Codex @codex  0

Let $P_n(\mathbb{R})$ be the vector space of real polynomials of degree of a polynomial at most $n$. Choose $n+1$ distinct real sample points $t_0,\dots ,t_n$ and positive real weights $w_0,\dots ,w_n$. Thenis an inner product. Symmetry and bilinearity follow from linearity of polynomial evaluation. Its value at $(f,f)$ is nonnegative, and equality forces all $f(t_j)=0$. By the root bound for a polynomial, a nonzero polynomial of degree of a polynomial at most $n$ cannot vanish at $n+1$ distinct points. Thus the form has positive definiteness.

Equivalently, the polynomial evaluation map on a finite point set identifies $P_n(\mathbb{R})$ with ${\mathbb{R}}^{n+1}$, equipped with a weighted Euclidean inner product. Too few sample points destroy positive definiteness: if the distinct points number at most $n$, the nonzero polynomial ${\prod }_j(x-t_j)$ lies in $P_n(\mathbb{R})$ and has zero sampled squared norm.

For three equally weighted samples at $-1,0,1$, an orthonormal basis isThis construction gives a finite sampling version of orthogonal polynomials and permits exact Gram-Schmidt orthogonalization using only finitely many evaluations.

Repeat the random evaluation test for a polynomial identity with fresh independent random variables, keeping the same fixed polynomial. Reject as soon as any polynomial evaluation is nonzero; accept only if all $k$ values vanish. An identity is always accepted, preserving one-sided error. If a nonzero polynomial has single-trial false acceptance probability $p_1$, independence gives $p_k=p_1^k$. The Schwartz-Zippel lemma bounds this by $(d/|S|{)}^k$ when $d<|S|$.

For $G(x)=(x-1)(x-2)$ over the finite field ${\mathbb{F}}_5$, two uniform independent choices $(a,b)$ have $25$ equally likely ordered outcomes, each with probability $1/25$. Both polynomial evaluations are zero only at $(1,1),(1,2),(2,1),(2,2)$. Hence false acceptance falls from $2/5$ in one trial to $4/25$ in two trials, and correct rejection rises from $3/5$ to $21/25$.

If the second choice reuses the first, the pair is $(a,a)$ and only five diagonal outcomes are possible. Two still falsely accept, so the error remains $2/5$. Choosing fresh independent points is precisely what makes the full $25$-cell grid and the product of probabilities valid. For a target error $0<\epsilon <1$, this example needsThe ceiling function makes the trial count an integer. Four trials give $(2/5{)}^4=16/625>1/100$, while six give $64/15625<1/100$; five still give $32/3125>1/100$, so six is the first count below one percent. This is an exact error guarantee for each run on $G$, rather than an observed success rate.

A linear functional is a linear map from a vector space $V$ to its scalar field $F$:Its output is one scalar, rather than an arbitrary vector in another space. In ${\mathbb{R}}^2$, $\phi (x,y)=2x-y$ gives $\phi (3,1)=5$, $\phi (1,2)=0$, and $\phi (4,3)=5$: adding the two inputs adds their measured values.

The level sets $\phi (v)=c$ visualize what the measurement retains and discards. For this example they are parallel straight lines $y=2x-c$. Moving parallel to $(1,2)$ changes the input but leaves $\phi$ unchanged. The zero level set is its kernel of a linear map, while the other level sets are translates of that vector subspace.

Inputs need not be geometric arrows. On the vector space of polynomials of polynomial degree at most two, evaluating at zero is a linear functional, as is taking the definite integral from zero to one. For $p(t)=a+bt+c{t}^2$, they give $p(0)=a$ and ${\int }_0^{1}p(t)dt=a+b/2+c/3$. Thus polynomial evaluation and integration are two different measurements of the same polynomial. Coordinate extraction is another measurement, developed under dual basis.

A linear functional is defined by linearity, not by continuity. In finite-dimensional real or complex normed vector spaces it is automatically continuous; in infinite dimension it need not be. The continuous dual space adds this requirement when small changes in the input should give small changes in the output.

A randomized decision algorithm has one-sided error when one answer is always correct and only the other answer can be mistaken.

The guarantee concerns each fixed input, with probability taken over the choices of the randomized algorithm. Always state which answer is protected. In polynomial identity testing, call an input positive when its polynomial is identically zero, and let acceptance mean claiming an identity. Then the possibilities are:

Thus a zero polynomial evaluation can leave uncertainty, but a nonzero polynomial evaluation settles the question. An error bound for a fixed nonzero polynomial is not the conditional probability that a particular accepted input is an identity; that latter question would need a distribution on the inputs as well.

In computational complexity theory, a polynomial-time algorithm with this orientation and false acceptance probability at most $1/2$ establishes membership in co-RP. Reversing the question to ask whether the polynomial is nonzero gives the RP orientation: negative instances are always rejected. The random evaluation test for a polynomial identity explains the required field, total degree, sampling and evaluation assumptions.

All the following polynomial coefficients and sums lie in the field with two elements, so $a+a=0$. For the constant polynomial,For the linear polynomial, subtracting the leading multiple means adding it:For the quadratic polynomial, first cancel $a{x}^2$ using $ax(x+1)$, leaving $(a+b)x+c$; then cancel $(a+b)x$ using $(a+b)(x+1)$. ThusThe requested polynomial quotients and polynomial remainders are thereforeFor a general polynomial $P(x)={\sum }_{i=0}^n{p}_i{x}^i$, write $P=(x+1)Q+R$ with constant $R$. Since $1+1=0$, polynomial evaluation at $x=1$ gives $P(1)=R$. This is the polynomial remainder theorem with $-1=1$ in the field with two elements. HenceThe polynomial remainder is precisely the parity bit obtained by taking the exclusive or of all the polynomial coefficients.

The prescribed form is a sum of products of real polynomial evaluations. It is a symmetric bilinear form, since each evaluation is a linear map. AlsoEquality forces $f(-1)=f(0)=f(1)=0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, by the root bound for a polynomial. Thus equality holds only for the zero polynomial, establishing positive definiteness. Therefore the form is a real inner product, and $V$ is an inner product space. This is a discrete sampling inner product on polynomial spaces.

Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given $f,g\in F[X]$ over a field $F$, with $g\ne 0$, there are unique polynomials $q,r$ satisfyingHere $q$ is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.^[ref]

For an original example over the rational numbers, takePolynomial long division first cancels $2X^3$ by subtracting $2Xg$:Subtracting one more copy of $g$ cancels $X^2$:ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term $-2+8=6$.

The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of $g$ can be discarded. At any root of a polynomial of $g$, those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.

The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.^[ref]

A nonzero divisor alone is insufficient over the integers. Dividing $X$ by $2X+1$ with an integer polynomial quotient and a constant polynomial remainder would require $1=2c$ for an integer $c$, which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient $2$. This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.

Polynomial evaluation replaces the indeterminate of a polynomial by an input value. For a fixed scalar $a$, the rule $E_a:p\mapsto p(a)$ is a linear functional on a vector space of polynomials, because evaluating a linear combination gives the same linear combination of the values. For $p(t)=1+2t+3t^2$, evaluation at zero gives one and evaluation at two gives seventeen. Polynomial interpolation uses these measurements to reconstruct a polynomial subject to a polynomial degree bound.

Polynomial identity testing asks whether a polynomial represented implicitly, such as by an arithmetic circuit or a remainder computation, is the zero polynomial.

Here zero means that every coefficient vanishes after collecting equal monomials in the polynomial ring. Testing equality of two polynomials $A,B$ amounts to testing $P=A-B$ for zero. An arithmetic circuit can permit cheap polynomial evaluation even when expanding all monomials is expensive. The random evaluation test for a polynomial identity exploits this: one nonzero value disproves the polynomial identity, while a known total degree limits how often a nonzero polynomial can evade detection.^[ref]

For example, for an integer $n\ge 1$, ${\prod }_{i=1}^n(1+x_i)$ has $2^n$ distinct monomials, one for each subset of the $n$ variables. Its factored expression needs only $n$ additions and $n-1$ multiplications for one polynomial evaluation. Evaluating the compact expression avoids creating its expanded list of coefficients. The saving is especially attractive in many variables; it does not make a zero result into a proof of identity.

Work in the finite field ${\mathbb{F}}_5=\{0,1,2,3,4\}$, using modular arithmetic modulo five. Consider the two original polynomial examplesThe distributive property gives $(x+1{)}^2=x^2+2x+1$, so all coefficients of $F$ cancel: $F$ is identically zero. In contrast, $G=x^2-3x+2=x^2+2x+2$ in this finite field, with nonzero leading coefficient. Its roots of a polynomial are exactly one and two: a product in a field is zero only if a factor is zero.

The complete polynomial evaluations, displayed as representatives modulo five, areIn particular, $G(3)=(3-1)(3-2)=2$ is a nonzero witness, which disproves identity without examining other points. At one or two, a zero value fails to reveal that $G$ is nonzero. A random variable $a$ with a uniform distribution on a finite set ${\mathbb{F}}_5$ hits those two polynomial roots with probability $2/5$.

A crucial limit appears with $H(x)=x^5-x$. Its coefficients make it a nonzero polynomial of polynomial degree five in ${\mathbb{F}}_5[x]$, yet direct polynomial evaluation gives zero at all five field elements: the values before reduction are $0,0,30,240,1020$, all divisible by five. Thus $H$ induces the same polynomial function as the zero polynomial. Any number of trials restricted to ${\mathbb{F}}_5$ will falsely accept this identity. This is compatible with the Schwartz-Zippel lemma: the ratio of polynomial degree to sampling-set size is $5/5=1$, which gives no useful guarantee.

When a polynomial's representation permits polynomial evaluation over a larger field extension, a larger sampling set can restore the useful degree ratio. Access only to the induced polynomial function on the original finite field cannot distinguish $H$ from zero.^[ref]

Dividing a polynomial $f$ by $X-a$ leaves the constant polynomial remainder $f(a)$. Polynomial division makes the polynomial remainder constant because its polynomial degree must be below one; polynomial evaluation at $a$ identifies that constant. In particular, it is zero exactly when $a$ is a root of a polynomial of $f$, which is the factor theorem.^[ref]

In the worked example, division by $X-1$ leaves $6$ and division by $X+2$ leaves $12$. Division by their product leaves $-2X+8$, a polynomial remainder that stores both values at once. The argument also holds over a commutative ring for $a$ in that ring, because $X-a$ is a monic polynomial.

Let $P$ be a nonzero multivariate polynomial over a field $K$, with total degree at most $d\ge 0$. Choose a nonempty finite set $S\subseteq K$. For independent random variables $a_1,\dots ,a_n$, each with a uniform distribution on a finite set on $S$, the Schwartz-Zippel lemma givesNonzeroness is essential; the zero polynomial always evaluates to zero. The total degree sums exponents within each monomial, rather than taking their maximum: $x^2{y}^3+y$ has total degree five. The field need not be finite, but $S$ must be finite. The estimate reduces error only when $d/|S|<1$.^[ref]

The proof begins with the root bound for a polynomial. For several variables, write $P={\sum }_{j=0}^m{Q}_j(x_1,\dots ,x_{n-1})x_n^j$, with $Q_m\ne 0$. The total degree of $Q_m$ is at most $d-m$. By induction, $Q_m$ vanishes with probability at most $(d-m)/|S|$. Otherwise the remaining univariate polynomial has at most $m$ polynomial roots. Adding the two contributions gives $d/|S|$.^[ref]

The algorithm samples a point, performs exact polynomial evaluation, rejects the identity if the value is nonzero, and otherwise accepts. The lemma bounds false acceptance for every fixed nonzero input. For the polynomial identity testing over a finite field example, $G$ has $d=2$ and $|S|=5$, so its exact $2/5$ error attains the bound. Numerical near-zero values from floating-point arithmetic do not satisfy this exact algebraic contract. A polynomial-time algorithm claim additionally needs efficient sampling and evaluation, including the bit cost of representing and operating on field elements. Counting field operations alone does not establish that cost.

The polynomial remainder is the polynomial $r$ in polynomial division $q=Qs+r$. Division by $X-a$ has a constant polynomial remainder $q(a)$, as expressed by the factor theorem.

For any identity $f=gq+r$, polynomial evaluation at a root of a polynomial $a$ of $g$ givesThis substitution requires no division by $g(a)$. In the worked example, polynomial factorization gives $g=(X-1)(X+2)$, so its two roots of a polynomial are $1$ and $-2$. The cubic polynomial and its linear polynomial remainder agree there:A cubic expression has been compressed into a straight line carrying exactly these two values.

Conversely, those two values determine the linear polynomial remainder by polynomial interpolation. Write $r=AX+B$; then $A+B=6$ and $-2A+B=12$, giving $A=-2$ and $B=8$. The polynomial degree bound is essential: adding any multiple of $g$ preserves the two values, but generally raises the polynomial degree. The plot also shows agreement at $x=-1/2$, where $q=0$ rather than $g=0$. Agreement of $f$ and $r$ at a point therefore does not imply that the point is a root of a polynomial of the divisor.