Polynomial Rings

10.1 Definition and Basic Properties

The polynomial ring $R[x]$ consists of all formal sums $\sum_{i=0}^{n} a_i x^i$ with $a_i \in R$ . It is a ring under the usual addition and multiplication of polynomials.

Theorem 10.1 (Division Algorithm). If $F$ is a field and $f, g \in F[x]$ with $g \neq 0$ Then There exist unique $q, r \in F[x]$ such that $f = qg + r$ with $\deg(r) \lt \deg(g)$ or $r = 0$ .

Theorem 10.2 (Factor Theorem). $a \in F$ is a root of $f \in F[x]$ if and only if $(x - a)$ divides $f$ .

Proposition 10.3. A polynomial of degree $n$ over a field has at most $n$ roots (counting multiplicity).

10.2 Irreducible Polynomials

A non-constant polynomial $f \in F[x]$ is irreducible if it cannot be factored as $f = gh$ With both $g$ and $h$ of degree less than $\deg(f)$ .

Proposition 10.4. Every polynomial in $F[x]$ factors uniquely into irreducible polynomials (up to Reordering and multiplication by units).

Theorem 10.5 (Eisenstein”s Criterion). Let $f(x) = a_n x^n + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]$ . If there exists a prime $p$ such that:

$p$ divides $a_0, a_1, \ldots, a_{n-1}$ .
$p$ does not divide $a_n$ .
$p^2$ does not divide $a_0$ .

Then $f$ is irreducible in $\mathbb{Q}[x]$ .

Proof. Suppose $f = gh$ with $g, h \in \mathbb{Z}[x]$ (by Gauss’s lemma), $\deg(g) = r \geq 1$ $\deg(h) = s \geq 1$ , $r + s = n$ . Modulo $p$ : $\bar{f} = \bar{g}\bar{h} = \bar{a}_n x^n$ in $(\mathbb{Z}/p\mathbb{Z})[x]$ . Since $\mathbb{Z}/p\mathbb{Z}[x]$ is an integral domain, $\bar{g} = bx^r$ And $\bar{h} = cx^s$ for some $b, c \in \mathbb{Z}/p\mathbb{Z}$ . In particular, the constant terms of $g$ and $h$ are both divisible by $p$ . But then $p^2$ divides $a_0$ Contradicting condition (3). $\blacksquare$

10.3 Worked Examples

Problem. Show that $x^2 + 1$ is irreducible in $\mathbb{R}[x]$ but reducible in $\mathbb{C}[x]$ .

Solution

Solution. In $\mathbb{R}[x]$ : if $x^2 + 1 = (x + a)(x + b)$ with $a, b \in \mathbb{R}$ Then $a + b = 0$ and $ab = 1$ Giving $-a^2 = 1$ Which has no real solution. So $x^2 + 1$ is irreducible In $\mathbb{R}[x]$ .

In $\mathbb{C}[x]$ : $x^2 + 1 = (x + i)(x - i)$ . $\blacksquare$

Problem. Use the Euclidean algorithm to compute $\gcd(x^4 + x^3 + x^2 + x + 1, x^3 + 1)$ in $\mathbb{Q}[x]$ .

Solution

Solution. Apply the division algorithm:

$x^4 + x^3 + x^2 + x + 1 = x(x^3 + 1) + (x^2 + x + 1)$

$x^3 + 1 = (x - 1)(x^2 + x + 1) + 2$

Since $2$ is a non-zero constant (a unit in $\mathbb{Q}[x]$ ), the polynomials are coprime: $\gcd(x^4 + x^3 + x^2 + x + 1, x^3 + 1) = 1$ . $\blacksquare$

Problem. Show that $f(x) = x^5 - 5x + 3$ is irreducible in $\mathbb{Q}[x]$ .

Solution

Solution. By the rational root theorem, possible rational roots are $\pm 1, \pm 3$ . $f(1) = -1$ , $f(-1) = 7$ , $f(3) = 216$ , $f(-3) = -269$ . No rational roots.

Since $\deg(f) = 5$ If $f$ is reducible, it must have an irreducible factor of degree $1$ or $2$ . No degree- $1$ factor means no rational root. We check for degree- $2$ factors by reducing modulo $2$ : $\bar{f} = x^5 + x + 1$ in $\mathbb{F}_2[x]$ . $\bar{f}(0) = 1$ , $\bar{f}(1) = 1$ So no roots in $\mathbb{F}_2$ . The only irreducible quadratic in $\mathbb{F}_2[x]$ is $x^2 + x + 1$ . Division gives $x^5 + x + 1 = (x^2+x+1)(x^3 + x^2) + 1$ So $x^2 + x + 1$ does not divide $\bar{f}$ .

Thus $f$ has no factor of degree $1$ or $2$ So $f$ is irreducible in $\mathbb{Q}[x]$ . $\blacksquare$

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