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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.

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.

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.

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.

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.

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

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

For an interval set avoiding one residue at each sieving prime, the Ramanujan sum at the forbidden Chinese remainder theorem residue gives a linear combination of Fourier samples equal to $\mu (d)|S|$. Cauchy-Schwarz inequality gives sample energy at least $|S{|}^2/\phi (d)$ for each squarefree modulus. Summing these energies and applying the analytic large sieve inequality gives the bound. If primes dividing $q$ are excluded, restrict the denominator to $(d,q)=1$; it is uniformly at least a constant times $(\phi (q)/q)\log (2Q)$.

A morphism of schemes between integral schemes can have a scheme-theoretic fibre that is neither reduced nor irreducible. For the map of affine lines given by $t=x^2(x-1{)}^2$, the fibre at $t=0$ has ring $k[x]/(x^2(x-1{)}^2)$. The Chinese remainder theorem identifies it with the product of two dual number rings, so it consists of two nonreduced double points. Its nilpotent structure is invisible to the set-theoretic fibre.