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
Since $x + 1$ divides both terms, that means it is a divisor of the expression on the left.
Q5
By induction assume $m = k$ is true, then
$$x^{k + 1} + 2 = x(x^k + 2) + (x + 2)$$$(x + 2)$ divides both sides and so is a divisor of $x^{k + 1} + 2$ in $\mathbb{Z}_3[x]$.
Likewise for $\mathbb{Z}_n[x]$
\begin{align*}
x^{k + 1} + (n - 1) &= x(x^k + (n - 1)) + (x + (n - 1)) \
&= x^{k + 1} + (n - 1)x + x + (n - 1) \
&= x^{k + 1} + nx + (n - 1) \
&= x^{k + 1} + (n - 1)
\end{align*}
and so $x + (n - 1) is a factor of $x^{k + 1} + (n - 1)$ in $\mathbb{Z}_n[x]$.
$$(2x + 2)^3 = 8x^3 + 24x^2 + 24x + 8 = 0 \text{ in } \mathbb{Z}_8[x]$$$$\implies 2x + 2 \text{ is a divisor of } 0$$
$$(1 - 4x)(1 + 4x) = 1 - 16x^2 = 1 \text{ in } \mathbb{Z}_8[x]$$$$\implies 1 + 4x \text{ and } 1 - 4x \text{ are invertible elements}$$
Q6
For any polynomial $b(x) \in A[x], \deg b(x) \geq 0$. If $\deg b(x) = 0$ then $xb(x) = 0$ because $b(x) = 0$. Otherwise $\deg [x \cdot b(x)] = \deg x + \deg b(x) = 1 + \deg b(x) \implies \deg [x \cdot b(x)] \geq 1$.
Since $x$ is in every non-zero polynomial domain, this means there are no polynomial fields.
Q7
Take $a(x) = x \in A[x]$, then $\deg a(x) = 1$ and $\deg [(a(x))^2] = 2$. In fact $\deg[(a(x))^n] = n$ in any ring and so there is no polynomial with a nonzero term that multiplied by $x$ produces $0$.
$$x(b_0 + b_1 x + \cdots + b_m x^m)$$
where $b_m \neq 0$ in the ring, then
$$\deg [a(x) \cdot b(x)] = m + 1 \neq 0$$
Q8
Idempotent: $(a(x))^2 = a(x)$
Nilpotent: $(a(x))^n = 0$ for some integer $n$.
Let $a(x) = x$, then $(a(x))^2 = x^2$, so $(a(x))^2 \neq a(x)$ and $a(x)$ is not idempotent.
Also $(a(x))^n = x^n \neq 0$ and so $a(x)$ is not nilpotent.
C. Rings $A[x]$ Where $A$ Is Not an Integral Domain
Q1
An integral domain is a commutative ring with unity having no divisors of 0.
Since $A[x]$ contains the elements from $A$, then if $A$ has zero divisors, so does $A[x]$ and hence $A[x]$ is not an integral domain.
Q2
Degree 0: $2 \times 2 = 0$ in $\mathbb{Z}_4[x]$
Degree 1: $2x \cdot 2x = 0$
Degree 2: $2x^2 \cdot 2x^2 = 0$
Q3
$5x^3(2x + 1) = 0$ in $\mathbb{Z}_10[x]$ lacks the cancellation property whereas the term $5x^3 = 0$ in $\mathbb{Z}_5[x]$ and disappears.
Q4
Any polynomials where the coefficient of the leading term is a multiple of the field size.
Thus the coefficient of x^{n + m} is 0 and so
$$\deg a(x) b(x) < \deg a(x) + \deg b(x)$$
Q6
In an integral domain
$$\deg a(x)b(x) = \deg a(x) + \deg b(x)$$
Non-constant polynomials have a degree greater than one. Let $a(x)$ be such a polynomial, while $b(x)$ is a non-zero polynomial such that $\deg b(x) \geq 1$. Then $\deg a(x)b(x) > 1$, while the degree of 1 is 1. So there are no non-constant invertible polynomials in integral domains.
In $\mathbb{Z}_4[x], (2x + 1)^2 = 1$, so $(2x + 1)$ is invertible and so are all powers of $(2x + 1)^k$ since $(2x + 1)$ is its own inverse.
Any polynomial of the form $1 + 4x + 4x^2 + \cdots + 4x^n$ when squared will equal 1, because every coefficient apart from the constant and leading term is greater than or equal to 2, and $4\times 2 = 8 = 0$, and the leading term is $16x^{2n} = 0$. So there are infinite polynomial square roots in $\mathbb{Z}_8[x]$.
D. Domains $A[x]$ Where $A$ Has Finite Characteristic
Q1
Every coefficient in $A[x]$ is a member of $A$. For all $a(x), b(x) \in A[x], c(x) = a(x) + b(x)$ then $c_i = a_i + b_i$, and therefore the characteristic is preserved since $\underbrace{1_A + 1_A + \cdots + 1_A}_{\char A} = 0$.
Q2
Consider the ring $\mathbb{Z}_n[x]$ of polynomials in one variable $x$ with coefficients in $\mathbb{Z}_n$. It is an infinite ring since $x^m \in \mathbb{Z}_n[x]$ for all positive integers $m$, and $x^{m_1} \neq x^{m_2}$ for $m_1 \neq m_2$. But the characteristic of $\mathbb{Z}_n[x]$ is clearly $n$.
Likewise the above applies for $(p - 1)$ in any domain of characteristic $p$.
Q4
By the cancellation property, the characteristic of every integral domain is prime, since if the characteristic was composite that would imply $rs = 0$ for some $r, s \in A$ which violates the zero divisor rule.
Thus the coefficients for all terms in the expansion $(x + c)^p$ except $x^p$ and $c^p$, by the binomial formula are equal to $\binom{p}{k} = \frac{p!}{k!(p - k)!}$. Since $p$ is prime and indivisible the coefficient becomes zero.
$$(x + c)^p = x^p + c^p$$
Q5
They aren't the same since $x \notin A$, and $\forall a \in A$, $a = a^2$ but $x \neq x^2$.
Q6
It is trivial to see that
\begin{align*}
[a_0 + (a_1x + \cdots + a_n x^n)]^p &= a_0^p + [a_1 x + (a_2 x^2 + \cdots + a_n x^n)]^p \
&= a_0^p + a_1^p x^p + [a_2x^2 + (a_3 x^3 + \cdots + a_n x^n)]^p \
&= a_0^p + a_1^p x_1^p + \cdots + a_n^p x^{np}
\end{align*}
E. Subrings and Ideals in A[x]
Q1
$B[x]$ contains all the polynomials with coefficients in $B$. Since $B$ is a subring of $A$, so $B[x]$ is a subring of $A[x]$.
Q2
Likewise $B$ absorbs all products with $A$, and hence so does $B[x]$,
Q3
Every coefficient $a_i$ with odd $i$ equal to zero, means the polynomial only has non-zero coefficients for even powers.
When adding polynomials, we add the coefficients. So the odd numbered powers remain zero, and even powers remain non-zero.
For multiplying two polynomials $a(x)b(x)$, the corresponding powers of each term are added together, $a_i b_j x^{i + j}$. Since both $i$ and $j$ are even, so is the resulting term and hence the result of $a(x)b(x)$ remains inside the set $S$ making it a subring.
The above statement does not apply when talking about odd non-zero coefficients, since multiplying two odd terms might result in an even power, for example $c(x) = a(x) b(x)$, $a_3 b_5 x^{3 + 5}$.
Q4
Let $b(x) \in A[x]$ and $a(x) \in J$, then the constant term in $b(x)$ is $b_0$. Since $b(x) a(x) = b_0 a(x) + b_1 x a(x) + \cdots + b_m x^n a(x)$, and the powers of all terms in $a(x)$ are $\geq 1$, so $b_0 a(x)$ has no constant term. So $\forall a(x) \in J$ absorbs products from $A[x]$ and is an ideal.
Q5
Let $a(x) = a_0 + a_1x + \cdots + a_n x^n \in J$ and $b(x) = b_0 + b_1x + \cdots + b_m x^m \in A[x]$. Then $a(x)b(x) = a_0 (b_0 + b_1x + \cdots + b_m x^m) + a_1x (b_0 + b_1x + \cdots + b_m x^m) + \cdots + a_n x^n (b_0 + b_1x + \cdots + b_m x^m)$. Then it can be seen plainly that the sum of cofficients for the result is $(a_0 + a_1 + \cdots + a_n)(b_0 + b_1 + \cdots + b_m) = 0$. Therefore $J$ is an ideal of $A[x]$.
Q6
Since $A$ is an integral domain, there are no divisors of zero. Therefore the values cannot be made to equal 0 unless one of the terms is zero. In the case of Q4, the polynomial is an ideal in $J$ with a zero constant coefficient and in Q5, the polynomial can be factorized into a polynomial where one of the terms has coefficients that sum to zero.
Let $a \in A$, then $a(x) \in J + a$, where $J$ is the ideal of $g$ (coefficients that sum to zero). Thus every value in $A$ is an image of an element in $A[x]$ and so $h$ is surjective.
The kernel of $g$ is described in 24E5: let $J$ consist of all the polynomials $a_0 + a_1 x + \cdots + a_n x^n$ in $A[x]$ such that $a_0 + a_1 + \cdots + a_n = 0$.
Since $A$ is an integral domain and there are no zero divisors, then $\ker h = { 0 }$.
Q6
Any polynomial $a(x) = a_0 + a_1 x + \cdots + a_n x^n$ can be produced by $h$ iff $c$ is invertible by setting the input to $a_0 + c^{-1} a_1 x + \cdots + c^{-n} a_n x^n$. Then the output of $h$ on this value will produce $a(x)$. Thus $h$ is an automorphism in this case.
G. Homomorphisms of Polynomial Domains Induced by a Homomorphism of the Ring of Coefficients
$$\forall a_i : 0 \leq i \leq n, a_i \in \ker h$$$$a(x) = a_0 + \cdots + a_n x^n$$
Q3
If $h$ is surjective, then every element of $B$ is of the form $h(a)$ for some $a$ in $A$. Thus, any polynomial with coefficients in $B$ is of the form $h(a_0) + h(a_1)x + \cdots + h(a_n)x^n = \bar{h}(a_0 + a_1 x + \cdots + a_n x^n)$.
Q4
Every coefficient of $A[x]$ maps to a distinct coefficient in $B[x]$ because $h$ is an injective function.
Every coefficient $a_i = qn$ and so $h(a_i) = 0$ because $n \mid a_i$. Thus $\bar{h}(a(x)) = 0$.
Q7
$\mathbb{Z}_n$ where $n$ is prime, means the domain of $\bar{h}$ is an integral domain.
$$\bar{h} : \mathbb{Z}[x] \underset{\ker \bar{h}}{\relbar\joinrel\twoheadrightarrow} \mathbb{Z}_n[x]$$
From 19F2, $J$ is a prime ideal iff $A/J$ is an integral domain. So in our case this means $\ker \bar{h}$ is a prime ideal.
An ideal $J$ of a commutative ring is said to be a prime ideal if for any two elements $a$ and $b$ in the ring,
$$\text{If } ab \in J \text{ then } a \in J \text{ or } b \in J$$
*Prove $A$ is an integral domain $\implies A[x]$ is an integral domain.$
Given any $A_i[x_{i + 1}]$ is an integral domain, we know that the leading term $a_k \neq 0$ (which includes the other non-zero $x$ values), multiplied by another $b_l \neq 0$, and so $a_k b_l \neq 0$ and therefore $A_i[x_{i + 1}]$ has a non-zero coefficient.
Q2
Degree of $p(x, y)$ is the greatest $n$ such that the coefficient $a_n$ is non-zero for the powers $x^i y^j$ such that $i + j = n$.
$A$ is a finite integral domain means it is a field with $\char(A)$ for $1_A$. The unity for $A(x)$ is $[1_A, 1_A]$ and $[a, b] + [c, d] = [ad + bc, bd]$.
$$[1_A, 1_A] + [1_A, 1_A] = [2_A, 1_A]$$$$[k_A, 1_A] + [1_A, 1_A] = [k_A + 1_A, 1_A]$$
\begin{align*}
\underbrace{[1_A, 1_A] + \cdots + [1_A, 1_A]}_{\char(A)} &= [\char(A), 1_A] \
&= [0_A, 1_A]
\end{align*}
Q2
$\mathbb{Z}_p$ is a finite field with characteristic $p$. Therefore the field of quotients $\mathbb{Z}_p(x)$ will have characteristic $p$ yet it is infinite because terms have any positive integer value (and indeed negative since $\mathbb{Z}_p$ has inverses because it is a field).
Because $h$ is isomorphic, each element of $B(x)$ is the image of no more than one element of $A(x)$, so $\bar{h}$ is injective.
Likewise every element of $B(x)$ is the image of an element in $A(x)$, so $\bar{h}$ is surjective.
$\therefore \bar{h}$ is an isomorphism.
J. Division Algorithm: Uniqueness of Quotient and Remainder
In the division algorithm, prove that $q(x)$ and $r(x)$ are uniquely determined. [HINT: Suppose $a(x) =
b(x)q_1 (x) + r_1 (x) = b(x)q_2 (x) + r_2 (x)$, and subtract these two expressions, which are both equal to $a(x)$.]
If $q_1(x) \neq q_2(x)$, then $\deg [q_1(x) - q_2(x)] > 0$ so $\deg [b(x)(q_1(x) - q_2(x)] > 0$.
But the entire expression is 0 and so its degree is zero. Hence $b(x)(q_1(x) - q_2(x))$ cannot have a degree higher than 0 so the term can only equal 0, which means $q_1(x) = q_2(x)$ since $b(x) \neq 0$.