Irreducibility and field construction

An irreducible polynomial is an indivisible relation over a specified coefficient field. Quotienting by it creates a field in which the relation has a root. To use that construction confidently, we need workable irreducibility tests and a clear distinction between a root test and a general factorization argument.

1 Low-degree tests and their limits

A positive-degree polynomial in $F[x]$ is irreducible if it cannot be factored into two positive-degree polynomials over $F$. Nonzero constants are units, so multiplying by one does not change irreducibility. A polynomial of degree 2 or 3 is reducible exactly when it has a root in $F$.

Proof. A root gives a linear factor by division by $x-a$. Conversely, a reducible polynomial of degree 2 or 3 has factors whose positive degrees sum to 2 or 3; at least one has degree 1, hence supplies a root in the field. $\square$

For integer-coefficient $f(x)=a_dx^d+\cdots+a_0$, a rational root $p/q$ in lowest terms has $p\mid a_0$ and $q\mid a_d$. Multiply the root equation by $q^d$ and reduce modulo $p$ or $q$, using coprimality. This yields a finite candidate list when $a_0\ne0$; a zero constant term already gives root zero.

The test does not extend unchanged to degree 4. For example,

$$x^4+4=(x^2-2x+2)(x^2+2x+2)$$

is reducible over $\mathbb Q$ but has no rational root because it is positive at every real number. A polynomial can factor without having any linear factor.

2 A justified higher-degree criterion

An integer polynomial is primitive when its coefficients have gcd 1. The product of primitive polynomials is primitive: if a prime divided every product coefficient, reduction modulo that prime would give a product of two nonzero polynomials equal to zero in $\mathbb F_p[x]$, impossible because a polynomial ring over a field is a domain.

Clearing denominators in a rational factorization gives Gauss's lemma: a primitive integer polynomial is reducible over $\mathbb Q$ exactly when it factors into positive-degree integer polynomials, up to units. For a detailed scalar check, write rational factors as $aG$ and $bH$, with $G,H$ primitive integer polynomials and $a,b\in\mathbb Q$. Their product $GH$ is primitive. If $ab=u/v$ is in lowest terms and $(u/v)GH$ has integer coefficients, $v$ divides every coefficient of $GH$, so $v=1$. If the resulting polynomial is primitive, its coefficient gcd is $|u|$, forcing $u=\pm1$. Thus the rational factorization really yields an integer factorization. This is the step needed before an integer divisibility argument can prove rational irreducibility.

Eisenstein criterion. Let $f(x)=a_dx^d+\cdots+a_0\in\mathbb Z[x]$ have positive degree $d\ge1$. Suppose a prime $p$ satisfies $p\nmid a_d$, $p\mid a_i$ for every $0\le i<d$, and $p^2\nmid a_0$. Then the primitive part of $f$ is irreducible over $\mathbb Q$.

Proof. If it factored into positive-degree integer factors, their leading coefficients would not be divisible by $p$. Reducing modulo $p$ makes their product a nonzero scalar times $x^d$. Over the field $\mathbb F_p$, each factor must therefore be a nonzero scalar times a power of $x$: any factor of a power of $x$ has this form, by repeatedly removing zero constant terms. Since the factor degrees remain positive on reduction, each has constant coefficient divisible by $p$. Their product's constant coefficient would be divisible by $p^2$, a contradiction. Gauss's lemma excludes rational factorizations. $\square$

For $x^5-2$, use $p=2$: all intermediate coefficients are zero, the leading coefficient is not divisible by 2, and the constant term is not divisible by 4. Thus it is irreducible over $\mathbb Q$. A list of failed rational roots alone would not prove the same degree-5 claim.

3 Irreducible quotient fields and degree

Let $p\in F[x]$ be irreducible of degree $d>0$. Every class in $K=F[x]/(p)$ has a unique representative of degree below $d$ by division. Thus $1,\alpha,\ldots,\alpha^{d-1}$ form an $F$-basis, where $\alpha=[x]$; spanning follows from remainders, independence from the fact that a nonzero multiple of $p$ cannot have degree below $d$.

If $[f]\ne0$, irreducibility gives $\gcd(f,p)=1$. Bézout yields $uf+vp=1$, so $[u][f]=[1]$. Hence the quotient is a field. Conversely, a reducible $p=ab$ creates nonzero zero-divisor classes, so no field results. A relation of degree $d$ therefore gives a field of vector-space dimension $d$ only when the polynomial is irreducible.

Worked finite construction. In $\mathbb F_2[x]$, $p=x^3+x+1$ is irreducible because it has no root: both 0 and 1 evaluate to 1. Its quotient has exactly eight elements, represented by $a+b\alpha+c\alpha^2$ with $a,b,c\in\{0,1\}$, and $\alpha^3=\alpha+1$. The product $\alpha(\alpha^2+1)=1$ gives $\alpha^{-1}=\alpha^2+1$. Its seven nonzero elements form a multiplicative group. Since 7 is prime and $\alpha\ne1$, $\alpha$ has order 7. The additive group has characteristic 2 and is not the cyclic group $\mathbb Z/8\mathbb Z$.

The precise Eisenstein boundary and the reduction step

Positive degree and the leading-coefficient condition are essential. A nonzero constant is a unit, not an irreducible polynomial. If the condition $p\nmid a_d$ were dropped, $2x^2+4x+2=2(x+1)^2$ with $p=2$ would satisfy the other divisibilities but have a reducible primitive part.

If $f=c f_0$ with $f_0$ primitive and $c$ the positive coefficient gcd, then $p\nmid a_d$ implies $p\nmid c$. Dividing by $c$ therefore preserves divisibility by $p$ of every nonleading coefficient, nondivisibility of the leading coefficient, and nondivisibility by $p^2$ of the constant term. Gauss's lemma can now be applied to $f_0$.

To justify the monomial claim in the reduction proof without a hidden factorization theorem, let nonzero polynomials $G,H\in\mathbb F_p[x]$ have highest degrees $r,s$ and lowest nonzero powers $i,j$. Both highest and lowest product coefficients are nonzero because the coefficient system is a field. If $GH$ is a nonzero scalar times $x^{r+s}$, its lowest power is also $r+s$, so $i+j=r+s$. Since $i\le r$ and $j\le s$, equality forces $i=r$, $j=s$. Thus both factors are single monomials. The integer factor leading coefficients survive reduction, so $r,s>0$; consequently both integer constant terms are divisible by $p$, giving the required contradiction to $p^2\nmid a_0$.

4 Graded practice

E1 Foundation. Prove $x^3-2$ is irreducible over $\mathbb Q$ using a justified low-degree test.

E2 Higher degree. Apply Eisenstein to $x^4+6x^3+12x^2+18x+3$.

E3 Quotient arithmetic. In the eight-element field above, reduce $\alpha^4$ and calculate $(1+\alpha)(1+\alpha^2)$.

E4 Dependence on the field. Is $x^2+1$ irreducible over $\mathbb Q$, $\mathbb R$ and $\mathbb F_2$?

E5 Proof. If $p$ has degree $d$ over the finite field $\mathbb F_q$ and is irreducible, prove its quotient field has $q^d$ elements.

5 Full solutions

S1. Rational-root candidates are $\pm1,\pm2$, none of which cubes to 2. A reducible cubic has a linear factor, so absence of rational roots proves irreducibility. Eisenstein with 2 is an alternative valid proof.

S2. Use $p=3$. All nonleading coefficients are divisible by 3, the leading coefficient is not, and the constant term 3 is not divisible by 9. The primitive polynomial is irreducible over $\mathbb Q$.

S3. $\alpha^4=\alpha(\alpha+1)=\alpha^2+\alpha$. The product is $1+\alpha+\alpha^2+\alpha^3=1+\alpha+\alpha^2+(\alpha+1)=\alpha^2$, since pairs cancel in characteristic 2.

S4. It is irreducible over $\mathbb Q$ and $\mathbb R$ because it has no root in those fields. Over $\mathbb F_2$, $x^2+1=(x+1)^2$, so it is reducible. The coefficient field is part of the assertion.

S5. Unique representatives have $d$ coefficients, each with $q$ choices. Thus there are $q^d$ classes. Irreducibility supplies the field property; the counting argument alone only counts a quotient ring and would still count the same number of classes for a reducible degree-$d$ polynomial.

Summary

Low-degree root tests, Gauss's lemma and Eisenstein provide different levels of irreducibility evidence. An irreducible degree-$d$ relation produces a field with basis of length $d$. Quotient arithmetic combines remainder reduction with Bézout inverses, while the chosen coefficient field determines all factorization claims.