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B. Short Questions Relating to Irreducible Polynomials
Q1
Polynomials of degree 1 in a field are irreducible. There are no zero divisors. Multiplying constant terms produces another constant term. Multiplying factors with degrees $m$ and $n$ respectively will produce a new factor with degree $m + n$, and since there are no zero divisors the product of two non-zero coefficients can never be zero.
Q2
If $a(x)$ and $b(x)$ have different degrees they cannot be associates since there are no zero divisors in a field and so no constant term can cancel factors in a polynomial.
Assume $a(x)$ and $b(x)$ have the same degree. Then their leading terms are both $x^n$. If there is some constant term $c > 1$ such that $a(x) = c \cdot b(x)$ then the leading term of $c \cdot b(x)$ will be $c x^n$ which is not the leading term in $a(x)$. Hence there is a contradiction and both polynomials are not associates.
Q3
$a(x)$ and $b(x)$ are distinct so $a(x) \neq b(x)$, and since $a(x)$ and $b(x)$ cannot be factored, then they share no factors and are relatively prime.
Q4
An associate of $a(x)$ is $b(x) = ka(x)$, and since $a(x)$ is irreducible, so is $b(x)$, hence all associates of $a(x)$ are irreducible if $a(x)$ is irreducible.
Q5
In a field, no constants multiplied can equal 0, so there is no $ka(x) = 0$.
Q6
$$|\mathbb{Z}_p| = \phi(p) = p - 1$$
There are $p - 1$ possible values for $k$ and hence $p - 1$ associates of non-zero $a(x)$.
Q7
Polynomial factors for $x^2 + 1$ are lower degree since it's reducible in $\mathbb{Z}_p$.
$a(x) = x^2 + 1 = kp_1(x)p_2(x)$ where $p_1(x)$ and $p_2(x)$ are monic irreducible polynomials.
$$x^2 + 1 = k(x + a)(x + b) \text{ in } \mathbb{Z}_p[x]$$
$$a + b \equiv 0 \mod p$$
$$ab \equiv 1 \mod p$$
Since $a, b \in \mathbb{Z}_p$, then $p = a + b$ (otherwise $a = ka_1, b = kb_1$ and $p = a_1 + b_1$)
C. Number of Irreducible Quadratics over a Finite Field
Q1
By theorem 4, each factorization is unique so $(x + a_1)(x + b_1) \neq (x + a_2)(x + b_2)$ for all $a_1, b_1, a_2, b_2 \in \mathbb{Z}_5$.
$(x + a)^2$, there are 5 possible values and $\binom{5}{2} = 10$, so there are 10 combinations for $(x + a)(x + b)$ (disregarding order since $\mathbb{Z}_5$ is commutative). There are 15 reducible monic quadratics in $\mathbb{Z}_5[x]$.
Q2
There are 15 reducible monics, and 0 is not an associate, 1 is unity, leaving $ka(x)$ different associates where $k \in { 2, 3, 4 }$ and $|k| = 5 - 2 = 3$.
Therefore the total number of reducible quadratics is $4 \times 15 = 60$.
The total number of quadradics in $\mathbb{Z}_5[x]$ is $5^3 = 125$ from all the possible combos of $ax^2 + bx + c$.
$a(x) \in J$ and $J = \langle m(x) \rangle \implies a(x) = b(x) m(x)$ for some $b(x) \in F[x]$
Q3
$$a(x)b(x) = kp_1(x)\cdots p_r(x)$$
$J = \langle m(x) \rangle$ where $m(x)$ is irreducible. Since $J$ is an ideal $a(x)b(x) \in J \implies a(x)b(x) = c(x)m(x)$ for some $c(x) \in F[x]$. But $m(x)$ is irreducible so $m(x) = p_i(x)$ where i is one of the factors from 1 to $r$.
Likewise if $J$ is a prime ideal then $a(x)b(x) \in J \implies a(x) \in J$ or $b(x) \in J$. Say $a(x) \in J$, then $a(x) = kp_1(x) \cdots p_r(x) \implies p_i(x) \in J$ for some $i \in {1, \dots, r } \implies J = \langle m(x) \rangle$ where $m(x)$ is irreducible.
Q4
$p(x)$ is irreducible.
$$J = \langle p(x) \rangle$$
Let there be an ideal $K$ such that $J \subset K$, and let $a(x) \in K$ such that $a(x) \notin J$.
Thus $a(x)$ is not a multiple of $p(x)$ and they share no common factors.
$$1 = r(x)a(x) + s(x)p(x)$$$$\implies r(x)a(x) = 1 - s(x)p(x)$$$$\implies r(a)a(x) \in J + 1$$
So $a(x)$ is invertible and so $a(x) \in K \implies K = A$.
Q5
Coefficient sum of $x - 1$ is $1 + (-1) = 0$ so $x - 1 \in S$. $x - 1$ is irreducible so it is a maximal ideal and $S = \langle x - 1 \rangle$.
Q6
$$g(a(x)) = g(a_0 + \cdots + a_n x^n) = a_0 + \cdots + a_n$$
$$k(x) \in K = \ker g \implies g(k(x)) = 0$$
From previous question $k = \langle x - 1 \rangle$.
$$ g : F[x] \underset{\langle x - 1 \rangle}{\relbar\joinrel\twoheadrightarrow} F $$
$$ F[x] / \langle x - 1 \rangle \cong F $$
Q7
$$a(x, y) = x \in J, b(x, y) = y \in J$$
But there is no polynomial in $F[x, y]$ such that $a(x, y) = r(x, y)b(x, y)$ and vice versa. Therefore $J$ cannot be principal and $F[x, y]$ can have non-principal ideals.
E. Proof of the Unique Factorization Theorem
Q1
Euclid's lemma for polynomials: let $p(x)$ be irreducible. If $p(x) \mid a(x)b(x)$ then $p(x) \mid a(x)$ or $p(x) \mid b(x)$.
If $p(x) \mid a(x)$ we are done. So lets assume $p(x) \nmid a(x)$. What integers are common divisors?
$p(x)$ is irreducible, so the only divisors are $\pm 1$ and $\pm p(x)$. But $p(x) \nmid a(x)$, so $\pm p(x)$ is not a common divisor of $p(x)$ and $a(x)$. So therefore that leaves $\pm 1$ as their common divisors. So by theorem 2
$$1 = k(x)p(x) + l(x)a(x)$$$$b(x) = k(x)p(x)b(x) + l(x)a(x)b(x)$$
But $p(x) \mid a(x)b(x)$ so there is an $h(x)$ such that $a(x)b(x) = p(x)h(x)$. Therefore
$$k(x)p(x)b(x) + l(x)p(x)h(x) = b(x)$$
that is $p(x)[k(x)b(x) + l(x)h(x)] = b(x)$ or $p(x) \mid b(x)$.
Q2
Grouping terms, we see that $p \mid a_1(x)$ or $p(x) \mid (a_2(x) \cdots a_n(x))$, and if $p(x) \nmid a_1(x)$, then
$p(x) \mid a_2(x)$ or $p(x) \mid (a_3(x) \cdots a_n(x))$ and so on.
For corollary 2, $p(x) \mid q_i(x)$ for one of the factors. Since the factors $q_1(x) \cdots q_r(x)$ are irreducible, so if $p(x) \mid q_i(x)$ for some integer, that means $p(x) = q_i(x)$.
Q3
Cancelling terms from both sides of
$$a(x) = kp_1(x) \cdots p_r(x) = lq_1(x) \cdots q_s(x)$$
Since for all $i \in { 1, \dots, r }, p_i(x) \mid (lq_1(x) \cdots q_s(x))$ then $p_i(x) = q_j(x)$ for some j in ${ 1, \dots, s }$. Cancelling terms we eventually end up with $k = l$.
Injective: $\forall a(x), b(x) \in F[x], h[a(x)] = h[b(x)] \implies a(x) = b(x)$. This is trivially visible.
Surjective: $\forall b(x) \in F[x], \exists a(x) \in F[x]: h[a(x)] = b(x)$. The value of $a(x)$ is simply $b(x)$.
Indeed $h(h(a_0 + a_1 x + \cdots + a_n x^n)) = h(a_n + a_{n - 1} x + \cdots + a_0 x^n) = a_0 + a_1x + \cdots + a_n x^n$. This means that $h \cdot h = \epsilon$.
Q3
Let $a(x) = a_0 + a_1x + \cdots + a_n x^n$ be irreducible but $b(x) = a_n + a_{n-1}x + \cdots + a_0 x^n$ be reducible so that $b(x) = c(x)d(x)$. But then $a(x) = h[b(x)] = h[c(x)d(x)] = h[c(x)]h[d(x)]$ meaning $a(x)$ is in fact reducible.