Alphabeta Math
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✓ 14 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 14 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Polynomial Rings and Roots: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Polynomial addition and multiplication computed from coefficient convolution

Example

In Z[x], let f=1+2x+3x2 and g=2−x. Then

f+g=3+x+3x2,fg=2+3x+4x2−3x3.

Facts & Assumptions

Given: The integer polynomials f=1+2x+3x2 and g=2−x.

[L1]

Polynomial addition is coefficientwise, and the coefficient of xn in a product is the finite convolution sum ∑i+j=naibj (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

The integers form a commutative ring (The integers form a commutative ring).

Verification

technique · direct
1.1

Coefficientwise addition over the commutative ring [L2] gives the coefficients (1+2,2−1,3+0)=(3,1,3).

givenL1L2algebra
2.1

Convolution gives coefficients 2, −1+4=3, −2+6=4, and −3 in degrees 0,1,2,3, respectively, proving the displayed product.

givenL1L2algebra∎
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Long division, the Euclidean algorithm and a Bézout identity in Q[x]

Example

In Q[x], put f=x3−2x+1 and g=x2−1. Long division and the Euclidean algorithm give

f=xg+(−x+1),g=(−x−1)(−x+1).

Thus gcd⁡(f,g)=x−1 and

x−1=−f+xg.

Facts & Assumptions

Given: The displayed polynomials f,g∈Q[x].

[L1]

Polynomial division over a field gives a unique remainder of degree below the divisor (Division algorithm for polynomials over a field).

[L2]

The last nonzero Euclidean remainder, normalized to be monic, is the gcd and is a polynomial linear combination of the inputs (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L3]

The rational numbers form a field (The rationals form a field).

Verification

technique · direct
1.1

Direct multiplication gives x(x2−1)+(−x+1)=x3−2x+1=f, and (−x−1)(−x+1)=x2−1=g, so these are the divisions licensed by [L1] over the field [L3].

givenL1L3algebra
2.1

The last nonzero remainder is −x+1, whose monic associate is x−1; negating the first remainder identity gives x−1=−f+xg, in agreement with [L2].

step 1.1L2algebra∎
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

Over Z/4, a nonconstant polynomial can be a unit and product degree can drop

Statement refuted

For every commutative ring R, every unit of R[x] is constant and degrees add under multiplication of nonzero polynomials.

Facts & Assumptions

Given: The ring R=Z/4 and the polynomial u=1+2x∈R[x].

[L1]

Over a commutative ring, product degree is at most the sum of the degrees, and the coefficient at that sum is the product of the leading coefficients (Degree inequalities for sums and products over a commutative ring).

[L2]

The ring Z/4 is the quotient ring Z/4Z, so 4=0 in it (For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ).

Counterexample

technique · direct
1.1

In R[x], u2=(1+2x)2=1+4x+4x2=1 by [L2], so the nonconstant polynomial u is a unit and is its own inverse.

givenL2algebra
2.1

Both factors u have degree 1, but their product has degree 0; [L1] permits this drop because the top coefficient is 2⋅2=0 in R, so both parts of the statement are refuted.

step 1.1L1L2∎
CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

Division with a degree-small remainder can fail over Z when the leading coefficient of the divisor is not a unit

Statement refuted

For every commutative ring R and every nonzero g∈R[x], each f∈R[x] can be written f=qg+r with r=0 or deg⁡r<deg⁡g.

Facts & Assumptions

Given: The polynomials f=x and g=2x+1 in Z[x].

[L1]

Division is guaranteed over a commutative ring when the divisor is monic (Division by a monic polynomial over a commutative ring).

[L2]

The integers form a commutative ring (The integers form a commutative ring).

Counterexample

technique · contradiction
1.1

Suppose for contradiction that x=q(2x+1)+r with r=0 or deg⁡r<1. If q had positive degree m and leading coefficient c≠0, then q(2x+1) would have nonzero coefficient 2c in degree m+1>1, which the constant remainder could not cancel. Thus q is a constant integer.

assume-contragivenL2algebra
2.1

Comparing coefficients of x then gives 1=2q, which has no integer solution because 2 is not a unit by [L3]. This does not contradict the monic-division theorem [L1], since 2x+1 is not monic; hence the claimed division statement fails.

step 1.1L1L2L3discharge-contradiction∎
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Quadratics can have four roots over Z/6 and Z/8

Statement refuted

Every nonzero quadratic over a commutative ring has at most two distinct roots.

Facts & Assumptions

Given: The polynomials f=x2−x in (Z/6)[x] and g=x2−1 in (Z/8)[x].

[L1]

The degree root bound holds for nonzero polynomials over integral domains (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

[L2]

Over any commutative ring, a is a root exactly when x−a divides the polynomial (Factor theorem over a commutative ring).

[L3]

The rings Z/6 and Z/8 are the corresponding quotient rings, so computations are modulo 6 and 8 (For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ).

Counterexample

technique · direct
1.1

Modulo 6, the four distinct residues 0,1,3,4 satisfy a(a−1)=0, so they are roots of the nonzero quadratic f.

givenL2L3algebra
2.1

Modulo 8, the four distinct odd residues 1,3,5,7 have square congruent to 1, so they are roots of the nonzero quadratic g; these rings have zero divisors, so the domain hypothesis in [L1] fails and the statement is refuted.

givenL1L2L3algebra∎
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Over Fp, xp−x and the zero polynomial induce the same function but are distinct polynomials

Statement refuted

Two formal polynomials over a field are equal whenever they induce the same function on that field.

Facts & Assumptions

Given: A prime p, the field Fp=Z/p, and the polynomials f=xp−x and 0 in Fp[x].

[L1]

Evaluation substitutes an element into a formal polynomial, while the formal polynomial itself is its coefficient sequence (Evaluation and roots of a polynomial in a commutative target ring).

[L5]

For prime p, the quotient ring Z/p is a field (For every prime p, the two operations on Z/p make it a field).

Counterexample

technique · direct
1.1

Fact [L5] licenses the field Fp=Z/p. For every a∈Fp, choose an integer representative; [L2] and [L3] give f(a)=ap−a=0, so f and the zero polynomial induce the same function.

givenL1L2L3L5
2.1

Because [L4] gives p>1, the degrees p and 1 are distinct; as a coefficient sequence, f has coefficient 1 in degree p and is therefore nonzero. Hence [L1] distinguishes f from the zero formal polynomial and refutes the statement.

givenL1L3L4∎
ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

The polynomial x2−2 is irreducible over Q

Example

The polynomial x2−2 is irreducible in Q[x].

Facts & Assumptions

Given: The polynomial f=x2−2∈Z[x]⊆Q[x].

[L1]

A reduced rational root r/s of an integer polynomial has numerator dividing the constant coefficient and denominator dividing the leading coefficient (Rational root theorem).

[L2]

A quadratic over a field is irreducible exactly when it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[L3]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L5]

Integer absolute value is given by the positive and negative cases (The absolute value ∣a∣ of an integer).

[L7]

The integers form an ordered ring (The integers form a totally ordered ring).

[L8]

The natural numbers embed in the integers preserving their arithmetic (The naturals embed in the integers).

[L9]

The natural-number order is discrete (Discreteness: σ(n) is the immediate successor).

[L10]

The rational numbers form a field (The rationals form a field).

Verification

technique · direct
1.1

Facts [L3] through [L9] show that the positive divisors of 2 are 1,2 and the only positive divisor of 1 is 1. Hence [L3] makes 2 prime, and [L1] forces a reduced rational root to have denominator 1 and numerator in the complete list 1,−1,2,−2.

givenL1L3L4L5L6L7L8L9
2.1

Evaluating gives −1,−1,2,2, respectively, so none is a root; [L10] supplies the field hypothesis and [L2] therefore makes the quadratic irreducible over Q.

step 1.1L2L10algebra∎
ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

The polynomial x2+1 is irreducible over R

Example

The polynomial x2+1 is irreducible in R[x].

Facts & Assumptions

Given: The polynomial f=x2+1∈R[x].

[L1]

A quadratic over a field is irreducible exactly when it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[L2]

The real numbers form an ordered field, so every square is nonnegative and 1>0 (The reals form a totally ordered field).

Verification

technique · direct
1.1

For every real a, [L2] gives a2≥0 and hence a2+1>0.

givenL2algebra
2.1

Thus f has no real root, and [L1] makes it irreducible in R[x].

step 1.1L1∎
ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

Reduction modulo 2 proves x3+17x+391 irreducible over Q

Example

The polynomial f=x3+17x+391 is irreducible in Q[x].

Facts & Assumptions

Given: The primitive integer polynomial f=x3+17x+391.

[L1]

If a primitive integer polynomial has leading coefficient nonzero modulo a prime and its reduction is irreducible, then it is irreducible over Q (Irreducibility after reduction modulo a prime implies irreducibility over Q when the leading coefficient survives).

[L2]

A cubic over a field is irreducible exactly when it has no root in the field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[L4]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L6]

Integer absolute value is defined by sign cases (The absolute value ∣a∣ of an integer).

[L8]

The integers form an ordered ring (The integers form a totally ordered ring).

[L9]

Natural numbers embed in the integers preserving arithmetic (The naturals embed in the integers).

[L10]
[L11]

For prime p, the ring Z/p is a field (For every prime p, the two operations on Z/p make it a field).

Verification

technique · direct
1.1

Facts [L4] through [L10] show that 2 is prime, so [L11] makes Z/2 a field. Using [L3], reduction gives fˉ=x3+x+1; its values at the only residues 0 and 1 are both 1, so [L2] makes fˉ irreducible.

givenL2L3L4L5L6L7L8L9L10L11algebra
2.1

The leading coefficient survives modulo 2, and f is primitive because it is monic, so [L1] proves that f is irreducible over Q.

step 1.1L1∎
ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

Eisenstein proves xn−2 irreducible over Q for every positive n

Example

For every positive natural number n, the polynomial xn−2 is irreducible in Q[x].

Facts & Assumptions

Given: A natural number n≥1.

[L1]

For every prime p and positive n, the polynomial xn−p is irreducible over Q (For every prime p and positive n, xn−p is irreducible over Q).

[L2]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L4]

Integer absolute value is defined by sign cases (The absolute value ∣a∣ of an integer).

[L6]

The integers form an ordered ring (The integers form a totally ordered ring).

[L7]

Natural numbers embed in the integers preserving arithmetic (The naturals embed in the integers).

[L8]

Verification

technique · direct
1.1

Facts [L2] through [L8] verify that 2>1 and that its only positive divisors are 1 and 2, so 2 is prime.

givenL2L3L4L5L6L7L8
2.1

Apply [L1] with p=2 and the given positive n to obtain the claimed irreducibility.

step 1.1L1∎
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11Open item page →

Translation turns x4+1 into an Eisenstein polynomial

Example

The polynomial x4+1 is irreducible over Q, although Eisenstein's criterion does not apply to it before a translation.

Facts & Assumptions

Given: The polynomial f=x4+1∈Q[x] and the substitution τ(f)(x)=f(x+1).

[L1]

The universal property makes substitution x↦x+1 a ring homomorphism; substitution x↦x−1 is its inverse (Universal property of R[x]: a coefficient homomorphism and the image of x determine a unique ring homomorphism).

[L2]

Eisenstein's criterion proves a primitive integer polynomial irreducible over Q from the stated prime-divisibility conditions (Eisenstein criterion over the integers).

[L3]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L4]

The rational numbers form a field (The rationals form a field).

Verification

technique · direct
1.1

No prime can make f=x4+1 Eisenstein before translation, because the criterion would require that prime to divide the constant coefficient 1. A direct divisor check using [L3] shows that 2 is prime. Binomial expansion gives τ(f)=x4+4x3+6x2+4x+2. It is monic and hence primitive, and it is Eisenstein at 2: every nonleading coefficient is even, the leading coefficient is odd, and 4 does not divide the constant coefficient 2. Thus [L2] makes τ(f) irreducible.

givenL2L3algebra
2.1

Over the field [L4], [L1] makes τ an automorphism; a nontrivial factorization of f would map to one of τ(f), so irreducibility of τ(f) implies irreducibility of f.

step 1.1L1L4∎
CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

A degree-four polynomial can be reducible over Q without having a rational root

Statement refuted

A polynomial over a field is irreducible whenever it has no root in that field.

Facts & Assumptions

Given: The polynomial f=(x2+1)(x2+2)=x4+3x2+2∈Q[x].

[L1]

A root is equivalent to divisibility by the corresponding linear polynomial (Factor theorem over a commutative ring).

[L2]

The rational numbers form a field (The rationals form a field).

[L3]

The rational numbers form an ordered field, so squares are nonnegative and positive constants remain positive when added (The rationals form a totally ordered field).

Counterexample

technique · direct
1.1

The displayed equality is a factorization in the field polynomial ring [L2] into two positive-degree nonunits, so f is reducible.

givenL2algebra
2.1

For every rational a, [L3] gives a2+1>0 and a2+2>0, so f(a)>0 and f has no rational root; by [L1] it has no rational linear factor, yet step 1.1 shows it reducible.

step 1.1L1L3algebra∎
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11Open item page →

In characteristic 2, x2+1=(x+1)2 has zero derivative and a repeated root

Example

In (Z/2)[x], the polynomial f=x2+1 has repeated root 1 and has formal derivative 0.

Facts & Assumptions

Given: The quotient field Z/2 and the polynomial f=x2+1.

[L1]

A root is repeated exactly when it is also a root of the formal derivative (A root is repeated exactly when it is also a root of the formal derivative).

[L2]

The power and Leibniz rules hold for formal derivatives over any commutative ring (Linearity, power rule, Leibniz rule and the degree bound for the formal derivative).

[L3]

The ring Z/2 is the quotient ring Z/2Z, so 2=0 and 1+1=0 (For every n∈N, the congruence-class ring Z/n is the quotient ring Z/nZ).

[L4]

An integer is prime when it exceeds 1 and has no positive divisors other than 1 and itself (Prime and composite integers: p is prime when p>1 and its only positive divisors are 1 and p).

[L5]

For prime p, the ring Z/p is a field (For every prime p, the two operations on Z/p make it a field).

Verification

technique · direct
1.1

A direct divisor check using [L4] shows that 2 is prime, so [L5] licenses the field Z/2. By [L3], −1=1 and (x−1)2=(x+1)2=x2+2x+1=x2+1=f, so 1 is a repeated root.

givenL3L4L5algebra
2.1

By [L2], f′=2x=0 in Z/2[x], so both f(1) and f′(1) vanish, agreeing with the criterion [L1].

step 1.1L1L2L3∎
CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-11Open item page →

For a field F, the ideal (x,y) in F[x,y] is not principal

Statement refuted

If F is a field, then every ideal of F[x,y] is principal.

Facts & Assumptions

Given: A field F, the iterated ring F[x,y]=F[x][y], and the ideal I=(x,y) generated by x and y.

[L1]

The bivariate polynomial ring is obtained by adjoining x and then y (Polynomial rings in finitely many commuting indeterminates by iteration).

[L2]

A polynomial ring in finitely many indeterminates over a field is an integral domain (A polynomial ring in finitely many indeterminates over an integral domain is an integral domain).

[L3]

The ideal generated by a set is the smallest ideal containing that set, and (h) denotes the ideal generated by one element h (The ideal generated by a subset and principal ideals).

[L4]

Counterexample

technique · contradiction
1.1

In a commutative ring the multiples of h form an ideal containing h and lie in every ideal containing h, so [L3] identifies (h) with the set of multiples of h. Suppose for contradiction that I=(h); since x,y∈I, the element h divides both. Treating the ring as F[x][y] by [L1], degree in y and the domain property [L2] show from h∣x that h∈F[x].

assume-contragivenL1L2L3algebra
2.1

From h∣y, comparison of the leading coefficient in y gives 1=hc for some c∈F[x], so h is a unit and (h) is the whole ring.

step 1.1L1L2L3algebra
3.1

Evaluation at x=0,y=0 exists by [L4] and has kernel an ideal containing x,y, so [L3] gives I inside that proper kernel; thus I is proper, contradicting step 2.1, and I is not principal.

step 2.1L3L4discharge-contradiction∎

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