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For functions $f$ and $g$ taking values in the same field, their quotient of functions is defined by $(f/g)(x)=f(x)/g(x)$ on the common domain of a function restricted to points where $g(x)\ne 0$. Thus $f(x)=1$ and $g(x)=x$ on the real line give a quotient of functions defined only for $x\ne 0$.

In polynomial division $q=Qs+r$ over a field, the polynomial quotient is the polynomial $s$. The polynomial remainder $r$ has smaller polynomial degree than the nonzero divisor $Q$ unless it is zero.

In the notation $f=gq+r$, the polynomial quotient is instead named $q$. It is a polynomial determined by the divisor and the polynomial degree bound. The pointwise ratio of the associated polynomial functions satisfiesFor the worked example, $f(0)/g(0)=6/(-2)=-3$, while $q(0)=1$; the correction is $r(0)/g(0)=8/(-2)=-4$. Thus a polynomial quotient need not equal the rational function $f/g$. At $x=1$ and $x=-2$, the pointwise ratio is undefined, but the polynomial quotient still has perfectly well-defined values. When $r=0$, the divisor is a polynomial factor of $f$ and the ratio agrees with $q$ wherever $g(x)\ne 0$.

For a finite-type integer algebra, a nonnilpotent element $r$ gives a nonzero localization of a ring $R[1/r]$. A maximal ideal there has finite residue field by the finite-field theorem for finitely generated integer algebras. The image of $R$ in that field is a finite integral domain, hence a field. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every maximal ideal, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are Jacobson rings. Merely contracting a localized maximal ideal without the finite-image argument would not prove maximality.

The Specht module $S^{(n-1,1)}$ is the augmentation subspace of the natural permutation module $F^n$, with the standard tabloid bilinear form. Its orthogonal complement in $F^n$ is the constant-vector line. That line lies in the augmentation subspace exactly when the field characteristic divides $n$, giving the displayed radical. Equivalently, the basis $v_i-v_n$ has Gram matrix $I_{n-1}+J_{n-1}$. The James submodule theorem makes every nonzero radical quotient simple, with dimension $n-2$ when $p\mid n$ and $n-1$ otherwise. The case $n=2,p=2$ gives the zero quotient rather than a nonzero simple module.

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.

A randomized algorithm makes some choices using random bits.

For a fixed input, the random bits can change the path taken by a randomized algorithm, its output, or its running time. Specify the distribution of those choices and which guarantee is required: one-sided error permits mistakes of only one kind, whereas ZPP requires a correct answer and bounds the expected running time. Polynomial identity testing makes this distinction concrete: choosing different field elements changes whether a nonidentity is detected.

A uniform distribution on a finite set must be implemented faithfully. To choose uniformly from $\{0,1,2,3,4\}$ using fair independent bits, read three bits as a number from zero to seven, discard five through seven, and repeat until a permitted value appears. Each retained value has the same chance. Taking the three-bit number modulo five instead would give zero, one and two twice as many preimages as three and four, invalidating a calculation that assumes equal chances.

Take independent uniform samples from a $d$-dimensional vector space over the two-element field. At current matrix rank $r$, the existing span contains $2^r$ of the $2^d$ possible vectors, so a new sample increases the matrix rank with probability $1-2^{r-d}$. The mean waiting time for that increase is $1/(1-2^{r-d})$ by the geometric distribution. Adding the waiting times gives $\mathbb{E}M=d+{\sum }_{j=1}^d(2^j-1{)}^{-1}<d+2$. For $d\ge 1$, the bound follows by separating $j=1$ and using $(2^j-1{)}^{-1}<2^{1-j}$ for $j\ge 2$; for $d=0$, no samples are needed.

After $m=d+t$ samples, failure to span implies that some nonzero linear functional vanishes on all samples. Each fixed nonzero functional does so with probability $2^{-m}$; the union bound over the $2^d-1$ possibilities gives failure less than $2^{-t}$. This provides both an expected sample count and a fixed-budget high-success version of Simon's algorithm.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

For a module $M$ over an integral domain $R$ with field of fractions $K$, its rank is the dimension of $K{\otimes }_{R}M$. The tensor product of modules here extends scalars to a field, turning the question into a dimension calculation. For a free module $R^r$, the result is $r$.

For a finitely generated module over a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain gives $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, with $d_i$ nonzero nonunits. Each torsion element of a module vanishes after extension of scalars, because its annihilating nonzero scalar becomes invertible in $K$. Thus $K{\otimes }_{R}M\cong K^r$, proving that the rank is the number of free summands. The rank does not record the torsion submodule: for example, $R\oplus R/(d)$ and $R$ have the same rank.

The rational canonical form of a linear operator on a finite-dimensional vector space over a field is a direct sum $C(f_1)\oplus \cdots \oplus C(f_r)$ of companion matrices, where the monic nonconstant invariant factors of a linear operator satisfy $f_1\mid \cdots \mid f_r$. Their degrees sum to the dimension of the space. It exists over any field and classifies similar matrices, without requiring their eigenvalues to lie in that field. It comes from the structure theorem for finitely generated modules over a principal ideal domain, applied to the module in which the indeterminate acts as the given operator.

Let $U=aT+bI$ on a nonzero finite-dimensional vector space over a field, with $a\ne 0$. An invariant block for $T$ is also invariant for $U$. If $v,Tv,\dots ,T^{d-1}v$ is its cyclic vector basis, then $v,Uv,\dots ,U^{d-1}v$ is another basis: the binomial theorem gives a triangular change-of-basis matrix with diagonal $1,a,\dots ,a^{d-1}$. The monic polynomial $g(t)=a^{d}f((t-b)/a)$ annihilates this block, and its degree is $d$, so it is the block's minimal polynomial. In the new cyclic basis the block is $C(g)$.

The substitution preserves divisibility of the monic invariant factors of a linear operator. Apply it to every block to obtain the rational canonical form of $U$. If $a=0$, then $U=bI$ is a scalar matrix instead: all invariant factors are $t-b$, one for each dimension. This exceptional case cannot be obtained by dividing by $a$.

Let $A$ act on a nonzero finite-dimensional vector space over a field $F$, with minimal polynomial an irreducible polynomial $f$ of polynomial degree $d$. The nonconstant invariant factors of a linear operator all divide $f$, so each equals $f$. Their degrees sum to the dimension of a vector space, giving $d\mid n$ and rational canonical form consisting of $n/d$ copies of the companion matrix $C_f$. Equivalently, make the vector space a module over the field $F[X]/(f)$ by letting $X$ act as $A$; a basis over this field gives the same block decomposition.

For an algebraic variety $X$ defined over a field $K$, a K-rational point is a morphism $\mathrm{Spec}K\to X$ over $K$. In affine coordinates it is a solution of the defining equations with every coordinate in $K$; in projective coordinates it is a nonzero coordinate vector over $K$, considered up to common nonzero scaling. The set is denoted $X(K)$. Over a number field, having points over every completion need not imply having a rational point; for genus-one coverings this failure is recorded by the Tate–Shafarevich group.

Tensoring a Tate module with the P-adic numbers gives its rational Tate module. For a CM elliptic curve with base-defined multiplication by $K$, it is free of rank one over $K\otimes {\mathbb{Q}}_{\ell }$, including when this algebra is a product of two fields. In the split case the two distinct roots of a nonrational CM endomorphism each have a one-dimensional eigenspace.

A real closure of an ordered field $K$ is a real closed algebraic extension whose order extends the chosen order of $K$. Existence is the real closure theorem; uniqueness holds up to ordered field isomorphism over $K$. A different ordering of a formally real field can produce a different ordered real closure.

Every ordered field has an ordered algebraic real closure. Combined with the Artin-Schreier ordering criterion, every formally real field embeds into a real closed field. This is one embedding direction in the model companion relation between their theories.

A real vector space is a vector space over the field $\mathbb{R}$ of real numbers.

Let the base be a smooth algebraic curve, the total surface be smooth, and the field be algebraically closed of characteristic different from two. At a rank-two singular fibre with kernel vector $v$, the fibre derivatives vanish at $[v]$. Smoothness requires $v^{\mathsf{T}}{M}^{\prime }v\ne 0$. The adjugate matrix is a nonzero multiple of $v{v}^{\mathsf{T}}$, so the determinant has a simple zero. At rank at most one, the projective kernel contains a line and the quadratic $v^{\mathsf{T}}{M}^{\prime }v$ has a zero on it; that point would make the total surface singular. Thus every singular fibre is a pair of distinct projective lines and its discriminant zero is simple.

For $1\le k<p$, choose $l$ with $kl\equiv 1(\mathrm{mod}p)$. The geometric sum $1+{\zeta }_p^k+\cdots +{\zeta }_p^{k(l-1)}$ is the inverse of $u_k$. Hence these are units, and their residues cover ${\mathbb{F}}_p^{\times }$. In the cyclotomic Milnor square, this makes the $K_1$ map onto the common field surjective and eliminates its connecting obstruction to $K_0$.

A representation over the rational numbers is a group representation on a finite-dimensional rational vector space. Choosing a basis gives rational matrices. It differs from a rational representation of an algebraic group, where rational refers to regular matrix-coefficient functions rather than the ground field. Scalar extension yields representations over every characteristic-zero field.