Unit 8: Polynomial Rings, Irreducibility Criteria & Field Extensions

1. Polynomial Rings over a Commutative Ring

Let $R$ be a commutative ring with unity $1_R$.

Definition 8.1 (Polynomial Ring $R[x]$): The polynomial ring in one indeterminate $x$ over $R$, denoted $R[x]$, is the set of all formal expressions:

where $a_i \in R$ and $n \ge 0$. If $a_n \ne 0$, then $n$ is called the degree of $f(x)$, denoted $\deg(f) = n$, and $a_n$ is the leading coefficient. If $a_n = 1$, $f(x)$ is called monic. The zero polynomial $0$ has $\deg(0) = -\infty$.

Degree Properties in an Integral Domain:

If $R$ is an integral domain, then for any non-zero $f, g \in R[x]$:

Moreover, the units of $R[x]$ are precisely the units of the coefficient ring: $R[x]^\times = R^\times$.

2. The Division Algorithm in $F[x]$ over a Field

When the coefficient ring is a field $F$, every non-zero element has a multiplicative inverse. This guarantees that long division of polynomials can always be performed without fractions in the coefficients.

Theorem 8.1 (Division Algorithm for Polynomials): Let $F$ be a field, and let $f(x), g(x) \in F[x]$ with $g(x) \ne 0$. There exist unique polynomials $q(x), r(x) \in F[x]$ (quotient and remainder) such that:

Complete Proof:

1. Existence:

Consider the set of all remainders:

• If $0 \in S$, then $f(x) = q(x) g(x)$ with $r(x) = 0$.
• Otherwise, consider the non-negative integer degrees of non-zero elements in $S$:

By the Well-Ordering Principle of $\mathbb{Z}_{\ge 0}$, $\Delta$ has a minimal element. Let $r(x) = f(x) - q(x) g(x) \in S$ have minimal degree. We claim $\deg(r) < \deg(g)$. Suppose for contradiction that $\deg(r) = m \ge n = \deg(g)$. Let $r(x) = a_m x^m + \dots + a_0$ and $g(x) = b_n x^n + \dots + b_0$ (with $b_n \ne 0$). Because $F$ is a field, $b_n^{-1} \in F$. Construct the polynomial:

The leading term of $(a_m b_n^{-1} x^{m-n}) g(x)$ is precisely $a_m b_n^{-1} b_n x^m = a_m x^m$, which cancels the leading term of $r(x)$! Thus $\deg(r_1) < m = \deg(r)$. Notice $r_1(x) = f(x) - [q(x) + a_m b_n^{-1} x^{m-n}] g(x) \in S$, which contradicts the minimality of $\deg(r)$! Therefore, $\deg(r) < \deg(g)$.

2. Uniqueness:

Suppose $f(x) = q_1(x) g(x) + r_1(x) = q_2(x) g(x) + r_2(x)$. Then:

If $q_1(x) \ne q_2(x)$, the left side has degree:

However, the right side has $\deg(r_2 - r_1) \le \max(\deg(r_2), \deg(r_1)) < \deg(g)$, which is impossible! Therefore, $q_1(x) = q_2(x)$, which forces $r_1(x) = r_2(x)$. $\blacksquare$

3. $F[x]$ is a Euclidean Domain and a PID

Theorem 8.2 ($F[x]$ is an ED, PID, and UFD): For any field $F$:

1. $F[x]$ is a Euclidean Domain with norm $d(f) = \deg(f)$.
2. $F[x]$ is a Principal Ideal Domain (PID). Every ideal is generated by a unique monic polynomial.
3. $F[x]$ is a Unique Factorization Domain (UFD). Every non-zero polynomial factors uniquely into irreducible polynomials up to non-zero constant multiples.

1. Evaluation Homomorphism and Roots

Let $F$ be a field and let $\alpha \in F$. The map:

is a surjective ring homomorphism called the evaluation homomorphism at $\alpha$.

Definition 8.2 (Root): An element $\alpha \in F$ is called a root (or zero) of $f(x) \in F[x]$ if $f(\alpha) = 0$, or equivalently, $f(x) \in \ker(\text{ev}_\alpha)$.

2. Remainder and Factor Theorems

Theorem 8.3 (Polynomial Remainder Theorem): If $f(x) \in F[x]$ is divided by the linear polynomial $(x - \alpha)$, the remainder is the constant $f(\alpha)$:

Proof:

By the Division Algorithm (Theorem 8.1), divide $f(x)$ by $(x - \alpha)$:

where $\deg(r) < \deg(x - \alpha) = 1$. Thus $r(x)$ is a constant $c \in F$. Evaluating both sides at $x = \alpha$:

Thus $r(x) = f(\alpha)$. $\blacksquare$

Corollary 8.3.1 (Factor Theorem): An element $\alpha \in F$ is a root of $f(x) \in F[x]$ if and only if $(x - \alpha)$ divides $f(x)$ in $F[x]$:

3. Bound on the Number of Roots

Theorem 8.4 (Number of Roots Bound): Let $F$ be a field, and let $f(x) \in F[x]$ be a non-zero polynomial of degree $n = \deg(f) \ge 0$. Then $f(x)$ has at most $n$ distinct roots in $F$.

Complete Proof (by induction on $n$):

• Base Case ($n = 0$): $f(x) = c \ne 0$ is a non-zero constant. It has $0$ roots. Since $0 \le 0$, the base case holds.
• Inductive Step: Assume the theorem holds for all polynomials of degree $n - 1$.

Let $\deg(f) = n \ge 1$.

• If $f(x)$ has no roots in $F$, then $0 \le n$ holds trivially.
• If $f(x)$ has a root $\alpha \in F$, then by the Factor Theorem:

Let $\beta \in F$ be any root of $f(x)$ distinct from $\alpha$ ($\beta \ne \alpha$). Then:

Because $F$ is a field (no zero divisors) and $\beta - \alpha \ne 0$, we must have:

Thus, every root of $f(x)$ other than $\alpha$ must be a root of $q(x)$. By our induction hypothesis, $q(x)$ has at most $n - 1$ roots in $F$. Therefore, the total number of roots of $f(x)$ is at most $1 + (n - 1) = n$. $\blacksquare$

1. Primitive Polynomials and Content

Let $f(x) = a_n x^n + \dots + a_1 x + a_0 \in \mathbb{Z}[x]$ be a polynomial with integer coefficients.

Definition 8.3 (Content): The content of $f(x)$, denoted $C(f)$, is the greatest common divisor of all its coefficients:

If $C(f) = 1$, $f(x)$ is called a primitive polynomial.

Every non-zero polynomial $f(x) \in \mathbb{Z}[x]$ can be factored uniquely as:

where $f^*(x)$ is primitive.

2. Gauss's Lemma

Theorem 8.5 (Gauss's Lemma): The product of two primitive polynomials in $\mathbb{Z}[x]$ is primitive:

Complete Proof:

Let $f(x) = \sum_{i=0}^n a_i x^i$ and $g(x) = \sum_{j=0}^m b_j x^j$ be primitive polynomials. Suppose for contradiction that $f(x) g(x)$ is NOT primitive. Then there exists a prime number $p$ such that $p \mid C(fg)$, meaning $p$ divides every coefficient of $fg$. Consider the canonical reduction homomorphism modulo $p$:

Since $p$ divides every coefficient of $f(x) g(x)$:

Since $\pi_p$ is a ring homomorphism:

Now, $\mathbb{Z}_p$ is a field (since $p$ is prime), so $\mathbb{Z}_p[x]$ is an integral domain! An integral domain has no zero divisors! Therefore:

• If $\pi_p(f(x)) = 0$, then $p \mid a_i$ for all $i$, meaning $p \mid C(f) = 1$, a contradiction!
• If $\pi_p(g(x)) = 0$, then $p \mid b_j$ for all $j$, meaning $p \mid C(g) = 1$, a contradiction!

Therefore, no such prime $p$ can divide all coefficients of $f(x) g(x)$. Hence $C(f \cdot g) = 1$. $\blacksquare$

3. The Equivalence of Irreducibility over $\mathbb{Z}$ and $\mathbb{Q}$

Theorem 8.6 (Gauss's Irreducibility Equivalence): Let $f(x) \in \mathbb{Z}[x]$ be a primitive polynomial. Then $f(x)$ is irreducible in $\mathbb{Z}[x]$ if and only if $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Complete Proof:

• $(\impliedby)$ If $f(x)$ is irreducible in $\mathbb{Q}[x]$, any non-trivial factorization in $\mathbb{Z}[x]$ would also be a factorization in $\mathbb{Q}[x]$, which is impossible.
• $(\implies)$ Suppose $f(x)$ is reducible in $\mathbb{Q}[x]$.

Then $f(x) = A(x) B(x)$ with $A(x), B(x) \in \mathbb{Q}[x]$ and $\deg(A), \deg(B) \ge 1$. Clear denominators by writing $A(x) = \frac{a}{b} A_0(x)$ and $B(x) = \frac{c}{d} B_0(x)$ where $A_0, B_0 \in \mathbb{Z}[x]$ are primitive, and $a, b, c, d \in \mathbb{Z}^+$. Then:

Take the content of both sides:

Therefore, $bd = ac$, which means $\frac{ac}{bd} = 1$! Substituting back:

with $A_0(x), B_0(x) \in \mathbb{Z}[x]$ and $\deg(A_0), \deg(B_0) \ge 1$. Thus $f(x)$ is reducible in $\mathbb{Z}[x]$. Taking the contrapositive yields the theorem. $\blacksquare$

1. Statement of Eisenstein's Criterion

Ferdinand Eisenstein published this celebrated irreducibility test in 1850.

Theorem 8.7 (Eisenstein's Irreducibility Criterion): Let $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \in \mathbb{Z}[x]$ with $n \ge 1$. If there exists a prime number $p$ such that:

1. $p \nmid a_n$ ($p$ does not divide the leading coefficient),
2. $p \mid a_i$ for all $0 \le i \le n - 1$ ($p$ divides every other coefficient),
3. $p^2 \nmid a_0$ ($p^2$ does not divide the constant term),

then $f(x)$ is irreducible over the field of rational numbers $\mathbb{Q}[x]$. (If $f(x)$ is also primitive, it is irreducible in $\mathbb{Z}[x]$.)

2. Complete Rigorous Proof

Assume for contradiction that $f(x)$ is reducible in $\mathbb{Q}[x]$. By Theorem 8.6 (Gauss's Theorem), $f(x)$ must factor in $\mathbb{Z}[x]$ into polynomials of lower degree:

where:

with $r + s = n$.

1. Constant Term Analysis:

The constant term of $f(x)$ is $a_0 = b_0 \cdot c_0$. By condition (2), $p \mid a_0$. Thus $p \mid (b_0 c_0)$. Since $p$ is prime, $p \mid b_0$ or $p \mid c_0$. However, by condition (3), $p^2 \nmid a_0$. Therefore, $p$ cannot divide both $b_0$ and $c_0$! Without loss of generality, assume:

2. Leading Coefficient Analysis:

The leading coefficient of $f(x)$ is $a_n = b_r \cdot c_s$. By condition (1), $p \nmid a_n$, so $p \nmid b_r$ and $p \nmid c_s$.

3. The Minimal Index of Non-Divisibility:

Since $p \mid b_0$ and $p \nmid b_r$, there must exist a smallest index $k \in \{1, 2, \dots, r\}$ such that:

Notice $k \le r < n$.

4. Examine the Coefficient $a_k$:

By polynomial multiplication, the coefficient of $x^k$ in $g(x) h(x)$ is:

Rearranging for $b_k c_0$:

Now evaluate divisibility by $p$:

• Since $k < n$, condition (2) states that $p \mid a_k$.
• By our choice of $k$, $p \mid b_j$ for all $j < k$, so $p$ divides every term in the bracketed sum:

Therefore, $p$ must divide the right-hand side, which implies:

Since $p$ is prime, this requires $p \mid b_k$ or $p \mid c_0$. However, by definition of $k$, $p \nmid b_k$, and by Step 1, $p \nmid c_0$! This is a complete contradiction!

Therefore, $f(x)$ cannot be factored into polynomials of lower degree. Hence $f(x)$ is irreducible in $\mathbb{Q}[x]$. $\blacksquare$

3. Application to Cyclotomic Polynomials

For any prime $p$, the $p$-th cyclotomic polynomial is:

Theorem 8.8 (Irreducibility of $\Phi_p(x)$): For every prime number $p$, the cyclotomic polynomial $\Phi_p(x)$ is irreducible over $\mathbb{Q}$.

Proof (via the shift $x \mapsto x + 1$):

Substitute $x = y + 1$:

Expanding $(y + 1)^p$ using the Binomial Theorem:

Dividing by $y$:

Now apply Eisenstein's Criterion with the prime $p$:

1. Leading coefficient is $1$, and $p \nmid 1$.
2. For all $1 \le k \le p - 1$, the binomial coefficients $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ are divisible by $p$.
3. Constant term is $\binom{p}{p-1} = p$. Clearly $p^2 \nmid p$.

By Theorem 8.7, $\Phi_p(y + 1)$ is irreducible over $\mathbb{Q}$. Since the substitution $x \mapsto y+1$ is an invertible ring automorphism of $\mathbb{Q}[x]$, $\Phi_p(x)$ is irreducible over $\mathbb{Q}$. $\blacksquare$

1. Field Extensions and Degree

Definition 8.4 (Field Extension): If $F$ and $K$ are fields with $F \subseteq K$, we say $K$ is an extension field of $F$ (written $K/F$). The larger field $K$ can be viewed as a vector space over the smaller field $F$, where addition is addition in $K$ and scalar multiplication is multiplication by elements of $F$. The dimension of $K$ as an $F$-vector space is called the degree of the extension, denoted:

If $[K : F] < \infty$, the extension is called finite.

2. Algebraic vs. Transcendental Elements

Definition 8.5 (Algebraic Element): Let $K/F$ be an extension and $\alpha \in K$.

1. $\alpha$ is algebraic over $F$ if there exists a non-zero polynomial $f(x) \in F[x]$ such that $f(\alpha) = 0$.
2. Otherwise, $\alpha$ is transcendental over $F$ (e.g., $\pi$ and $e$ over $\mathbb{Q}$).

Definition 8.6 (Minimal Polynomial): If $\alpha \in K$ is algebraic over $F$, the unique monic polynomial $m_{\alpha, F}(x) \in F[x]$ of minimal degree having $\alpha$ as a root is called the minimal polynomial of $\alpha$ over $F$.

• $m_{\alpha, F}(x)$ is irreducible over $F$.
• For any $g(x) \in F[x]$, $g(\alpha) = 0 \iff m_{\alpha, F}(x) \mid g(x)$.
• The simple extension field is isomorphic to:

and its degree is $[F(\alpha) : F] = \deg(m_{\alpha, F})$. A basis is $\{1, \alpha, \alpha^2, \dots, \alpha^{n-1}\}$.

3. The Tower Law (Degree Multiplication Formula)

The Tower Law is the field-theoretic analogue of Lagrange's Theorem for groups.

Theorem 8.9 (The Tower Law): Let $F \subseteq L \subseteq K$ be a tower of field extensions. Then $K/F$ is finite if and only if both $K/L$ and $L/F$ are finite, and in that case:

``` K | [K : L] = n L | [L : F] = m F ---------------- [K : F] = m * n ```

Complete Constructive Proof:

Let $[L : F] = m$ and $[K : L] = n$. Let $\mathcal{B}_L = \{u_1, u_2, \dots, u_m\}$ be a basis of $L$ as an $F$-vector space. Let $\mathcal{B}_K = \{v_1, v_2, \dots, v_n\}$ be a basis of $K$ as an $L$-vector space. Consider the set of all $m \cdot n$ products:

We will prove that $\mathcal{S}$ is a basis of $K$ over $F$.

Step 1: $\mathcal{S}$ spans $K$ over $F$

Let $\theta \in K$. Since $\mathcal{B}_K$ is a basis for $K$ over $L$, there exist coefficients $\lambda_j \in L$ such that:

Since each $\lambda_j \in L$ and $\mathcal{B}_L$ is a basis for $L$ over $F$, each $\lambda_j$ can be written as:

Substituting this back into $\theta$:

Since $a_{ij} \in F$, this proves that $\theta$ is an $F$-linear combination of the elements in $\mathcal{S}$. Thus $\text{span}_F(\mathcal{S}) = K$.

Step 2: $\mathcal{S}$ is linearly independent over $F$

Suppose there exist scalars $c_{ij} \in F$ such that:

Regrouping terms:

Let $\mu_j = \sum_{i=1}^m c_{ij} u_i \in L$. Then $\sum_{j=1}^n \mu_j v_j = 0$. Since the set $\{v_1, \dots, v_n\}$ is linearly independent over $L$, all coefficients $\mu_j$ must be zero:

Now, each equation $\sum_{i=1}^m c_{ij} u_i = 0$ is a linear combination of $\{u_1, \dots, u_m\}$ with coefficients $c_{ij} \in F$. Since $\{u_1, \dots, u_m\}$ is linearly independent over $F$, all coefficients must be zero:

Therefore, $\mathcal{S}$ is linearly independent over $F$.

Since $\mathcal{S}$ spans $K$ over $F$ and is linearly independent over $F$, it is a basis for $K$ over $F$. The number of elements in $\mathcal{S}$ is $m \cdot n$. Therefore: