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In a Noetherian ring, an ideal $I$ disjoint from a nonempty multiplicative subset $S$ extends to an ideal $P$ maximal among those disjoint from $S$, by the ascending chain condition. It is proper. If $ab\in P$ but neither factor belongs to $P$, both $P+(a)$ and $P+(b)$ meet $S$. Multiplying such representatives puts an element of $S$ in $P$, a contradiction. Thus $P$ is a prime ideal. For a general commutative ring the same argument follows after applying the Zorn lemma to disjoint ideals; the union of a chain is disjoint from $S$.

The subset of the spectrum of a ring where the localization at a prime ideal is reduced. Since the nilradical commutes with localization, its complement is the support of a module $N=\mathrm{Nil}(R)$. When $N$ is a finitely generated module, vanishing of a finitely generated module near a prime givesso the locus is a Zariski-open subset. In particular this holds for every Noetherian ring.

The Rees module of an $R$-module $M$ with respect to an ideal $I$ is ${\mathcal{R}}_I(M)={⨁}_{j\ge 0}{I}^{j}M{t}^j$, a graded module over the Rees ring $\mathcal{R}(I)$. Generators of $M$ in degree zero generate this module. For $N\subseteq M$, the graded submodule ${⨁}_j(I^{j}M\cap N)t^j$ is finitely generated when $R$ is a Noetherian ring; a bound on its generators' degrees yields the Artin-Rees lemma.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.

A finitely generated module over a Noetherian ring admits a possibly infinite free resolution whose every term is finitely generated. Surject a finite free module onto it; the kernel is finitely generated because a finite free module is Noetherian. Repeat this construction on the successive kernels. If the original module has finite projective dimension $n$, its $n$th syzygy is projective by dimension shifting of the Ext functor, giving a length-$n$ resolution with finite projective terms.

A ring homomorphism from a ring to itself. Surjectivity need not imply injectivity in general, but a surjective ring endomorphism of a Noetherian ring is a ring automorphism.

For a commutative ring $k$ and a set $X$, the set $k^X$ of all functions $X\to k$ is a commutative ring under pointwise addition and multiplication. It is the direct product of rings with one copy of $k$ for each $x\in X$.

If $k$ is a field and $X$ is finite, every ideal is generated by some of the singleton indicator functions $e_x$. Indeed, whenever $f$ in the ideal has $f(x)\ne 0$, multiplication by $e_x/f(x)$ puts $e_x$ in the ideal. Every function in that ideal is a finite sum of its coordinate multiples of these $e_x$.

If $X$ is infinite and $k$ is a nonzero commutative ring, the functions with finite support form an ideal without a finite generating set of an ideal. A finite collection of such functions vanishes outside one finite subset, as does the ideal it generates, whereas a singleton indicator function outside that subset belongs to the ideal of all finitely supported functions. Thus, for a field $k$, $k^X$ is a Noetherian ring exactly when $X$ is finite.

For an additive semigroup $D$ with identity zero, the semigroup algebra $k[D]$ has a $k$-basis of formal symbols $t^q$, $q\in D$, and multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. For $D=\mathbb{Z}[1/2{]}_{\ge 0}$ it is the integral domain ${\bigcup }_{h\ge 0}k[t^{1/2^h}]$. Its positive-exponent ideal is a nonzero idempotent ideal, and the strictly ascending chain $(t)⊊(t^{1/2})⊊\cdots$ shows that the algebra is not a Noetherian ring.

For a Noetherian ring $A$, an injective module $I$ gives a flasque sheaf on $\mathrm{Spec}A$. If $U=\mathrm{Spec}A\setminus V(\mathfrak{a})$, clearing denominators on finitely many principal opens identifies $\Gamma (U,\tilde{I})$ with ${\underrightarrow{\lim }}_n{\mathrm{Hom}}_A({\mathfrak{a}}^n,I)$. Each map extends to $A$ by injectivity. Thus every section on $U$ extends globally, and all restriction maps are surjective. This concerns sheafification of an injective module, rather than the different assertion that an injective sheaf of modules is flasque.

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.

For a surjective ring endomorphism $f$ of a Noetherian ring, the ascending kernel chain $\ker f\subseteq \ker f^2\subseteq \cdots$ stabilizes. If $\ker f^m=\ker f^{m+1}$ and $f(a)=0$, choose $b$ with $f^m(b)=a$. Then $b\in \ker f^{m+1}=\ker f^m$, so $a=0$. Thus the endomorphism is a ring automorphism.

Let $M$ be a nonzero finitely generated module over a commutative Noetherian ring. Take a length-$n$ resolution with finite projective terms. If its top Ext functor with coefficients in $A$ vanished, the dual of its last injection would be surjective. The dual of a finite projective module is projective, so that surjection splits. Finite projective modules are isomorphic to their double duals, so dualizing gives a splitting of the original injection. The resolution then shortens, contradicting minimality of $n$. When $n=0$, a nonzero coordinate of a finite free module containing $M$ as a summand gives a nonzero map $M\to A$.