.
Note, this means every polynomial of degree $n$ which has a root in $K$ makes $K$ its splitting field. The converse does not hold. An irreducible polynomial of degree $n$ does not necessarily have a splitting field of degree $n$. See this answer.
Q9
First we prove this for an irreducible polynomial $p(x)$ with $n = \deg p(x)$ roots of the form $c_1, \dots, c_n$. Inductively adjoining $c_1$ to $F$ forms a field $F(c_1)$ such that $[F(c_1):F] = n$ with a basis ${ 1, c_1, \dots, c_1^{n - 1} }$. Dividing $p(x)$ by $(x - c_1)$ leaves a polynomial $q(x)$ with degree $n - 1$ and adjoining the second root to $F(c_1)$ forms a field extension with degree $[F(c_1, c_2):F(c_1)] = n - 1$. Proceeding in this way, we obtain a maximum order of $n!$.
The polynomial $a(x)$ is reducible to $n$ irreducible factors $a(x) = p_1(x) p_2(x) \cdots p_n(x)$ where $\deg p_i(x) = k_i$. Each of these fields forms a simple extension $F(c_1, \cdots, c_n)$, where $[F(c_1, \cdots, c_n):F] = [F(c_1, \cdots, c_n):F(c_1, \cdots, c_{n - 1})]\cdots[F(c_1):F] = k_n \cdot k_{n - 1} \cdots k_1$.
Since $\deg a(x) = N$ and $k_1 + \cdots k_n$, so $[K:F] \mid k_1! \cdots k_n!$.
From the binomial formula, given $n = k + l$, then there's an integer $z$ such that
$$z = \frac{n!}{k!l!} \implies n! = z k! l!$$
So therefore given $n = k_1 + \cdots + k_n$, we can see that $x \mid k_1! \cdots k_n! \implies x \mid n!$
D. Reducing Iterated Extensions to Simple Extensions
The roots of $x^2 - 2x - 1$ by completing the square are $\pm \sqrt{2} + 1$.
For a cubic with real coefficients in a field, it either has all real roots or 2 complex roots. By differentiating and sketching the curve where it's increasing or decreasing, we see that this cubic has two complex roots.
According to the Complex Conjugate Theorem, if $x = a + ib$ is a solution to a polynomial with real coefficients, then so is $x = a - ib$.
Thus we conclude that $\mathbb{Q}(a, b) = \mathbb{Q}(a + b)$ where $a$ is a complex root of $x^3 - x - 1$.
Q3
These factors are all linearly independent so $c = \sqrt{2} + \sqrt{3} + \sqrt{-5}$.
Q4
From C6, because $\sqrt{2}$ and $\sqrt{3}$ are independent, the basis is ${ 1, \sqrt{2}, \sqrt{3}, \sqrt{6} }$ and so the minimum polynomial has degree 4.
sage: # We let x = sqrt(2) + sqrt(3), so now we square it on both sidessage: (sqrt(2) +sqrt(3))^2
(sqrt(3) +sqrt(2))^2sage: ((sqrt(2) +sqrt(3))^2).expand()
2*sqrt(3)*sqrt(2) +5sage: # so now (x^2 - 5) = 2*sqrt(3)*sqrt(2)sage: # lets square again both sidessage: (2*sqrt(3)*sqrt(2))**224sage: ((x^2-5)^2).expand()
x^4-10*x^2+25sage: ((x^2-5)^2).expand() -24x^4-10*x^2+1
The roots of $x^n - 1$ are $1, \omega, \dots, \omega^{n - 1}$ which is the basis for $\mathbb{Q}(\omega)$ generated by $\omega$ since it is primitive.
Q2
Define a substitution function $\sigma_\omega$$$\sigma_\omega(a(x)) = a(\omega)$$$\sigma_\omega$ is a homomorphism because
\begin{align*}
\sigma_\omega(a(x)b(x)) &= a(\omega)b(\omega) \
&= \sigma_\omega(a(x))\sigma_\omega(b(x))
\end{align*}
Which has a kernel of
\begin{align*}
\ker \sigma_\omega &= { a(x) : \sigma_\omega(a(x)) = a(\omega) = 0 } \
&= J
\end{align*}
The kernel of any homomorphism is an ideal. In $F[x]$ every ideal is a principal ideal so $J = \langle p(x) \rangle$. So $p(x)$ is a polynomial of lowest degree among all nonzero polynomials in $J$. Hence it is irreducible.
When $n$ is prime, then
$$x^{n - 1} + x^{n - 2} + \cdots + x + 1$$
is irreducible. Therefore $p(x) = x^{n - 1} + \cdots + 1$ and $p(\omega) = 0$.
Since
$$\mathbb{Q}(\omega) \cong \mathbb{Q}[x] / \langle p(x) \rangle$$
Then $$[\mathbb{Q}(\omega):\mathbb{Q}] = \deg p(x) = n - 1$$
The roots are $1, s, s^2$ and $-1, -s, -s^2$ respectively where $s$ is the third root of unity.
But from above we know that $s^2 + s + 1 = 0 \implies s^2 = -(s + 1)$ and $s^2 \in \mathbb{Q}(s)$, so $\mathbb{Q}(\omega) = \mathbb{Q}(s)$ with basis ${ 1, s }$.
With roots $-1, 1$ and $i, -i$ for $(x^2 - 1)$ and $(x^2 + 1)$ respectively.
The 4th roots of $-1$ are $\pm \frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}}$.
$\mathbb{Q}(\omega) = \mathbb{Q}(\sqrt{2}, i)$ with a basis ${ 1, \sqrt{2}, i, \sqrt{2}i }$ and
$$[\mathbb{Q}(\omega):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, i) : \mathbb{Q}] = 4$$
Q5
$$\forall r \in { 1, 2, \dots, n - 1 }, (\sqrt[n]{a} \omega^r)^n = 1$$
$$| { \sqrt[n]{a} \omega^r : r \in { 0, 1, \dots, n - 1 } } | = n$$
Q6
The basis of $\mathbb{Q}(\omega, \sqrt[n]{a})$ is the set ${ \omega^i (\sqrt[n]{a})^j }$ where $i, j \in { 0, 1, \dots, n - 1 }$, which contains the ideal for $\sigma(c) = x^n - 1$, which is $J = { \sqrt[n]{a}, \sqrt[n]{a} \omega, \dots, \sqrt[n]{a} \omega^{n - 1} }$.
Q7
The degree of $[\mathbb{Q}(\omega, \sqrt[3]{2}):\mathbb{Q}] = [\mathbb{Q}(\omega, \sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}:\mathbb{Q}] = 2 \times 3 = 6$.
You can calculate $\cos (\pi / 3)$ and $\sin (\pi / 3)$ by splitting an equilateral triangle with unit sides in half. The sum of a triangle's angles will always be $\pi$, and so each corner of the equilateral triangle will be $(\pi / 3)$. Use pythagoreas theorem to calculate the midline as $o^2 = 1^2 - (\frac{1}{2})^2$.
Using wolfram alpha, we see that the cube roots of 1 are $1, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i$.
Let $s$ be the nth root of $a$ in $K$, then it is the root of $x^n - a$. Then $\sqrt[n]{a} \omega^i$ is also a root of this polynomial. For every $n$ there is an irreducible cyclotomic polynomial $p(x)$ with roots that consist of $\phi(n)$ nth roots of $a$ that have a multiplicative order of $n$. By theorem 7, since $p(x)$ has one root in $K$ and is irreducible, therefore all its roots are in $K$.
For $n$ is not prime, $p(x) \mid x^n - a$ so all $\phi(n)$ primitive roots of $p(x)$ are also roots of $x^n - a$ and so generates all roots. We use the irreducible polynomial $p(x)$ to prove there's an isomorphism permuting roots of $p(x)$, therefore showing both fields are equivalent and contain all roots.
Fixing $\mathbb{Q}$, there is an isomorphism $h: \mathbb{Q}(s) \rightarrow \mathbb{Q}(\sqrt[n]{a} \omega^i)$. Note that $p(h(c)) = (h(c))^n - a = h(c^n - a) = 0$ as the function is homomorphic so $h(c)^n = h(c^n)$ and fixes $\mathbb{Q}$ leaving $a$ unchanged. This function is an isomorphism mapping one to one and onto, we can say that $h(x)$ permutes the roots $c$ of $x^n - a$. So we can see that $\mathbb{Q}(s) = \mathbb{Q}(\sqrt[n]{a} \omega^i)$, and that $K$ contains all roots of $x^n - a$.
Since the spltting field contains all powers of $\sqrt[n]{a} \omega^i$ by the isomorphism $h(x)$ as well as $\sqrt[n]{a} \omega^0 = \sqrt[n]{a}$ and also its inverse $(\sqrt[n]{a})^{-1}$ (because it's a field), so it also contains all nth roots of unity since $(\sqrt[n]{a})^{-1} (\sqrt[n]{a} \omega^i) = \omega^i$.
F. Separable and Inseparable Polynomials
Q1
From theorem 1, $F$ has characteristic 0 $\implies$ irreducible polynomials never have multiple roots.
Q2
Each power of $x$ in $a(x)$ is independent, so $a_m x^m + a_n x^n \neq 0$ when $m \neq 0$ unless $a_m = a_n = 0$.
Therefore $a'(x) = \sum m_i a_i x^{m_i - 1} = 0 \implies p \mid m_i$ for all $m_i$ and all nonzero terms of $a(x)$ are of the form $a_{mp} x^{mp}$.
Q3
$$a(x) = (x - c)^2 q(x)$$$$a'(x) = 2(x - c)q(x) + (x - c)^2 q'(x)$$
Both $a(x)$ and $a'(x)$ have $c$ as a root, but $a(x)$ is irreducible, so $a(x) \mid a'(x)$, but this cannot be true since
$\deg a'(x) < \deg a(x)$
unless $a'(x) = 0$.
From the previous answer $a'(x) = 0$ means the nonzero terms of $a(x)$ are of the form $a_{mp} x^{mp}$, and $a(x)$ is a polynomial in powers of $x^p$.
Q4
$a(x)$ is a polynomial in powers of $x^p \implies a'(x) = 0 \implies a(x) \mid a'(x) \implies$ share common factor of $a(x) \implies a(x)$ has a multiple root.
Q5
This follows from $[a(x) + b(x)]^p = a(x)^p + b(x)^p$ in any field of characteristic $p$ and is proved in 24D6.
The frobenius automorphism for a finite field of characteristic $p$ is bijective since the function is injective, and any injective function from a finite set to itself is also surjective.
This implies the coefficients of $a(x^p)$ all have $p$th roots, and so
\begin{align*}
a(x^p) &= c_0^p + c_1^p x^p + \cdots + c_n^p x^{np} \
&= (c_0 + c_1 x + \cdots + c_n x^n)^p
\end{align*}
Thus $b(x) = c_0 + c_1 x + \cdots + c_n x^n$ and $a(x^p) = [b(x)]^p$.
Q7
Assume there is an irreducible polynomial $a(x)$ that is inseperable. Then it is a polynomial in powers of $x^p$.
Then there is a polynomial $[b(x)]^p = a(x)$. Thus $a(x)$ is reducible, which is a contradiction.
Thus every irreducible polynomial is separable.
G. Multiple Roots over Infinite Fields of Nonzero Characteristic
Q1
There are infinite powers of $y$, so $\mathbb{Z}_p[y]$ is an infinite ring with characteristic $p$.
Thus
$$\mathbb{Z}_p(y) = { a(y) / b(y) : a(y), b(y) \in \mathbb{Z}_p[y] }$$
is an infinite field, and so is $\mathbb{Z}_p(y^p)$
Q2
By binomial theorem, all coefficients for the terms of $(x - y)^p$ are a factor of $p$. $E[x]$ has characteristic $p$ so
$$x^p - y^p = (x - y)^p$$
However $y \notin K[x]$ so $x^p - y^p$ is irreducible in $K[x]$.
Q3
$$\repr{i}(a_0 + \cdots + a_n x^n) = i(a_0) + \cdots + i(a_n) x^n$$
but for all $a_i \in F, i(a_i) = a_i$ so $\repr{i}(p(x)) = p(x)$.
Q4
Expanding out $(x - a)^m$, the coefficients in $F$ and $x$ remain fixed, but $a$ is mapped to $b$, so
$$\repr{i}((x - a)^m) = (x - b)^m$$
Q5
Since $\repr{i}$ leaves $p(x)$ fixed, so
\begin{align*}
\repr{i}(p(x)) &= \repr{i}((x - a)^m s(x)) \
&= (x - b)^m \repr{i}(s(x)) \
&= p(x)
\end{align*}
so $a$ and $b$ have the same multiplicity in $p(x)$.
H. An Isomorphism Extension Theorem (Proof of Theorem 3)
Q1
$$F_1(a) \cong F_1[x] / \langle p(x) \rangle$$
Where $p(x)$ is the minimum polynomial with $a$ as a root. The homomorphism $\phi : F_1[x] \rightarrow F_1(a)$ by $\phi_c(a(x)) = a(c)$ has the kernel $J = \langle p(x) \rangle$ since in $F[x]$ every ideal is a principal ideal.
Thus since $s(x) = c(x) - d(x)$ has a root $a$ since $s(a) = 0$, so $s(x) \in J$ and it is a multiple of $p(x)$.
Observing that $h(c(x) - d(x)) = h(p(x)q(x))$, we easily see that $h(p(x)q(x)) = hp(x) hq(x)$ and also that $h(c(x) - d(x)) = hc(x) - hd(x)$ since
We start with $F_1(u) = K_1$ and want to prove that this means $F_2(v) = K_2$. This will also automatically prove the converse statement if shown to be true.
Let $p(x)$ be the minimum polynomial $p(x)$ for $u$ such that $p(u) = 0$.
$h: F_1(u) \rightarrow F_2(v)$ but $F_1(u) = K_1$ so $h: K_1 \rightarrow F_2(v)$, but $h$ is surjective and so $\deg p(x) = \deg hp(x)$ since both $p(x)$ and $hp(x)$ are irreducible.
Because there are $\deg hp(x) = \deg p(x)$ such roots $v_i$ which correspond to $u_i$ roots of $p(x)$.
Q3
$$a(x) = p(x) q(x)$$$$p(u) = 0$$
\begin{align*}
h(p(u)) &= h(p(x))(v) = 0 \
&= hp(x) = hp(v)
\end{align*}
Since both are equivalent.
We see that $v$ is a root of $hp(x)$.
$F_1(u) = K_1$
If $F_1(u) = K_1$, then $F_1(u)$ contains all roots of $p(x)$ and then
$$F_1(u) = K_1 \iff F_2(v) = K_2$$
Recalling that $p(x)$ is an irreducible factor of $a(x)$,
$$u \in K_1$$$$p(u) = 0$$$$v \in K_2$$$$hp(v) = 0$$$$\implies F_1(u) \cong F_2(v)$$
Putting both together
$$K_1 \cong K_2$$
$F_1(u) \neq K_1$
See that we can extend $h$ fixing the base field.
$$h(u) = v$$$$h : F_1(u) \rightarrow F_2(v)$$
In $F_1(u)[x]$, $a(x) = (x - u)a_1(x)$.
In $F_2(v)[x], ha(x) = (x - v)ha_1(x)$.
And $\deg a_1(x) = \deg ha_1(x) = n - 1$
Now let there be a new $u' \in K_1, u' \notin F_1(u'), p(u') = 0$ and likewise for $hp(x)$ and $v$.
$\deg a(x) = 1$
Lastly when $n = \deg a(x) = 1$, then $K_1 = F_1$ and $K_2 = F_2$, since the basis are simply scalars and $a(x)$ is of the form $(x - a)$.
The root of $a(x)$ is in $F_1$ itself, and $h$ is an ismomorphism from $F_1 \rightarrow F_2$ so
$$K_1 \cong K_2$$
Q4
$$u \in K_1, v \in K_2$$
But $F_1 = F_2$$$h : F[x] \rightarrow F[x]$$$$\forall a \in F, h(a) = a$$$$h = \textrm{id}_F$$$$h(u) = v$$$$\implies K_1 \cong K_2$$
Since isomorphisms preserve roots, we deduce they permute roots. There are $p - 1$ roots for the minimum polynomial of $\omega$ which has degree $p - 1$.
Q2
$p(x)$ is irreducible in $F[x]$. $c \in \mathbb{C} : p(c) = 0$.
Let $h : F \rightarrow \mathbb{C}$ be a monomorphism (injective homomorphism), then $h : F \rightarrow h(F)$ is an isomorphism and
$$F \cong h(F)$$
The minimum polynomial are $p(x)$ and $hp(x)$ respectively with $\deg p(x) = \deg hp(x) = n$. Since $h$ permutes roots and by theorem 7 contains all roots, there are $n$ possible monomorphisms.
Q3
$$h : F \rightarrow h(F)$$$$\phi : K \rightarrow \mathbb{C}$$$$[K : F] = n$$
So $K$ forms a splitting field over $F$ for a minimum polynomial $p(x) \in F[x]$ of degree $n$.
From the previous question we see that there are $n$ monomorphisms $F(c) = K \rightarrow \mathbb{C}$.
Thus all monomorphisms $h : \mathbb{Q}(a) \rightarrow \mathbb{C}$ fix $\mathbb{Q}$.
Q5
$$\mathbb{Q}(\sqrt[3]{2}) \rightarrow \mathbb{C}$$
Minimum polynomial for $c = \sqrt[3]{2}$ is $p(x) = x^3 - 2$
Three roots of $p(x)$ are $\sqrt[3]{2}, \sqrt[3]{2} \omega, \sqrt[3]{2} \omega^2$.
$\mathbb{Q}$ remains fixed. Roots are permuted (see above questions).
Three monomorphisms are
$$\sqrt[3]{2} \rightarrow \sqrt[3]{2}$$$$\sqrt[3]{2} \rightarrow \sqrt[3]{2} \omega$$$$\sqrt[3]{2} \rightarrow \sqrt[3]{2} \omega^2$$
K. Normal Extensions
Q1
$K$ is a finite extension of $F \implies K$ is a simple extension and so that $K = F(c)$.
There is a minimum polynomial $p(x)$ for $c$ in $F$.
So by the question $K$ is a normal extension.
Q2
$K$ is a finite extension of $F \implies K = F(c)$.
Let $p(x)$ be the minimum polynomial for $c$, so $p(c) = 0$.
Let $h$ be an isomorphism fixing $F$$$h: K \rightarrow h(K)$$
Then by the question $h(K) \subseteq K$ and since $h$ is an isomorphism where $K \cong h(K)$, so $h(K) = K$, and $h : K \rightarrow K$ is an automorphism fixing $F$ and permuting roots of $p(x)$.