ch27.md

title	author	header-includes	abstract
Abstract Algebra by Pinter, Chapter 27	Amir Taaki	- \usepackage{mathrsfs} - \usepackage{mathtools} - \usepackage{extpfeil} - \DeclareMathOperator\ker{ker} - \DeclareMathOperator\ord{ord} - \DeclareMathOperator\gcd{gcd} - \DeclareMathOperator\lcm{lcm} - \DeclareMathOperator\char{char} - \DeclareMathOperator\max{max} - \DeclareMathOperator\ran{ran} - \DeclareMathOperator\deg{deg} - \newcommand{\mod}[1]{\ (\mathrm{mod}\ #1)} - \newcommand{\repr}[1]{\overline{#1}} - \newcommand{\leg}[2]{\left(\frac{#1}{#2}\right)}	Chapter 27 on Extensions of Fields

A. Recognizing Algebraic Elements

Q1

a.

$$p(x) = x^2 + 1 \implies p(i) = 0$$

b.

$$p(\sqrt{2}) = 0 \implies p(x) = x^2 - 2$$

c.

$$a = 2 + 3i \qquad (a - 2)^2 = -9$$ $$p(x) = a^2 - 4a + 13$$

d.

$$p(\sqrt{1 + \sqrt[3]{2}}) = 0 \implies p(x) = (x^2 - 1)^3 - 2$$

e.

$$p(x) = (x^4 - 1)^2 - 8$$

f.

$$p(x) = (x^2 - 5)^2 - 24$$

g.

Let $x = \sqrt[3]{2}$, then $y = \sqrt[3]{2} + \sqrt[3]{4} = x + x^2$.

$$y^3 = x^6 + 3x^5 + 3x^4 + x^3 = 4 + 6x^2 + 6x + 2 = 6 + 6y$$

$$\implies p(y) = y^3 - 6y - 6$$

Q2

a.

$$p(x) = x^2 - \pi$$

b.

$$p(x) = x^4 - \pi^2$$

c.

$$p(x) = \pi^3 x - \pi^6 + \pi^3$$

B. Finding the Minimum Polynomial

Q1

a.

$$a = 1 + 2i$$ $$(a - 1)^2 = -4$$ $$p(x) = x^2 - 2x + 5$$

Reducing the equation from $\mathbb{Q}$ to $\mathbb{Z}_3$ then $\bar{p}(x) = x^2 + x + 2$ which has no roots in the field and so is irreducible.

b.

sage: p = lambda x: (x - 1)**2 - 2
sage: p(x + 1)
x^2 - 2
sage: p(x + 2)
x^2 + 2*x - 1
sage: p(x + 3)
x^2 + 4*x + 2

By Eisenstein’s criterion with $p = 2$, then this polynomial is irreducible.

c.

sage: p = lambda x: (x - 1)**4 - ((2*I)**(1/2))**4
sage: p(x)
x^4 - 4*x^3 + 6*x^2 - 4*x + 5

Let $h: \mathbb{Q} \rightarrow \mathbb{Z}_3$ then $h(p(x)) = x^4 + 2x^3 + 2x + 2$ which by Eisenstein’s criterion means the polynomial is irreducible.

d.

sage: p = lambda x: (x^2 - 2)**3 - 3
sage: p(x)
x^6 - 6*x^4 + 12*x^2 - 11

TODO: finish this

e.

sage: p = lambda x: (x**2 - 3 - 5)**2 - 4*3*5
sage: p(x)
x^4 - 16*x^2 + 4

$$a + c = 0$$ $$ac + b + d = -16$$ $$bc + ad = 0$$ $$bd = 4$$

$$\implies b = \pm 1, \pm 2, \pm 4$$ $$a + c = 0 \implies a = -c$$ $$bc + ad = bc - dc = 0 \implies b = d \implies b = \pm 2$$ $$ac + b + d = -c^2 \pm 4 = -16$$ $$\implies c^2 = 16 \pm 4$$ $$\implies c^2 = 12, 20$$ which has no roots in $\mathbb{Z}$.

f.

sage: p = lambda x: (x^2 - 1)^2 - 2
sage: p(x)
x^4 - 2*x^2 - 1
sage: p(x + 1)
x^4 + 4*x^3 + 4*x^2 - 2

By Eisenstein’s criterion with $p = 2$, this polynomial is irreducible.

Q2

a.

$$a = \sqrt{2} + i$$ $$(a - \sqrt{2})^2 = -1$$ $$x - 2\sqrt{2}x + 3$$

b.

$$a = \sqrt{2} + i$$ $$a^2 = 1 + 2\sqrt{2}i$$ $$(a^2 - 1)^2 = a^4 - 2a^2 + 1 = -8$$ $$x^4 - 2x^2 + 9$$

c.

$$a = \sqrt{2} + i$$ $$(a - i)^2 = a^2 - 2ai - 1 = 2$$ $$x^2 - 2ix - 3$$

Q3

$\sqrt{3} + i$

$\mathbb{R}$

sage: ((x - 3**(1/2))**2 + 1).expand()
x^2 - 2*sqrt(3)*x + 4

$\mathbb{Q}$

sage: (x^2 - 2)**2 + 2*3
x^4 - 4*x^2 + 10

$\mathbb{Q}(i)$

sage: ((x - I)**2 - 3).expand()
x^2 - 2*I*x - 4

$\mathbb{Q}(\sqrt{3})$

sage: ((x - 3**(1/2))**2 + 1).expand()
x^2 - 2*sqrt(3)*x + 4

$\sqrt{i + \sqrt{2}}$

$\mathbb{R}$

sage: ((x^2 - 2**(1/2))**2 + 1).expand()
x^4 - 2*sqrt(2)*x^2 + 3

$\mathbb{Q}(i)$

sage: ((x^2 - I)^2 - 2).expand()
x^4 - 2*I*x^2 - 3

$\mathbb{Q}(\sqrt{2})$

sage: ((x^2 - 2**(1/2))**2 + 1).expand()
x^4 - 2*sqrt(2)*x^2 + 3

$\mathbb{Q}$

sage: ((x^4 - 1)^2 + 8).expand()
x^8 - 2*x^4 + 9

Q4

a.

$$(x + 1)^2 - 8 = 0$$ $$x = \pm sqrt{8} - 1$$

b.

$$(x^2 + 1)^2 - 2 = 0$$ $$x^2 = \pm sqrt{2} - 1$$ $$x = \pm \sqrt{\pm \sqrt{2} - 1}$$

c.

$$(x^2 - 5)^2 - 24 = 0$$ $$x^2 = \pm 2 \sqrt{6} + 5$$ $$x = \pm \sqrt{\pm 2 \sqrt{6} + 5}$$

Q5

a.

$$\sigma_{\sqrt{2}}(a(x)) = a(\sqrt{2})$$ $$J = \langle p(x) \rangle \implies p(\sqrt{2}) = 0$$ $$p(x) = x^2 - 2$$

b.

Same as 27B1b:

$$x^2 + 4x + 2$$

c.

Same as 27B1f:

$$x^4 + 4x^3 + 4x^2 - 2$$

C. The Structure of Fields $F[x] / \langle p(x) \rangle$

Q1

$$t(x) \in F[x], t(x) = p(x)q(x) + r(x) : \deg r(x) < \deg p(x)$$ $$p(c) = 0 \implies t(c) = 0 + r(c) = r(c)$$

Q2

$s(c) = t(c) \implies J + s(x) = J + t(x)$, $J = \langle p(x) \rangle$, but $\deg s(x) &lt; \deg p(x)$ and $\forall a(x) \in J + s(x), a(x) = p(x)q(x) + s(x)$. Since $\deg t(x) &lt; \deg p(x)$, then $$t(x) = 0 + s(x) = s(x)$$

Q3

Every element in $F(c)$ can be written as $r(c)$ where $\deg r(x) &lt; \deg p(x)$, which is unique since for any $s(c) = t(c)$ where the degree $&lt; n$, 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 $\deg r(x) &lt; \deg p(x) = x^2 + x + 1$/ $$0, 1, c, c + 1$$

$$c^2 + c + 1 = 0$$ $$\implies c^2 = c + 1$$ $$(c + 1)^2 = c^2 + 1 = c$$ $$c(c + 1) = c^2 + c = 1$$

$$J = { 0, x^2 + x + 1 }$$ $$J + 1 = { 1, x^2 + x }$$ $$J + x = { x, x^2 + 1 }$$ $$J + x + 1 = { x + 1, x^2 }$$

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*}

sage: x = PolynomialRing(IntegerModRing(2, is_field=True), 'x').gen()
sage: (x^3 + x^2)%(x^3 + x + 1)
x^2 + x + 1
sage: (x^3 + x^2 + 1)%(x^3 + x + 1)
x^2 + x
sage: (x^3 + x^2 + x)%(x^3 + x + 1)
x^2 + 1
sage: (x^3 + x^2 + x + 1)%(x^3 + x + 1)
x^2

\begin{align*} J + x^2 &= { x^2, x^3 + x^2 + x + 1 } \ J + x^2 + x &= { x^2 + x, x^3 + x^2 + 1 } \ J + x^2 + 1 &= { x^2 + 1, x^3 + x^2 + x } \ J + x^2 + x + 1 &= { x^2 + x + 1, x^3 + x^2 } \end{align*}

Q6

sage: x = PolynomialRing(IntegerModRing(3, is_field=True), 'x').gen()
sage: rem = lambda px: px % (x^3 + x^2 + 2)
sage: rem(x), rem(2*x)
(x, 2*x)
sage: rem(x^2)
x^2
sage: rem(x^2 + x), rem(x^2 + 2*x)
(x^2 + x, x^2 + 2*x)
sage: rem(x^2 + 1), rem(x^2 + 2)
(x^2 + 1, x^2 + 2)
sage: rem(x^2 + x + 1)
x^2 + x + 1
sage: rem(x^3)
2*x^2 + 1
sage: rem(x^3), rem(x^3 + 1), rem(x^3 + 2)
(2*x^2 + 1, 2*x^2 + 2, 2*x^2)
sage: rem(x^3 + x), rem(x^3 + 2*x)
(2*x^2 + x + 1, 2*x^2 + 2*x + 1)
sage: rem(x^3 + x + 1), rem(x^3 + x + 2)
(2*x^2 + x + 2, 2*x^2 + x)
sage: rem(x^3 + 2*x + 1), rem(x^3 + 2*x + 2)
(2*x^2 + 2*x + 2, 2*x^2 + 2*x)
sage: rem(x^3 + x^2), rem(x^3 + 2*x^2)
(1, x^2 + 1)
sage: rem(x^3 + x^2 + 1), rem(x^3 + x^2 + 2)
(2, 0)
sage: rem(x^3 + 2*x^2 + 1), rem(x^3 + 2*x^2 + 2)
(x^2 + 2, x^2)
sage: rem(x^3 + x^2 + x), rem(x^3 + x^2 + 2*x)
(x + 1, 2*x + 1)
sage: rem(x^3 + 2*x^2 + x), rem(x^3 + 2*x^2 + 2*x)
(x^2 + x + 1, x^2 + 2*x + 1)
sage: rem(x^3 + x^2 + x + 1), rem(x^3 + x^2 + 2*x + 2)
(x + 2, 2*x)
sage: rem(x^3 + 2*x^2 + x + 1), rem(x^3 + 2*x^2 + 2*x + 2)
(x^2 + x + 2, x^2 + 2*x)

\begin{align*} J &= { 0, x^3 + x^2 + 2, 2x^3 + 2x^2 + 1 } \ J + 1 &= { 1, x^3 + x^2, 2x^3 + 2x^2 + 2 } \ J + 2 &= { 2, x^3 + x^2 + 1, 2x^3 + 2x^2 } \ J + x &= { x, x^3 + x^2 + x + 2, 2x^3 + 2x^2 + x + 1 } \ J + x + 1 &= { x + 1, x^3 + x^2 + x, 2x^3 + 2x^2 + x + 2 } \ J + x + 2 &= { x + 2, x^3 + x^2 + x + 1, 2x^3 + 2x^2 + x } \ J + 2x &= { 2x, x^3 + x^2 + 2x + 2, 2x^3 + 2x^2 + 2x + 1 } \ J + 2x + 1 &= { 2x + 1, x^3 + x^2 + 2x, 2x^3 + 2x^2 + 2x + 2 } \ J + 2x + 2 &= { 2x + 2, x^3 + x^2 + 2x + 1, 2x^3 + 2x^2 + 2x } \ ... \end{align*}

D. Short Questions Relating of Field Extensions

Q1

$c$ 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 $p(x) = \bar{p}(x)$.

Q6

sage: (x^5 + 2*x^3 + 4*x^2 + 6).find_root(-100,100)
-1.5236546776809101

$\mathbb{Z}(-1.5236546776809101)$

Q7

$$a = 1 \pm i$$ $$(a - 1)^2 = (\pm i)^2$$ $$a^2 - 2a + 1 = -1$$ $$a^2 - 2a + 2 = 0$$ $$\implies \mathbb{Q}(1 + i) \cong \mathbb{Q}(1 - i)$$

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 $\sqrt{3}$.

Q8

$$\frac{F[x]}{\langle p(x) \rangle} \cong F(\alpha)$$

$$(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta$$ 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 $$F[x] / \langle p(x + c) \rangle \cong F[x] / \langle p(x) \rangle$$

Q5

$$F(a) = F(ca)$$ $$F[x] / \langle p(cx) \rangle \cong F[x] / \langle p(x) \rangle$$

Q6

a.

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 $p(x)$.

b.

$$p(x) = x^2 - 2, p(x - 2) = x^2 - 4x + 2$$

c.

$$p(x) = x^2 - 2, p(2x) = 4(x^2 - 1/2)$$

F. Quadratic Extensions

Q1

$$x^2 + bx + c = 0$$ $$(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c = 0$$ $$x = \pm \sqrt{(\frac{b}{2})^2 - c} - \frac{b}{2}$$

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 $p(cx)$.

Q3

$g(x) = p(cx), g(\sqrt{b}) = 0 \implies F(\sqrt{b}) \cong F[x] / \langle g(x) \rangle \implies F(\sqrt{b}) \cong F[x] / \langle p(cx) \rangle$, but $F[x] / \langle p(cx) \rangle \cong F[x] / \langle p(x) \rangle$ and $F(\sqrt{a}) \cong F[x] / \langle p(x) \rangle \implies F(\sqrt{a}) = F(\sqrt{b})$.

Q4

$F(\sqrt{a}) \cong F(\sqrt{b}) \implies$ 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.

I. Derivatives and Their Properties

Q1

\begin{align*} [a(x) + b(x)]' &= [a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_nx^n + b_0 + b_1x + b_2x^2 + b_3x^3 + \cdots + b_n x^n]' \ &= a_1 + b_1 + 2 a_2x + 2 b_2x + 3a_3 x^2 + 3b_3 x^2 + \cdots + na_n x^{n - 1} + nb_n x^{n - 1} \ \end{align*} $$[a(x) + b(x)]' = a'(x) + b'(x)$$

Q2

$$a(x)b(x) = a_0 b_0 + (a_0 b_1 + b_0 a_1) x + (a_0 b_2 + a_1 b_1 + a_2 b_0)x^2 + \cdots + a_n b_n x^{2n}$$

\begin{align*} [a(x)b(x)]' &= (a_0 b_1 + b_0 a_1) + 2(a_0 b_2 + a_1 b_1 + a_2 b_0) x + \cdots + 2n a_n b_n x^{2n - 1} \ &= c_0 + c_1 x + \cdots + c_{2n - 1} x^{2n - 1} \end{align*} $$\text{where } c_k = \sum_{i + j = k + 1}[(k + 1)(a_i + b_j)] = (k + 1)\sum_{i + j = k + 1}(a_i + b_j)$$ Now by definition we have $a'(x) = a_1 + 2a_2 x + \cdots + na_n x^{n - 1}$ and likewise for $b(x)$ giving us \begin{align*} a'(x)b(x) &= a_1 b_0 + (a_1 b_1 + 2a_2 b_0) x + \cdots + na_n b_n x^{2n - 1} \ &= d_0 + d_1 x + \cdots + d_{2n - 1}x^{2n - 1} \end{align*} $$d_k = \sum_{(i - 1) + j = k} i a_i b_j$$ \begin{align*} a(x)b'(x) &= a_0 b_1 + (a_1 b_1 + 2a_0 b_2) x + \cdots + na_n b_n x^{2n - 1} \ &= e_0 + e_1 x + \cdots + e_{2n - 1}x^{2n - 1} \end{align*} $$e_k = \sum_{i + (j - 1) = k} j a_i b_j$$ $$a'(x)b(x) + a(x)b'(x) = (a_0 b_1 + b_0 a_1) + 2(a_0 b_2 + a_1 b_1 + a_2 b_0)x + \cdots + 2na_n b_n x^{2n - 1} = \sum_{k = 0}^{2n - 1} (d_k + e_k)x^k$$ \begin{align*} d_k + e_k &= \sum_{(i - 1) + j = k} i a_i b_j + \sum_{i + (j - 1) = k} j a_i b_j \ &= \sum_{i + j = k + 1} (i + j)(a_i + b_j) \ &= (k + 1) \sum_{i + j = k + 1} (a_i + b_j) \ &= c_k \end{align*}

Q3

$$a(x) = a_0 + a_1x + a_2x^2 + \cdots + a_n x^n$$ $$ka(x) = ka_0 + ka_1x + ka_2x^2 + \cdots + ka_n x^n$$ $$[ka(x)]' = ka_1 + k2a_2 x + \cdots + kna_n x^{n - 1}$$

$$a'(x) = a_1 + 2a_2 x + \cdots + na_n x^{n - 1}$$ $$ka'(x) = ka_1 + k2a_2 x + \cdots + kna_n x^{n - 1}$$

Q4

There does not exist an $n \in \mathbb{Z}$ such that $n \cdot 1 = 0$, so $k a_k x^{k - 1}$ for values of $k \geq 0$ can only be zero when $k = 0$.

Otherwise if the characteristic is nonzero then two positive values in the ring can be 0 and the above does not hold.

Q5

$$[x^6 + 2x^3 + x + 1]' = x^6 + x^2 + 1$$ $$[x^5 + 3x^2 + 1]' = x$$ $$[x^15 + 3x^10 + 4x^5 + 1]' = 0$$

Q6

$\char F = 0 \implies p \cdot 1 = 0 \implies \forall a \in F, p \cdot a = 0$. 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 $a_{mp} x^{mp}$.

J. Multiple Roots

Q1

$a(x) = (x - c)^m$ for some $m &gt; 1 \implies a(x) = (x - c)^2[(x - c)^{m - 2}q(x)] = (x - c)^2 q'(x)$. Since $c \in K$, thus $a(x) \in K[x]$.

Q2

\begin{align*} a(x) &= (x^2 - 2cx + c^2)q(x) \ &= x^2 q(x) - 2cxq(x) + c^2 q(x) \ a'(x) &= 2xq(x) + x^2 q'(x) - 2cq(x) - 2cxq'(x) + c^2q'(x) \end{align*}

Q3

\begin{align*} a'(x) &= 2q(x)(x - c) + q'(x)(x - c)^2 \ &= (x - c)[2q(x) + q'(x)(x - c)] \end{align*}

Thus $a(x)$ and $a'(x)$ share a common factor in $F[x]$.

Q4

\begin{align*} {(x - c_1)[(x - c_2) \cdots (x - c_n)]}' &= (x - c_1)'[(x - c_2) \cdots (x - c_n)] + (x - c_1)[(x - c_2) \cdots (x - c_n)]' \ &= (x - c_2)\cdots (x - c_n) + (x - c_1)[(x - c_2)'(x - c_3) \cdots (x - c_n) + (x - c_2)[(x - c_3) \cdots (x - c_n)]'] \ &= (x - c_2) \cdots (x - c_n) + (x - c_1)(x - c_3) \cdots (x - c_n) + (x - c_1)(x - c_2)[(x - c_3)'(x - c_4) \cdots (x - c_n) + (x - c_3)[(x - c_4) \cdots (x - c_n)]'] \ &= (x - c_2) \ cdots (x - c_n) + (x - c_1)(x - c_3) \cdots (x - c_n) + (x - c_1)(x - c_2)(x - c_4) \cdots (x - c_n) + (x - c_1) \cdots (x - c_{i - 1})(x - c_{i + 1}) \cdots (x - c_n) + (x - c_1) \cdots (x - c_{n - 1}) \end{align*}

Q5

$a(x)$ does not have multiple roots and no term in $a'(x)$ repeats.

Q6

No common roots, hence no common factors.

Q7

Using polynomial long division, we see the derivatives do not factor the equations:

$$(2x - 8) \nmid (x^2 - 8x + 8)$$

$$(x + 3) \nmid (x^2 + x + 1)$$

$$2x^{99} \nmid x^{100} - 1$$