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To divide $q$ by a nonzero polynomial $Q$ over a field, subtract the multiple of $Q$ that cancels the leading term of the current remainder, and add that multiple to the quotient. Each step strictly decreases the remainder degree, so the process terminates with $q=Qs+r$ and $\mathrm{deg}r<\mathrm{deg}Q$. If two such pairs existed, $Q(s-s\prime )=r\prime -r$ would have degree both at least $\mathrm{deg}Q$ and less than $\mathrm{deg}Q$, unless both differences vanished.

For a prime ideal $P$ of a polynomial ring over a field, the quotient's Krull dimension is the transcendence degree of its fraction field, and its height is $n$ minus that degree. This dimension theorem identifies algebraic codimension with the length of prime chains. It must not be applied indiscriminately to every Noetherian ring without its polynomial-ring hypotheses.

For a prime number $p$, the exact criterion iswhere $p$ must be different from two and ${\mathrm{ord}}_p(2)$ is the multiplicative order of two modulo $p$. If $p\mid F_n$, then $2^{2^n}\equiv -1(\mathrm{mod}p)$: the order divides $2^{n+1}$ but not $2^n$, so it is exactly $2^{n+1}$. Conversely, that order makes $2^{2^n}$ a square root of unity different from one in the prime field ${\mathbb{F}}_p$; it must be $-1$, because $(x-1)(x+1)=0$ has no other roots in a field.

For example, modulo seventeen the powers of two are $2,4,8,16,15,13,9,1$. Their first return to one occurs after eight steps. The half-way value is $16=-1(\mathrm{mod}17)$, proving that $17\mid F_2$. Likewise, ${\mathrm{ord}}_{641}(2)=64$ identifies $641$ as a prime factor of $F_5$. These are statements about divisibility, not evidence that the entire Fermat number is prime.

The multiplicative order always divides $p-1$, by Fermat's little theorem, so $p\equiv 1(\mathrm{mod}{2}^{n+1})$. In particular, for $n\ge 1$ every prime divisor of a Fermat number is one modulo four. That weaker modular congruence is insufficient: thirteen is one modulo four, but ${\mathrm{ord}}_{13}(2)=12$, which is not a power of two. It divides no Fermat number at any index. The exact order criterion explains the whole family, instead of testing finitely many terms.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

A nonzero central idempotent is primitive central if it is not a sum of two nonzero orthogonal central idempotents. For an Artinian ring, it determines a block of an Artinian algebra. This differs from a primitive idempotent in the whole ring: for a matrix algebra of size greater than one over a field, $1$ is primitive central but is a sum of diagonal idempotents.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

A quadratic polynomial in one variable over a field has polynomial degree two and the displayed form. Its roots of a polynomial satisfy a quadratic equation. The polynomial and its associated polynomial function are distinct objects: over a finite field, different polynomials can define the same function.

Over a field with $q\ne 0$ and $q^2\ne 1$, this Hopf algebra has generators $E,F,K,K^{-1}$ and relations $KE=q^{2}EK$, $KF=q^{-2}FK$ and $EF-FE=(K-K^{-1})/(q-q^{-1})$. One comultiplication convention is $\Delta E=E\otimes K+1\otimes E$, $\Delta F=F\otimes 1+K^{-1}\otimes F$ and $\Delta K=K\otimes K$. Its counit takes $E,F$ to zero and $K$ to one. The antipode takes $E$ to $-E{K}^{-1}$, $F$ to $-KF$ and $K$ to $K^{-1}$.

For a field $k$ and $q\in k^{\times }$, the quantum plane is $k_q[X,Y]=k⟨X,Y⟩/(YX-qXY)$. Its ordered monomials $X^i{Y}^j$, $i,j\ge 0$, are a basis, and their multiplication is $X^i{Y}^j{X}^u{Y}^v=q^{ju}{X}^{i+u}{Y}^{j+v}$. Leading monomials show it is a noncommutative domain. Some sources interchange $q$ with $q^{-1}$ in the defining relation.

For a field of characteristic other than two and nonzero $a,b$, the quaternion algebra has basis $1,i,j,ij$ with the displayed relations. It is a four-dimensional central simple algebra. After adjoining a square root of $a$, diagonalizing $i$ and taking $j=\left(\begin{array}{cc}0& b\\ 1& 0\end{array}\right)$ identifies it with $M_2$. Quaternion conjugation fixes scalars and negates $i,j,ij$, giving norm $x_0^2-a{x}_1^2-b{x}_2^2+ab{x}_3^2$. An element is invertible exactly when its norm is nonzero. Thus the quaternion algebra is either a division algebra or a split matrix algebra. The familiar quaternion algebra is $(-1,-1{)}_{\mathbb{R}}$.

If $f$ is an irreducible polynomial over a field $F$, then a nonzero residue $[g]$ modulo $(f)$ has $\gcd (f,g)=1$. The Euclidean algorithm supplies $uf+vg=1$, giving inverse $[v]$ for $[g]$. Thus the quotient ring $F[X]/(f)$ is a field. If $\mathrm{deg}f=d$ and $F$ has $q$ elements, polynomial division supplies exactly $q^d$ residue classes.

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.

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 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 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.

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.