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If $I$ is an $\mathfrak{m}$-primary ideal generated by $n>0$ elements in a Noetherian local ring and $L=\mathrm{length}(R/I)$, then $I^j/I^{j+1}$ is a quotient of $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $R/I$. Summing their lengths gives the displayed bound. Since $I\subseteq \mathfrak{m}$, it also bounds $\mathrm{length}(R/{\mathfrak{m}}^t)$. Together with the prime-chain lower bound for local length, localization at a minimal prime proves the Krull height theorem.

The height of a prime ideal is the supremum of lengths of strict prime chains ending at it, equivalently the Krull dimension of its localization at a prime ideal. The Krull height theorem bounds the height of a minimal prime over an $n$-generated ideal by $n$.

For an algebraically closed field $k$ and an ideal $I\subseteq k[x_1,\dots ,x_n]$, the strong Hilbert Nullstellensatz isIts weak form says that an ideal with empty affine zero set is the unit ideal. Equivalently, if finitely many polynomials have no common zero, a polynomial combination of them equals one.

An ideal in a graded ring is homogeneous when it is generated by homogeneous elements, equivalently when every homogeneous component of each of its elements also belongs to the ideal.

In $K=\mathbb{Q}(\sqrt{-14})$, the Minkowski bound for ideal classes is less than five. The ideals $\mathfrak{p}=(2,\sqrt{-14})$ and $\mathfrak{q}=(3,1+\sqrt{-14})$ generate all classes, with ${\mathfrak{p}}^2=(2)$ and $(2-\sqrt{-14})=\mathfrak{p}{\mathfrak{q}}^2$. No element has field norm two, so $[\mathfrak{p}]$ has order two and $[\mathfrak{q}]$ has order four.

For ideals $I,J$ in a commutative ring, the ideal quotient is $(I:J)=\{r\in R:rJ\subseteq I\}$. Closure under subtraction and multiplication by arbitrary ring elements proves it is an ideal. If $I\subseteq J=(a)$, then $I=a(I:J)$: write $x=ar\in I$ and observe that $rJ=xR\subseteq I$. Consequently, if also $(I:J)=(b)$, then $I=(ab)$ is a principal ideal.

An ideal $I$ is idempotent when $I^2=I$. Then $I^j=I$ for every $j\ge 1$, so a nonzero idempotent ideal gives a nonzero intersection of all its powers. A semigroup algebra with arbitrarily divisible positive exponents supplies examples in a non-Noetherian integral domain. If $I$ is finitely generated, the determinant trick applied to $I=I^2$ produces $r\in I$ such that $(1+r)I=0$.

The polynomial ring $\mathbb{Z}[X]$ is a unique factorization domain by Gauss lemma for polynomials, but its ideal $(2,X)$ is not principal. Any generator would divide both 2 and $X$, hence be a unit: a divisor of 2 has degree zero, and a constant dividing $X$ is $\pm 1$. Yet evaluation at zero followed by reduction modulo 2 annihilates the ideal and not 1. It embeds in the principal ideal domain $\mathbb{Q}[X]$, illustrating that the property need not pass to a subring.

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

A proper ideal $I$ is irreducible if $I=J\cap K$ implies $I=J$ or $I=K$. In a Noetherian ring, every irreducible ideal is a primary ideal; every proper ideal is a finite intersection of irreducible ideals.

The kernel of a ring homomorphism $\phi :R\to S$ is the ideal $\{r\in R:\phi (r)=0\}$. The ring homomorphism is injective exactly when this ideal is zero.

Every proper ideal in a Noetherian ring is a finite intersection of primary ideals. This existence theorem does not give uniqueness of the individual components: embedded primary components may vary.

For a Noetherian ring, injectivity of a module can be tested at all prime localizations. By the Baer criterion, test ${\mathrm{Ext}}_R^1(R/J,M)$ for every ideal $J$. Localization of Ext over a Noetherian ring identifies its localization with the corresponding ideal test over $R_P$. Every ideal of $R_P$ is extended from its contraction, and localization detects zero elements applies to the resulting Ext module even when it is not finitely generated. These observations prove both directions.

An endomorphism ring is local when its nonunits form a proper two-sided ideal; commutativity is not required. The Fitting lemma gives this property for an indecomposable finite-length module. If two nonunits had an invertible sum, multiplication by its inverse would give $I=f+g$ with noninvertible $f,g$. Fitting makes $f$ nilpotent, so $I-f=g$ would be invertible, a contradiction. Products with a nonunit are nonunits by finite-length injectivity/surjectivity. The quotient by the nonunit ideal is a division ring.

For a Noetherian ring and a proper ideal $I$, every minimal prime $P$ over $I$ is an associated prime of a module $R/I$. The localized quotient has one prime, so its finitely generated maximal ideal is nilpotent and its socle is nonzero. An element with annihilator $P{R}_P$ can be lifted to the quotient. Clearing denominators for a finite generating set of $P$ gives a nonzero element with annihilator exactly $P$.