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$$a + c = 0$$$$ac + b + d = -16$$$$bc + ad = 0$$$$bd = 4$$
$$\implies b = \pm 1, \pm 2, \pm 4$$$$a + c = 0 \implies a = -c$$$$bc + ad = bc - dc = 0 \implies b = d \implies b = \pm 2$$$$ac + b + d = -c^2 \pm 4 = -16$$$$\implies c^2 = 16 \pm 4$$$$\implies c^2 = 12, 20$$
which has no roots in $\mathbb{Z}$.
Every element in $F(c)$ can be written as $r(c)$ where $\deg r(x) < \deg p(x)$, which is unique since for any $s(c) = t(c)$ where the degree $< n$, then $s(x) = t(x)$.
$$\forall t(x) \in F[x], t(x) = p(x)q(x) + r(x) \implies t(x) \equiv r(x) \mod {p(x)}$$
Q4
Every element in $F(c)$ can be written as $r(c)$ where $\deg r(x) < \deg p(x) = x^2 + x + 1$/
$$0, 1, c, c + 1$$
$$c^2 + c + 1 = 0$$$$\implies c^2 = c + 1$$$$(c + 1)^2 = c^2 + 1 = c$$$$c(c + 1) = c^2 + c = 1$$
$$J = { 0, x^2 + x + 1 }$$$$J + 1 = { 1, x^2 + x }$$$$J + x = { x, x^2 + 1 }$$$$J + x + 1 = { x + 1, x^2 }$$
Q5
\begin{align*}
J &= { 0, x^3 + x + 1 } \
J + 1 &= { 1, x^3 + x } \
J + x &= { x, x^3 + 1 } \
J + x + 1 &= { x + 1, x^3 }
\end{align*}
$c$ is algebraic over $F$, means there is a polynomial $p(x) \in F[x] : p(c) = 0$. Let $a(x) = p(x - 1)$, then $a(c + 1) = p(x) = 0$, and so $c + 1$ is algebraic over $F$.
Likewise since $F$ is a field then every nonzero $k \in F$ has an inverse $k^{-1}$. Let $a(x) = p(k^{-1}x)$, then $a(kc) = p(k^{-1}kx) = 0$ and so $kc$ where $k \in F$ is algebraic over $F$.
Q2
See 25G5.
Q3
$g(x) = p(xd) \implies g(c) = 0$, so $c$ is algebraic over $F(d)$. Likewise with $g(x) = p(x + d)$.
Q4
$\deg p(x) = 1 \implies p(x) = x - b$ where $b \in F$, but $p(a) = a - b = 0 \implies a = b \implies a \in F$.
Q5
$p(a) = 0 \implies p(x) \in J$, but $J$ is generated by a monic polynomial $\bar{p}(x)$, so $p(x) = \bar{p}(x)q(x)$, but $p(x)$ is irreducible so $p(x) = \bar{p}(x)$.
For the second part, there is no values $a, b \in \mathbb{Q}$ such that $(\sqrt{2})^2 = (a\sqrt{3} + b)^2$.
All the elements of $\mathbb{Q}(\sqrt{3})$ are of the form $a \sqrt{3} + b$ because $(\sqrt{3})^2 \in \mathbb{Q}$, so any higher power of $\sqrt{3}$ is either in $\mathbb{Q}$ or a multiple of $\sqrt{3}$.
$$(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta$$
Then $p(x) = x^2 - bx + c$, with $b \in F$ where $b = \alpha + \beta$. Since $b \in F, \alpha \in F(\alpha)$, then also $\beta \in F(\alpha)$.
E. Simple Extensions
Q1
$$c \implies F \implies -c \in F \implies (a + c) - c \in F(a + c) \implies a \in F(a + c) \implies F(a + c) = F(a)$$
Likewise $F$ is a field, and $c \in F \implies c^{-1} \in F$.
Q2
From 27D4, the minimum polynomial is degree 2 or higher. Let the minimum polynomial be
$$p(x) = \cdots + a_2 x^2 + a_1 x + a_0$$
and
$$a_2 a^2 + a_1 a + a_0 = 0$$
so $a^2 \in F(a)$. The reverse is not true as $F(i) \neq F(i^2) = F(-1)$.
$F(a, b)$ forms an extension field containing both $a$ and $b$, so includes $a + b$. The converse isn't true since if $a$ is not in $F$, and $a^2$ is the root of a polynomial in $F(a^2)$ then $a$ is not necessarily in $F(a^2)$. Likewise for $F(a + b)$.
Q3
$p(a + c) = 0$ so $a + c$ is a root of $p(x)$, and $a$ is a root of $g(x) = p(x + c)$. Likewise let $g(x) = p(cx)$, then $g(a) = 0$ and $p(ca) = 0$.
Q4
From 27E1, $F(a) = F(a + c)$ so
$$F[x] / \langle p(x + c) \rangle \cong F[x] / \langle p(x) \rangle$$
Let $p(x) = x^2 + 1$, then $p(x + 6) = x^2 + 12x + 36 + 1 = x^2 + x + 4$ in $\mathbb{Z}{11} \implies \mathbb{Z}{11}(\alpha) = \mathbb{Z}_{11}(\alpha + 6)$ where $\alpha$ is the root of $p(x)$.
Both $b, c \in F$, so $\frac{b}{2} \in F$ and $(\frac{b}{2})^2 - c \in F$, thus $a = (\frac{b}{2})^2 - c \in F$, and $\pm \sqrt{a} - \frac{b}{2}$ is a root of $x^2 + bx + c$.
Since $F(\sqrt{a} - \frac{b}{2}) = F(\sqrt{a})$, any quadratic extension of $F$ is of the form $F(\sqrt{a})$.
Q2
$p(x)$ and $q(x)$ are irreducible, so there is no $\sqrt{a}$ or $\sqrt{b}$ in $F$. If there was, then $p(x)$ could be factored as $(x - \sqrt{a})(x + \sqrt{a})$ and likewise for $q(x)$.
Thus $a$ and $b$ are non-squares, so by the theorem $a/b$ is square.
Lastly $c = \sqrt{a} / \sqrt{b}$, so $\sqrt{a} = c \sqrt{b}$, and $p(\sqrt{a}) = p(c \sqrt{b}) = 0 \implies \sqrt{b}$ is a root of $p(cx)$.
$F(\sqrt{a}) \cong F(\sqrt{b}) \implies$ there exists an isomorphism $h: F(\sqrt{a}) \rightarrow F(\sqrt{b})$. This comes automatically from the fundamental isomorphism theorem.
Q5
For any number in the field of reals $\mathbb{R}$ that is not a square (does not have a square root in $\mathbb{R}$), then $a/b$ is a square by the theorem since $\mathbb{R}$ is a field. Therefore for any number $a \in \mathbb{R}$, such that $\sqrt{a} \notin \mathbb{R} \implies \sqrt{a} \in \mathbb{C}$, then
$$F(\sqrt{a}) \cong F(\sqrt{b}) \cong F(\sqrt{c}) \cong \cdots$$$$\implies F(\sqrt{a}) \cong \mathbb{C}$$
G. Questions Relating to Transcendental Elements
Q1
$c$ is transcendental so the ideal is $J = { 0 } \implies F(c) = { a(c) : a(x) \in F[x] } \cong F[x]$.
Q2
$Q$ is a field of quotients of $F(c) = { a(c): a(x) \in F[x] }$ but $F(c)$ contains every possible polynomial so $Q \subseteq F(c)$, but since $F(c)$ by definition is the minimum field containing both $F$ and $c$, then $F(c) \subseteq Q$, so $F(c) = Q$.
Since $c$ is transcendental and $F(c)$ contains all quotients of $a(c)$, thus $F(c) \cong F(x)$.
Q3
$c$ is transcendental, so there is no $p(x) \neq 0 : p(c) = 0$, so there is no $q(x)$ such that $q(c + 1) = 0$ or $q(kc) = 0$, because then $p(x) = q(x - 1)$ or $p(x) = q(k^{-1} x)$ would make $c$ a root and algebraic.
If $c^2$ is algebraic over $F[x]$, then there is a $p(x) = a_n x^n + \cdots + a_0$ such that $p(c^2) = 0$. Let $g(x) = p(x^2)$, then $g(c) = p(c^2)$ and hence $c$ is algebraic - a contradiction.
Q4
Every element of $F(c)$ can be written as $a_0 + a_1 c + \cdots + a_n c^n$.
Generalizing the argument previously, for any $n \in \mathbb{Z}$, $c$ is transcendental over $F \iff$$c^n$ is transcendental. Likewise for $kc : k \in F$ and $c + k$.
So every polynomial of degree 1 or more containing $c$ is transcendental over $F$.
H. Common Factors of Two Polynomials: Over $F$ and over Extensions of $F$
Q1
$a(c) = 0 = b(c) \implies a(x), b(x) \in J$ but $J = \langle p(x) \rangle$ where $p(x)$ is a monic irreducible polynomial in $F[x]$. So $a(x)$ and $b(x)$ are both multiples of $p(x)$ and share $p(x)$ as a common factor.
Q2
$a(x), b(x) \in F[x]$ and
$$s(x)a(x) + t(x)b(x) = 1$$
remains true in $K[x]$. Likewise the converse holds.
$\char F = 0 \implies p \cdot 1 = 0 \implies \forall a \in F, p \cdot a = 0$. The derivative of $a'(x)$ consists of terms of the form $k a_k x^{k - 1}$. So $a'(x) = 0 \implies a(x)$ consists of terms of the form $a_{mp} x^{mp}$.