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Cosets and Lagrange. A left coset is $gH=\{gh:h\in H\}$. The left cosets of a subgroup partition a group, each with $|H|$ elements. Lagrange's theorem states $\boxed{|G|=[G:H]|H|}$ for a finite group, so every subgroup order divides the group order.

The 64-element group. Work in the commutative ring $R=\mathbb{Z}/4\mathbb{Z}$, which has zero divisors. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct matrix multiplication givesAll arithmetic is modulo four, so the set is closed. Associativity comes from matrix multiplication, the identity element is $A(0,0,0)$, andThe formula works in either multiplication order and requires no division. Hence these matrices form a group of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the Heisenberg group over a commutative ring.

The abelian subgroup. If $a=c$, write $h(a,b)=A(a,b,a)$. ThenThe identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an abelian subgroup. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with Lagrange's theorem. This is the abelian plane subgroup of the Heisenberg group modulo four.

More explicitly, $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of $R$, with kernel $H$. The four left cosets are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a normal subgroup.

Matrices squaring to the identity. The product formula givesThe first and third coordinates vanish only when $a,c\in \{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in \{0,2\}$. Therefore the complete set isIt is a subgroup: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the direct product of groups $C_2\times C_2\times C_2$. The seven nonidentity elements are involutions.

Finally, the last subset in the PDF is defined by $a\ne 0\ne b$, with no restriction on $c$. It has $3\cdot 3\cdot 4=36$ elements, which cannot be a subgroup because $36$ does not divide $64$. It also excludes the identity element. Thus

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.

If the positive characteristic were composite, say $r=uv$ with $1<u,v<r$, then $(u{1}_K)(v{1}_K)=0$ while neither factor is zero, by minimality of $r$. This contradicts a field having no zero divisors. Thus the characteristic is a prime $p$.

For a ring $R$, its ring characteristic is the least positive integer $n$ with $n{1}_R=0$, or zero if no positive such integer exists. Equivalently it is the least positive integer satisfying $nr=0$ for every $r\in R$, because $nr=(n{1}_R)r$. This additive formulation also defines the characteristic of a rng when an identity is not assumed.

Let $R$ be an integral domain. Since $1_R\ne 0$, its characteristic is not one. If its positive characteristic were a composite integer $n=ab$ with $1<a,b<n$, minimality would give $a{1}_R\ne 0$ and $b{1}_R\ne 0$, butThis contradicts the absence of zero divisors. ThereforeThis is the characteristic of an integral domain theorem. If an integral domain is defined without requiring an identity, the same conclusion follows from the additive definition: for a hypothetical composite characteristic $n=ab$, choose $r$ with $ar\ne 0$, which exists by minimality of $n$. Then $(ar)(bs)=n(rs)=0$ for every $s$, forcing $bs=0$ for every $s$ and contradicting minimality again.

Take $R=\mathbb{Z}/4\mathbb{Z}$, a quotient ring with four residue classes. The class $\overline{2}$ is nonzero but $\overline{2}\overline{2}=\overline{0}$. Thus $R$ has a zero divisor and is not an integral domain.

The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

There are six functions: the four power maps $x\mapsto x^k$ for $k=1,2,3,4$, and the two constant maps $0,1$. All calculations in this part are in the finite field ${\mathbb{Z}}_5$.

Put $e=f(1)$. Multiplicativity gives $e^2=e$, so the absence of zero divisors forces $e=0$ or $e=1$. If $e=0$, then $f(x)=f(x)f(1)=0$ for every $x$, giving only the constant-zero map.

Suppose $e=1$. For nonzero $x$, the equality $f(x)f(x^{-1})=1$ shows $f(x)\ne 0$. The nonzero elements form a cyclic group generated by $2$, in the order $1,2,4,3$. Choosing $a=f(2)$ determines every value on this group by $f(2^j)=a^j$. Each $a\in \{1,2,3,4\}$ satisfies $a^4=1$, so all four choices give consistent group homomorphisms.

Finally, $z=f(0)$ satisfies $z^2=z$ and $z=zf(x)$ for every $x$. The choice $z=0$ works for all four group maps. The choice $z=1$ forces every nonzero value to be $1$, giving one extra map, the constant-one map. Thus the count is $1+4+1=6$. The four zero-preserving possibilities are exactly the four displayed power maps. This is the special case with five elements of multiplicative self-maps of a finite field.

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.

The localization of a ring $A$ at the powers of $f$ is generated over $A$ by $1/f$. The presentation $A_f\cong A[t]/(ft-1)$ follows from the universal property of localization: both sides universally receive a ring homomorphism from $A$ making $f$ invertible. This proof works with zero divisors and includes the zero ring when $f$ is nilpotent. In particular, a principal open subscheme of an affine scheme has an inclusion which is a morphism of finite type.

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.

For a formal group law over any commutative ring $R$, the invariant differential of a formal group law satisfies $[p{]}^{\ast }\omega =p\omega$. Its coefficient is a unit in a ring, so $[p{]}^{\prime }(T)\in pR[[T]]$. If $[p](T)=\sum a_m{T}^m$, this means $m{a}_m\in pR$. When $p\nmid m$, the integer Bézout's identity implies $a_m\in pR$, even if $p$ is a zero divisor. Put those coefficients into $pf(T)$ and the exponents divisible by $p$ into $g(T^p)$.

A principal ideal ring is a ring whose left ideals and right ideals each have one generator. For a commutative ring, this says that every ideal is a principal ideal. The examples here use commutative rings. Unlike a principal ideal domain, it may have zero divisors: $\mathbb{Z}/4\mathbb{Z}$ has this property, with ideals $(0)$, $(2)$, and $(1)$. A finite direct product of rings that are principal ideal rings is again a principal ideal ring. Indeed, the coordinate idempotent elements show that every ideal is a product of coordinate ideals; a tuple of their generators generates the product.

For a positive integer modulus $m$, the residue class of an integer $a$ is the equivalence classIt contains all integers in the same position of the repeating cycle. For example,The residue class is this whole infinite collection; its canonical representative is the least nonnegative residue. Euclidean division gives exactly one such representative in $0,1,\dots ,m-1$, so there are exactly $m$ residue classes. When $m=1$, there is just one, $[0{]}_1=\mathbb{Z}$.

Define addition and multiplication of residue classes byThese definitions must give the same answer whichever integers represent the classes. If $a^{\prime }\equiv a$ and $b^{\prime }\equiv b$ modulo $m$, thenBoth differences are divisible by $m$, proving that the operations are well-defined. For example, $[9{]}_{12}+[5{]}_{12}=[2{]}_{12}$ and $[9{]}_{12}[5{]}_{12}=[9{]}_{12}$, since $45=3\cdot 12+9$. Replacing nine by $-3$ gives the same answers with smaller intermediate integers.

With these operations the residue classes form the finite commutative ring $\mathbb{Z}/m\mathbb{Z}$, a quotient ring. It need not be a field: $[6{]}_{12}[2{]}_{12}=[0{]}_{12}$ although neither factor is the zero class. These zero divisors explain why division requires a unit modulo n. Reducing after each multiplication is especially useful in modular exponentiation, where the unreduced powers can become enormous.

A square root of unity is an element $x$ of a ring with identity satisfying $x^2=1$. In a field of characteristic different from two, the only possibilities are $1$ and $-1$: the difference of squares identity $(x-1)(x+1)=0$ has no other roots because a field has no zero divisors. In a general ring more solutions may occur; modulo eight the four residue classes $1,3,5,7$ all have square one.

For a module $M$ over an integral domain $R$, its torsion submodule isThis is a submodule. If $rm=0$ and $sn=0$ with $r,s\ne 0$, then $rs(m+n)=0$ and $rs\ne 0$; also $r(am)=a(rm)=0$ for every $a\in R$. The integral domain hypothesis supplies both the nonzero product and commutativity. These are the conventions of ^[ref].

A mixed example is the direct sum $M={\mathbb{Z}}^2\oplus \mathbb{Z}/6\mathbb{Z}$. If a nonzero integer kills $(a,b,\overline{c})$, then $a=b=0$. Conversely, $6$ kills every $(0,0,\overline{c})$. ThusThe torsion submodule isolates the finite-order part while the quotient module keeps the independent integer directions. In general $M/T(M)$ is a torsion-free module: if $r(m+T(M))=0$ with $r\ne 0$, some $s\ne 0$ kills $rm$, so $sr\ne 0$ kills $m$.

The hypothesis matters. In the ring $R=\mathbb{Z}/6\mathbb{Z}$ viewed as a module over itself, the elements $\overline{2}$ and $\overline{3}$ are killed by nonzero scalars $\overline{3}$ and $\overline{2}$, but their sum $\overline{5}$ is a unit and is killed by no nonzero scalar. Thus using arbitrary nonzero scalars in a ring with zero divisors need not give a submodule. A common broader convention uses regular elements of a ring as annihilators; the integral domain convention here avoids that ambiguity.

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]