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For a two-sided ideal $I$ of a ring $R$, the powers $I^n$ form a neighborhood base at zero for the adic topology. For an $R$-module $M$, use $I^{n}M$ on the left or $M{I}^n$ on the right. Convergence means that for every $n$ the differences eventually belong to the corresponding submodule. The formal power series ring $k[[t]]$ and formal power series module $M[[t]]$ are complete for $I=(t)$; convergence fixes each finite list of coefficients eventually. This is formal convergence, without an analytic condition on the size of coefficients.

An affine algebraic set is the common zero set $V(I)$ of an ideal in a polynomial ring over a field.

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.

For a regular local ring of embedding dimension $e$, a minimal maximal-ideal generating set induces the displayed polynomial-algebra isomorphism. It is surjective in every degree. A nonzero homogeneous kernel relation would bound the cumulative Hilbert function by $O(t^{e-1})$, contradicting the prime-chain lower bound for local length at $\dim R=e$. The graded ring is a domain; the Krull intersection theorem then shows the original ring is a domain by multiplying nonzero initial forms.

A central idempotent is an idempotent in the center of an associative algebra. It gives a decomposition $A=Ae\oplus A(1-e)$ into two-sided ideals, with products between the summands zero. The identities of these summands are $e$ and $1-e$.

For an ideal $H\subseteq R[[X]]$, let $A_n$ consist of coefficients of $X^n$ in members of $H$ with all lower coefficients zero. These are ideals of $R$ and $A_n\subseteq A_{n+1}$, by multiplication by $X$. If $R$ is Noetherian, this chain stabilizes. Choose series whose leading coefficients generate the finitely many distinct coefficient ideals, then cancel coefficients successively. The accumulated multipliers are formal power series, giving an ordinary finite ideal generating set for $H$, rather than merely a dense subideal.

For an ideal $I\subseteq R$, its conormal module is $I/I^2$, naturally an $R/I$-module. It records generators and relations to first order along the quotient $R/I$. If $I$ is the maximal ideal of a local ring, it is the cotangent space of a local ring over the residue field. For the augmentation ideal $(X_1-1,\dots ,X_r-1)$ of $A[X_1^{\pm 1},\dots ,X_r^{\pm 1}]$, reduction modulo $I^2$ gives $I/I^2\cong A^r$, with basis the classes of $X_i-1$.

A family of ideals satisfies the descending chain condition when every descending sequence of ideals eventually becomes constant. A ring is Artinian precisely when all its ideals satisfy this condition.

Every proper ideal of a unital commutative ring is contained in a minimal prime over that ideal. A maximal ideal supplies a prime containing it, and an intersection of a decreasing chain of primes is still prime. Zorn's lemma, with reverse inclusion, therefore supplies a minimal member. No Noetherian assumption is needed.

A ring homomorphism transports ideals forward by extension and backward by inverse image, or contraction.

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.

In a Noetherian ring, every radical of an ideal is an intersection of finitely many prime ideals. A maximal counterexample ideal $J$ cannot be prime. Choose $a,b\notin J$ with $ab\in J$; the strictly larger ideals $J+(a)$ and $J+(b)$ satisfy the result, and $\sqrt{J}=\sqrt{J+(a)}\cap \sqrt{J+(b)}$. This contradicts maximality. The unit ideal uses an empty intersection.