Polynomial Rings and Roots: Examples and Counterexamples

Given: The integer polynomials $f=1+2x+3x^2$ and $g=2-x$.

Polynomial addition is coefficientwise, and the coefficient of $x^n$ in a product is the finite convolution sum $\sum_{i+j=n}a_i b_j$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

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

Given: The displayed polynomials $f,g\in\mathbb Q[x]$.

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

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).

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

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

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).

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

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

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

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

The only units of $\mathbb Z$ are $1$ and $-1$ ($(\mathbb{Z}, \cdot, 1)$ is a commutative monoid whose group of units is $\{1, -1\}$; equivalently $u \mid 1$ holds exactly for $u = 1$ and $u = -1$).

Given: The polynomials $f=x^2-x$ in $(\mathbb Z/6)[x]$ and $g=x^2-1$ in $(\mathbb Z/8)[x]$.

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).

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

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

Given: A prime $p$, the field $\mathbb F_p=\mathbb Z/p$, and the polynomials $f=x^p-x$ and $0$ in $\mathbb F_p[x]$.

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).

Fermat's little theorem gives $a^p\equiv a\pmod p$ for every integer $a$ (Fermat's little theorem: for prime $p$, $p\nmid a$ implies $a^{p-1}\equiv1\pmod p$, and always $a^p\equiv a\pmod p$).

The ring $\mathbb Z/p$ is the quotient $\mathbb Z/p\mathbb Z$ (For every $n\in\mathbb N$, the congruence-class ring $\mathbb Z/n$ is the quotient ring $\mathbb Z/n\mathbb Z$).

A prime integer is greater than $1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$).

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

Given: The polynomial $f=x^2-2\in\mathbb Z[x]\subseteq\mathbb Q[x]$.

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).

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).

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 $p$).

A divisor of a nonzero integer has absolute value at most the absolute value of that integer (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$; hence the set of divisors of a nonzero integer is bounded above by $|a|$).

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

Integer absolute value is nonnegative and vanishes exactly at zero (Absolute value in $\mathbb{Z}$: $|a| \ge 0$; $|a| = 0$ exactly when $a = 0$; $|-a| = |a|$; $|ab| = |a|\,|b|$; $-|a| \le a \le |a|$; and $|a| \le c$ exactly when $-c \le a \le c$).

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

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

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

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

Given: The polynomial $f=x^2+1\in\mathbb R[x]$.

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).

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

Given: The primitive integer polynomial $f=x^3+17x+391$.

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

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).

The ring $\mathbb Z/2$ is the quotient $\mathbb Z/2\mathbb Z$ (For every $n\in\mathbb N$, the congruence-class ring $\mathbb Z/n$ is the quotient ring $\mathbb Z/n\mathbb Z$).

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 $p$).

Integer divisors of a nonzero integer are bounded in absolute value (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$; hence the set of divisors of a nonzero integer is bounded above by $|a|$).

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

Integer absolute value is nonnegative and detects zero (Absolute value in $\mathbb{Z}$: $|a| \ge 0$; $|a| = 0$ exactly when $a = 0$; $|-a| = |a|$; $|ab| = |a|\,|b|$; $-|a| \le a \le |a|$; and $|a| \le c$ exactly when $-c \le a \le c$).

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

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

The natural order is discrete (Discreteness: $\sigma(n)$ is the immediate successor).

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

Given: A natural number $n\ge1$.

For every prime $p$ and positive $n$, the polynomial $x^n-p$ is irreducible over $\mathbb Q$ (For every prime $p$ and positive $n$, $x^n-p$ is irreducible over $\mathbb Q$).

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 $p$).

A divisor of a nonzero integer is bounded in absolute value (If $d \mid a$ and $a \ne 0$ then $d \ne 0$ and $|d| \le |a|$; hence the set of divisors of a nonzero integer is bounded above by $|a|$).

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

Integer absolute value is nonnegative and detects zero (Absolute value in $\mathbb{Z}$: $|a| \ge 0$; $|a| = 0$ exactly when $a = 0$; $|-a| = |a|$; $|ab| = |a|\,|b|$; $-|a| \le a \le |a|$; and $|a| \le c$ exactly when $-c \le a \le c$).

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

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

The natural order is discrete (Discreteness: $\sigma(n)$ is the immediate successor).

Given: The polynomial $f=x^4+1\in\mathbb Q[x]$ and the substitution $\tau(f)(x)=f(x+1)$.

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

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

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 $p$).

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

Given: The polynomial $f=(x^2+1)(x^2+2)=x^4+3x^2+2\in\mathbb Q[x]$.

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

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

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).

Given: The quotient field $\mathbb Z/2$ and the polynomial $f=x^2+1$.

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).

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).

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

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 $p$).

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

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

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

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).

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).

Chosen images of the indeterminates determine evaluation homomorphisms (Universal property of $R[x]$: a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).