You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
Note that $4^{\frac{1}{3}} \notin \mathbb{Q}(2^{\frac{1}{2}})$, otherwise $4^{1/3} = a + b2^{\frac{1}{2}}$ which is impossible, since squaring both sides would lead to a contradiction. So $\mathbb{Q}(2^{\frac{1}{2}}) = \mathbb{Q}(2^{\frac{1}{2}}, 4^{\frac{1}{3}})$/
Basis for $\mathbb{Q}(2^{\frac{1}{2}} + 4^{\frac{1}{3}})$
Let $\sqrt{7} \in \mathbb{Q}(\sqrt{5})$, then
$$\sqrt{7} = a + b \sqrt{5} : a, b \in \mathbb{Q}$$
Squaring both sides we have
$$7 = a^2 + 2ab \sqrt{5} + 5b^2$$
This is a contradiction since re-arranging terms would mean $\sqrt{5} \in \mathbb{Q}$ and hence a rational number for $a, b \neq 0$.
If $b = 0$, then $\sqrt{7} = a$ which is rational and if $a = 0$ then $\sqrt{7} = b \sqrt{5}$ or $\sqrt{7} \cdot \sqrt{5} = 5b$, again a contradiction.
$$\implies \sqrt{7} \notin \mathbb{Q}(\sqrt{5})$$$$x^2 - 7 = 0 \implies [\mathbb{Q}(\sqrt{7}): \mathbb{Q}] = 2$$$$\implies \mathbb{Q}(\sqrt{5}, \sqrt{7}) = {a + b\sqrt{5} + c\sqrt{7} + d\sqrt{35} : a, b, c, d \in \mathbb{Q}}$$
Q6
$$\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) = {a \sqrt{2} + b \sqrt{3} + c \sqrt{5} + d \sqrt{6} + e\sqrt{10} + f \sqrt{15}: a, b, c, d, e, f \in \mathbb{Q} }$$
Q7
$\pi$ is algebraic meands it is the root of some polynomial in the field. Of degree 3 means the polynomial has degree 3 which is also the degree of the field.
Suppose $\pi \in \mathbb{Q}(\pi^3)$, then
$$\pi = a + b \pi^3$$
but this is impossible since $\pi$ is transcendental over $\mathbb{Q}$ and $\pi \neq \pi^3$.
This $\pi$ is algebraic over $\mathbb{Q}(\pi^3)$ with
$$x^3 - \pi^3 = 0$$
$p(x) = x^2 + x + 1$ is irreducible because $p(0) = 1$ and $p(1) = 1$.
Quotient field formed by $p(x)$ consists of all 1 degree polynomials of the form $a_0 + a_1 x$, where $a_i \in \mathbb{Z}_2$.
There is a $c$ st $p(c) = 0$, and
$$\mathbb{Z}_2(c) \cong \mathbb{Z}_2[x] / \langle p(x) \rangle$$
\begin{align*}
p(c) &= c^2 + c + 1 = 0 \
\implies c^2 &= c + 1
\end{align*}
\noindent\begin{tabular}{c | c c c c c}
+ & 0 & 1 & c & c + 1 \
\cline{1-5}
0 & 0 & 1 & c & c + 1 \
1 & 1 & 0 & c + 1 & c \
c & c & c + 1 & 0 & 1 \
c + 1 & c + 1 & c & 1 & 0 \
\end{tabular}
\noindent\begin{tabular}{c | c c c c c}
$\times$ & 0 & 1 & c & c + 1 \
\cline{1-5}
0 & 0 & 0 & 0 & 0 \
1 & 0 & 1 & c & c + 1 \
c & 0 & c & c + 1 & 1 \
c + 1 & c + 1 & c & 1 & c \
\end{tabular}
Q3
$$p(x) = x^3 + x^2 + 1$$$p(0) = 1, p(1) = 1 \implies p(x)$ is irreducible and has no roots in $\mathbb{Z}_2$. $\deg p(x) = 3 \implies B = { 1, x, x^2 }$
Let there be a $c$ such that $p(c) = $, then $\mathbb{Z}_2(c) \cong \mathbb{Z}_2[x] / \langle p(x) \rangle$.
Q4
$a$ is algebraic over $F$ of degree $n$$$\implies F(a) = { a_0 + \cdots + a_{n - 1} x^{n - 1} : a_i \in F }$$
There are q possible values for $a_0, a_1, \dots, a_{n - 1}$ each and so $|{(a_0, \dots, a_{n - 1}) : a_i \in F }| = q^n$$$\implies |F(a)| = q^n$$
Q5
Let $p(x) = x^2 - k$ where $k \in \mathbb{Z}_p$ then if $p(x)$ is reducible then $c^2 = k$.
From 23H, let $h : \mathbb{Z}_p^* \rightarrow \mathbb{Z}_p^*$ be defined by $h(\repr{a}) = \repr{a}^2$, then the range of $h$ has $(p - 1) / 2$ and so is non-injective and non-surjective.
This means there exists $k \in \mathbb{Z}_p$, such that there is no $c \in \mathbb{Z}_p : c^2 = k$, and so $p(x) = x^2 - k$ has no roots in $\mathbb{Z}_p$.
D. Degrees of Extensions
Q1
$K$ forms an extension field over $F$ with basis of dimension 1 $\iff K = F$.
Q2
$L \subset K \implies \dim L < \dim K$.
Dimensions cannot be the same or that would imply they are the same.
So $L$ divides the order of $K$.
$$[K : F] = [K : L][L : F]$$
But $[K : F]$ is prime so $L$ cannot exist.
Q3
$$a \in K - F \implies [F(a):F] \leq [K : F]$$
But there are subfields of $K$ except $F$ since the extension order is prime so $K = F(a)$.
E. Short Questions Relating to Degrees of Extensions
Q1
$$\frac{1}{a} \in F(a) \textrm{ and } a \in F(\frac{1}{a}) \implies F(a) = F(\frac{1}{a})$$
$p(x)$ is the minimum polynomial for $a$, then substitute $a + c$ or $ac$ and the degree of the polynomial doesn't change.
Q2
$$p(x) = x - a, \deg p(x) = 1$$
Q3
If $c \in \mathbb{Q}$, then $\deg p(x) = 1$, thus
$$\deg p(x) > 1 \implies c \notin \mathbb{Q}$$
Q4
$$b(c) = x^2 - \frac{m}{n} = 0$$$$p \mid m, p^2 \nmid m \implies b(x) \textrm{ is irreducible}$$$$\implies \sqrt{m/n} \notin \mathbb{Q}$$
Q5
$$b(x) = x^q - \frac{m}{n}$$
and Eisenstein's criteria still holds.
Q6
$F(a)$ is a finite extension of $F$, and $F(a, b)$ is a finite extension of $F(a)$.
$(r \cdot s)(x) = r(x)s(x)$, $(r\cdot s)(a) = 0$ and $(r \cdot s)(b) = 0$, so $F(a, b)$ is a finite extension of $F$ since the degree of $r \cdot s$ is finite.
F. Further Properties of Degrees of Extensions
Q1
$K$ is a finite extension of $F$, so all elements of $K$ are also algebraic over $F$. So all algebraic extensions of $K$ are also finite algebraic extensions of $F$.
The degree of the minimum polynomial does not divide the degree of $p(x)$, so when applying polynomial long division there will be a remainder left over, which $p(x)$ itself. So $p(x)$ is not divided by $q(x)$.
G. Fields of Algebraic Elements: Algebraic Numbers
Q1
$F(a, b)$ is algebraic extension, and $a + b, a - b, ab, a / b \in F(a, b)$
Q2
Every element of the set forms a closed field over $F$, so the set is a subfield of $K$ which contains $F$.
Q3
All the coefficients belong to $\mathbb{A}$ which are algebraic over $\mathbb{Q}$ and hence form a finite extension of $\mathbb{Q}$.
Q4
$\mathbb{Q}_1(c)$ is a finite extension of $\mathbb{Q}_1$ and $\mathbb{Q}_1$ is a finite extension of $\mathbb{Q} \implies \mathbb{Q}_1(c)$ is a finite extension of $\mathbb{Q}$.
Q5
$c$ is the root of a finite polynomial whose coefficients are in finite extensions of $\mathbb{Q}$, and so $c$ forms a finite extension over $\mathbb{Q} \implies c \in \mathbb{A}$.