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Every commutative Artinian ring is Noetherian. Its nilradical $N$ is nilpotent by the nilradical of a commutative Artinian ring, and the Chinese remainder theorem identifies the reduced quotient $R/N$ with the finite product of the residue fields at the finitely many maximal ideals. Each $N^i/N^{i+1}$ is an Artinian module over that product of fields and hence a finite-dimensional vector space. The finite filtration by these layers makes every ideal finitely generated.

Every nonzero commutative Artinian ring is a finite product of Artinian local rings. The primary components of zero have distinct maximal radicals and are pairwise comaximal ideals, so the Chinese remainder theorem gives the product. Each local factor has nilpotent maximal ideal and finite composition length. These factors and the components of zero are unique.

An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

A Carmichael number is a composite number $N>1$ for whichEvery unit modulo $N$ satisfies the same modular congruence supplied by Fermat's little theorem for a prime number. The name refers to Robert Daniel Carmichael. The coprime qualification is essential: a nonunit can expose a prime factor through a greatest common divisor calculation. A Carmichael number is therefore a Fermat pseudoprime to every coprime base, rather than to every nonzero residue class.^[ref]

For example,The factors are distinct prime numbers. Their predecessor numbers $2,10,16$ all divide $560$, so Korselt criterion applies. More explicitly, for any integer $a$ coprime to $561$, Fermat's little theorem givesThe Chinese remainder theorem combines these into $a^{560}\equiv 1(\mathrm{mod}561)$. This proves all coprime bases pass from three short divisibility checks, rather than from a long table of sampled bases.

Equivalently, a composite $N$ is a Carmichael number exactly when $a^N\equiv a(\mathrm{mod}N)$ for every integer $a$, including nonunits. The equivalence follows from Korselt criterion: modulo each prime factor, either $a$ is zero or its nonzero power returns to one, and squarefreeness combines those congruences. This does not say a nonunit satisfies $a^{N-1}\equiv 1$. For instance $a=3$ gives $3^{560}\equiv 375(\mathrm{mod}561)$, while $3^{561}\equiv 3(\mathrm{mod}561)$; the greatest common divisor already supplies the factor three.^[ref]

The Carmichael function supplies another compact description: a composite $N$ is Carmichael exactly when $\lambda (N)\mid N-1$. For $561$, $\lambda (561)=\mathrm{lcm}(2,10,16)=80$ and $560=7\cdot 80$. In contrast, Euler's totient function gives $\phi (561)=320$, which does not divide $560$. The least common return exponent, rather than the number of units, controls this test.

The obstruction is useful to know when interpreting a Fermat primality test: even checking every coprime base cannot certify primality. The passing-base fraction for the Fermat test makes this failure quantitative. Stronger primality tests require additional structure, such as the squaring conditions in a Miller-Rabin primality test.

The historical record predates the name. Václav Šimerka's 1885 paper lists $561,1105,1729,2465,2821,6601,8911$ on printed page 224, together with factorizations. This is an early published list of examples, independently of their later name.^[ref]

Choose a primitive root $g$ modulo $p$ and write the nonzero coefficients $3,5,7$ as $g^a,g^b,g^c$. The Chinese remainder theorem gives an exponent $r$ congruent to $a,b,c$ modulo $3,5,7$, respectively. Then $n=g^r$ and the three roots $g^{(r-a)/3}$, $g^{(r-b)/5}$, $g^{(r-c)/7}$ give the displayed common nonzero residue. This technique converts multiplicative power constraints into simultaneous additive modular congruences for exponents.

The least divisor of a character's modulus through whose unit-group reduction the Dirichlet character factors. The Chinese remainder theorem reduces its construction to choosing the least exponent at each prime. Surjectivity of reduction on units makes the descended primitive Dirichlet character unique.

A congruence of squares says that two integers $a,b$ have the same square modulo a positive integer $N$. Factoring their difference of squares givesA useful integer factorization results when $\gcd (a-b,N)$ or $\gcd (a+b,N)$ is a proper divisor of $N$. Divisibility of a product does not generally force divisibility of either factor by all of $N$: different prime factors of $N$ can divide different parts of that product.

For $N=77$, $20^2-13^2=231=3\cdot 77$. ThereforeThus $77=7\cdot 11$. The prime seven divides the difference, while eleven divides the sum. These computations already give a complete prime factorization because seven and eleven are prime numbers.

The Chinese remainder theorem explains the hidden sign choices. The four congruence classes solving $z^2\equiv 15(\mathrm{mod}77)$ have representatives $13,20,57,64$. Modulo seven each is $1$ or $-1$; modulo eleven each is $2$ or $-2$. Two independent sign choices produce four roots. The inputs thirteen and twenty agree modulo seven and disagree by sign modulo eleven, allowing their sum and difference to separate the two prime factors.

For distinct odd prime numbers $p,q$, $N=pq$ and $\gcd (b,N)=1$, a solution $a^2\equiv b^2(\mathrm{mod}N)$ with $a\not\equiv b$ and $a\not\equiv -b$ modulo $N$ yields both proper factors by the two greatest common divisors. For a general composite integer, a proper divisor need not itself be prime; further factoring can remain. If $a\equiv b$ or $a\equiv -b(\mathrm{mod}N)$ in the coprime case, the two calculations return $N$ and one instead. A factor base helps generate relations, but does not ensure every resulting relation is useful.

For historical context, Pomerance describes Kraitchik's 1920s enhancement of the difference-of-squares approach associated with Fermat: seek a difference that is a multiple of $N$, rather than exactly $N$. His account also notes antecedents in Gauss and Seelhoff, so this is not an attribution of the whole idea to a single inventor. ^[ref].

Let $q_1,\dots ,q_r$ be distinct prime numbers and $a_1,\dots ,a_r$ positive integers. There are infinitely many positive integers $n$ with $q_i^{a_i}\mid n+i$ for every $i$. Indeed the moduli $q_i^{a_i}$ are pairwise coprime, so the Chinese remainder theorem solves $n\equiv -i(\mathrm{mod}{q}_i^{a_i})$ simultaneously. Adding multiples of their product gives arbitrarily large positive solutions. This allows long runs in which every integer has a large prescribed prime power divisor, even though neighboring integers are coprime.

For positive integers $m,n$, two finite cyclic groups combine into one cyclic group precisely when their group orders are coprime:The reason is that both cyclic counters must return to their identity elements simultaneously. If $a,b$ are generators of a group for the two factors, the order of a direct-product element gives $\mathrm{ord}(a,b)=\mathrm{lcm}(m,n)$. The pair generates all $mn$ elements exactly when this least common multiple is $mn$.

For $C_2\times C_3$, repeated addition of $(1,1)$ visitsAll six pairs appear before the return, so $(1,1)$ is a generator. In contrast, in $C_2\times C_2$ the same step gives only $(0,0),(1,1),(0,0)$. Every nonidentity element has order of a group element two, so none can generate all four pairs. This is the Klein four-group; there is no group isomorphism to $C_4$ even though both have four elements.

For a brief proof of the converse, each component order of a group element divides the corresponding factor group order. Every pair's order therefore divides $\mathrm{lcm}(m,n)$. If $\gcd (m,n)>1$, this is smaller than $mn$, so no pair can generate the whole product. The Chinese remainder theorem gives the explicit isomorphism in the coprime case: $t\mathrm{mod}mn\mapsto (t\mathrm{mod}m,t\mathrm{mod}n)$.

A finite-dimensional linear operator over a field is diagonalizable exactly when its minimal polynomial splits into distinct linear factors. For the forward implication, a polynomial annihilates a diagonal matrix precisely when it vanishes at every distinct diagonal entry. Conversely, if $m_T={\prod }_j(t-{\lambda }_j)$ with distinct roots, the Chinese remainder theorem yields a direct sum of the kernels of $T-{\lambda }_{j}I$. Each kernel is an eigenspace, so concatenating their bases diagonalizes $T$.

For a reduced ring with a nonempty disconnected spectrum of a commutative ring, choose a separation $V(I)\bigsqcup V(J)$. Disjointness gives $I+J=R$; covering the spectrum gives $IJ\subseteq \sqrt{(0)}=0$. For comaximal ideals, $I\cap J=IJ$. The Chinese remainder theorem then supplies a product of two nonzero quotient rings. Equivalently, if $i+j=1$ with $i\in I$, $j\in J$, then $i,j$ are nontrivial orthogonal idempotents. This connects topology of the spectrum to ring product decomposition by an idempotent without a Noetherian assumption.

Every odd composite integer has an Euler witness. For a nonsquarefree integer, a base congruent to $1+p$ modulo a repeated prime factor $p^2$ fails the Euler congruence by the binomial theorem. For a squarefree integer, use the Chinese remainder theorem to choose a base that is a quadratic nonresidue modulo one prime factor and $1$ modulo another. Its Jacobi symbol is $-1$, while its power is $1$ modulo the second factor.

The graph of the factorial function has a fixed-length first-order formula in the language of ordered rings interpreted on the natural numbers. A remainder code records $a_0=1$, $a_n=y$, and $a_{i+1}=(i+1)a_i$ for every $i<n$. Uniqueness of remainders proves $y=n!$; sufficiently large pairwise-coprime moduli and the Chinese remainder theorem supply a code for every input. Formula length is independent of the input variable's value, although substituting a numeral adds its encoding length.

For the Fermat primality test, a passing base satisfies $a^{n-1}\equiv 1(\mathrm{mod}n)$. Passing proves only this modular congruence, not that $n$ is a prime number. An odd composite number that passes a coprime base is a Fermat pseudoprime to that base. A Carmichael number passes every coprime base, so even testing all the units would not certify that $n$ is a prime number. ^[ref]

This small table uses only the bases $2,3,5,11$, each reduced modulo $n$ first. A numeric entry is $a^{n-1}\mathrm{mod}n$: one passes, while any other numeric entry gives a Fermat witness. A "factor" entry records a proper greatest common divisor and stops before modular exponentiation. The prime/composite classification is independently established by trial division or the displayed prime factorization, not by the passing entries. These four bases are illustrative, with no asserted universal coverage or error probability.

For $341=11\cdot 31$, $\gcd (2,341)=1$ and $2^{10}=1024=3\cdot 341+1$, so $2^{340}=(2^{10}{)}^{34}\equiv 1(\mathrm{mod}341)$. Yet $\gcd (3,341)=1$ and $3^{340}\equiv 56(\mathrm{mod}341)$, so three is a Fermat witness. One coprime base passes and another fails: being a Fermat pseudoprime depends on the base. Choosing eleven instead reveals the proper factor through $\gcd (11,341)=11$.

For $561=3\cdot 11\cdot 17$, every coprime base passes: each of $2,10,16$ divides $560$, so Fermat's little theorem gives $a^{560}\equiv 1$ modulo each prime factor; the Chinese remainder theorem combines the three modular congruences. Equivalently, the Carmichael function gives $\lambda (561)=80\mid 560$. Bases two and five have greatest common divisor one and pass, but $\gcd (3,561)=3$, $\gcd (11,561)=11$ and $\gcd (17,561)=17$ expose factors. The qualification "coprime" is therefore crucial even for a Carmichael number.

The Fermat primality test is useful because a Fermat witness can prove compositeness using repeated squaring without finding a factor. Its failure to certify a passing candidate motivates stronger primality testing methods such as the Miller-Rabin primality test.

A commutative Artinian ring has finite composition length as a module over itself. It has finitely many maximal ideals, and its Jacobson radical $J$ is nilpotent. The Chinese remainder theorem makes $R/J$ a finite product of fields; each Artinian module $J^i/J^{i+1}$ has finite-dimensional components over those fields. Adding their lengths along the finite radical filtration proves the assertion, and in particular the ascending chain condition.

A linear congruential generator has period $M$ for every seed if and only if $\gcd (c,M)=1$, every prime dividing $M$ divides $a-1$, and $4\mid M$ implies $4\mid a-1$. The coprimality of $c$ prevents the orbit from being trapped in a residue class; the multiplier conditions permit full-period lifting through each prime power. Combining the prime-power periods by the Chinese remainder theorem gives period $M$. For $M=12^k=2^{2k}{3}^k$, the multiplier criterion reduces to $a\equiv 1(\mathrm{mod}12)$, together with $\gcd (c,6)=1$. A positive increment alone is insufficient.

The Gödel beta function codes finite sequences by remainders:Choosing $c$ sufficiently large and divisible by the relevant small integers makes the moduli pairwise coprime, so the Chinese remainder theorem codes any prescribed finite sequence.

These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

For finitely many distinct prime ideals, one can prescribe finite valuation data simultaneously by the Chinese remainder theorem.

For an odd positive nonsquare integer $a$, there are infinitely many primes $p$ with Legendre symbol $(a/p)=-1$. Choose a residue $n$ modulo $a$ with Jacobi symbol $(n/a)=-1$ by making it a quadratic nonresidue at one prime occurring to odd multiplicity and a square at every other prime. The Chinese remainder theorem also imposes $n\equiv 1(\mathrm{mod}4)$ and $n\equiv 1$ at any finite list of previously found primes. Then Jacobi reciprocity gives $(a/n)=-1$, so a prime factor of $n$ has $(a/p)=-1$ and is outside that list.