Irreducible polynomials

Overview

Relevant theorems:

• (Theorem) Unique factorization in polynomial rings
• (Theorem) Chinese remainder
• (Theorem) Rational roots test (and generalizations)

Related notes:

• Factorization on polynomial rings
• Factorization on integral domains
• Extension fields

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Basic definition and properties

Irreducible polynomial

A polynomial $p\in F[x]$ is irreducible if it satisfies the following equivalent statements:

• (i) The polynomial $p$ is neither $0$ nor a unit (i.e., $p$ is a non-constant polynomial), and whenever $p=fg$ for some $f,g\in F[x]$, then either $f=c\in F^{\ast }$ (hence $g=c^{-1}p$) or $g=c\in F^{\ast }$ (hence $f=c^{-1}p$).
• (ii) The polynomial $p$ is not a product of two polynomials $fg$ such that both $\mathrm{deg}f<\mathrm{deg}p$ and $\mathrm{deg}g<\mathrm{deg}p$ hold.

Hence an irreducible polynomial is a nonconstant polynomial that does not factor into a product of polynomials with strictly smaller degrees.

Modern Algebra II 2.10: Relation between irreducibility and relatively prime polynomials

Let $p\in F[x]$ be irreducible.

• (i) For any polynomial $f\in F[x]$, either $p|f$ or $p,f$ are relatively prime.
• (ii) For all $f,g\in F[x]$, if $p|fg$ then either $p|f$ or $p|g$. In particular, if $f_1,\dots ,f_n\in F[x]$ and $p|f_1\cdots f_n$, then there exists $1\le i\le n$ such that $p|f_i$.

Proof from Modern Algebra II.

• (i) Suppose $d=\gcd (p,f)$. Then $d|p$, so by definition of an irreducible polynomial, we have either $d\in F^{\ast }$ as a unit or $d=c^{-1}p$, a multiple of $p$ by some unit $c\in F^{\ast }$. In the first case, we can take $d=1$ and conclude that $p,f$ are relatively prime. In the second case, we can take $d=p$ and conclude that $p|f$.
• (ii) Suppose $p|fg$ and $p$ does not divide $f$. Then by (i), $p$ and $f$ are relatively prime, it follows from the earlier proposition that $p|g$ as claimed. $\square$

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Tests for irreducibility

Theorem

Let $R$ be a unique factorization domain, $F$ be its field of quotients, and

$$f=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0\in R[x]$$

be a polynomial of degree $n\ge 1$. Suppose $I\subseteq R$ is an ideal in $R$ for which $a_n\notin I$. If the reduction $\bar{f}$ is not the product of two polynomials in $(R/I)[x]$ with degrees $d,e$, respectively, satisfying $0<d,e<n$, then $f$ is irreducible in $F[x]$.

Eisenstein criterion

Let $f\in R[x]$ be a polynomial of degree $n\ge 1$, and let $M\subseteq R$ be a maximal ideal in $R$. If the following conditions hold:

• (i) The leading coefficient $a_n$ is not in $M$;
• (ii) For $i<n$, we have $a_i\in M$;
• (iii) We have $a_0\notin M^2$, meaning there do not exist $b,c\in M$ such that $a_0=bc$;

Then $f$ is not the product of two polynomials of strictly smaller degree in $R[x]$, and hence $f$ is irreducible as an element of $F[x]$.

Irreducibility modulo prime $p$ implies irreducibility in $\mathbb{Q}[x]$

Let $f\in \mathbb{Z}[x]$ be a nonconstant polynomial and $p$ be a prime number which does not divide the leading coefficient of $f$. If $f$ is irreducible over the field ${\mathbb{F}}_p$, then $f$ is irreducible over $\mathbb{Q}$.

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The quotient by an (irreducible) polynomial

Given a field $F$ and a polynomial $f=c_d{x}^d+\cdots +c_0\in F[x]$ that has $\mathrm{deg}>0$ (hence $c_d\ne 0$), the quotient ring $F[x]/(f)$ is quite similar to the ring of Integers modulo n.

	$F[x]/(f)$	$\mathbb{Z}/n\mathbb{Z}$
Elements	Cosets $g+(f)$	Cosets $a+n\mathbb{Z}$
Description of elements	Uniquely described by $a_0+a_1\alpha +\cdots +a_{d-1}{\alpha }^{d-1}$, where $a_i\in F$ and $\alpha =x+(f)$	Written as $0,1,\dots ,n-1$
Computations	Addition is computed by adding coefficients, and multiplication by the rule $a^d=-c_d^{-1}(c_{d-1}{\alpha }^{d-1}+\cdots +c_0)$	Both addition and multiplication are computed by setting multiples of $n$ to be $0$
When the quotient is a field	When $f$ is irreducible in $F[x]$	When $n$ is prime
Chinese remainder theorem	If $f,g\in F[x]$ are relatively prime, then the quotient by the product $F[x]/(fg)$ is isomorphic to the product of quotients $(F[x]/(f))\times (F[x]/(g))$	If $n,m\in \mathbb{Z}$ are relatively prime, then the modulo of the product $\mathbb{Z}/nm\mathbb{Z}$ is isomorphic to the product of modulos $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}$

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Examples

Irreducibility depends on the underlying field

• A linear (degree 1) polynomial is always irreducible.
• A quadratic (degree 2) or cubic (degree 3) polynomial is reducible if and only if it has a linear factor in $F[x]$, i.e., it has a root in $F$. For example, $x^2-2$ and $x^3-2$ are both irreducible in $\mathbb{Q}[x]$, but not in $\mathbb{R}[x]$; $x^2+1$ is irreducible in $\mathbb{R}[x]$ but not in $\mathbb{C}[x]$.
• The polynomial $f=x^2+x+1$ is irreducible in ${\mathbb{F}}_2$, the field of integers modulo $2$, since $f(0)=f(1)=1$ and $f$ does not have a root in ${\mathbb{F}}_2$.
• The polynomial $x^4-4$ is reducible in $\mathbb{Q}[x]$ even though it does not have a root in $\mathbb{Q}$: we have a non-trivial factorization $x^4-4=(x^2+2)(x^2-2),$ but the roots $\pm \sqrt{2}$, $\pm i\sqrt{2}$ are certainly not in $\mathbb{Q}$.

Irreducibility for polynomials of degree $2$ in $\mathbb{Q}$

• Checking the discriminant: A quadratic polynomial of the form $f=a{x}^2+bx+c$ is reducible in $\mathbb{Q}$ (or in $\mathbb{Z}$) if and only if its discriminant $b^2-4ac$ is a perfect square in $\mathbb{Q}$ (or in $\mathbb{Z}$).

Factoring polynomials of degree $4$ in $\mathbb{Q}$

• Linear factors: Linear factors, corresponding to roots of the polynomial, can be checked using the rational root test: if $f(a/b)=0$ for some $a/b\in \mathbb{Q}$, the $(bx-a)$ is a factor.

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Notes

• Einstein’s criterion?