Ideal

An ideal is an additive subgroup of a ring that absorbs multiplication by ring elements.

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.

The whole ring $R$, viewed as an ideal. Elements $f_1,\dots ,f_r$ generate the unit ideal exactly when coefficients $g_i\in R$ give ${\sum }_i{g}_i{f}_i=1$.

If ${\sum }_i{g}_i{f}_i=1$, expand its power $r(N-1)+1$. By the pigeonhole principle, each monomial contains some $f_i^N$. Collecting terms expresses $1$ as a linear combination of these powers. This lets one use a common denominator exponent in localization arguments.

In ${\mathbb{F}}_2[u,v]/(u,v{)}^2$, the ideals $(u)$, $(v)$ and $(u+v)$ are pairwise incomparable and nonprime. Their union is the ideal $(u,v)$: the three lines cover its two-dimensional vector space over ${\mathbb{F}}_2$. This shows why a primality hypothesis matters in prime avoidance.

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 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.

In a Noetherian ring, an irreducible ideal is a primary ideal. In its quotient, if $xy=0$ and $x\ne 0$, choose $r$ after the annihilators of powers of $y$ stabilize. Then $(x)\cap (y^r)=0$, so irreducibility forces $y^r=0$. Combined with finite decomposition into irreducible ideals, this proves the Lasker–Noether theorem.

An ideal $I$ is radical when $f^m\in I$ for some positive integer $m$ implies $f\in I$. Over an algebraically closed field, the Hilbert Nullstellensatz identifies the vanishing ideal of $V(I)$ with the radical $\sqrt{I}$.

For an ideal $I$ and an element $a$ of a commutative ring,

An ideal $I$ is nilpotent when $I^r=0$ for some positive integer $r$.

An Oka family of ideals is a family $\mathcal{G}$ such that $I+(a)\in \mathcal{G}$ and $(I:a)\in \mathcal{G}$ imply $I\in \mathcal{G}$.

Every ideal maximal outside an Oka family is prime. If $xy\in I$ while $x,y\notin I$, both $I+(x)$ and $(I:x)$ properly contain $I$, contradicting the Oka condition.

A two-sided ideal $I$ of a noncommutative ring satisfies $RI\subseteq I$ and $IR\subseteq I$.

A square-zero ideal is an ideal $I$ satisfying $I^2=0$, so every product of two elements of $I$ vanishes.

For a unital associative algebra $A$ and prescribed $A$-bimodule $M$, an extension is $0\to M\to E\to A\to 0$ where $E$ is unital, $E\to A$ is unital, $M$ is a square-zero ideal, and the induced actions are the prescribed ones. Equivalences induce the identity on $A$ and $M$. A unital linear section produces a normalized Hochschild cocycle $\mu (a,b)=s(a)s(b)-s(ab)$. Conversely, $(a,m)(b,n)=(ab,an+mb+\mu (a,b))$ on $A\oplus M$ defines the extension. Changing section changes $\mu$ by a coboundary, giving classification by $H{H}^2(A,M)$.

If $I^2=0$, then $1+I=\{1+a:a\in I\}$ is an abelian subgroup of the unit group. Multiplication satisfies $(1+a)(1+b)=1+a+b$, and $(1+a{)}^{-1}=1-a$, so $a\mapsto 1+a$ identifies the additive group of $I$ with $1+I$.

An ideal $I$ is primary when $ab\in I$ and $a\notin I$ imply $b^n\in I$ for some positive integer $n$.

The radical $\sqrt{I}=\{x:x^n\in I\text{ for some }n\ge 1\}$ of a primary ideal is prime. In a number ring this forces a nonzero primary ideal to have only one prime in its unique factorization.

A minimal primary decomposition of an ideal $I$ is an irredundant finite intersection $I={\bigcap }_i{Q}_i$ of primary ideals whose radicals are pairwise distinct.

The distinct radicals in a minimal primary decomposition are exactly the associated primes of a module $R/I$, and hence are invariant. The injection into the sum of primary quotients proves one inclusion; an element supported on one component, supplied by irredundancy, proves the other. This does not imply uniqueness of embedded primary components.

A component in a minimal primary decomposition is embedded if its radical of an ideal strictly contains another radical occurring in that decomposition. Such components need not be unique. For example,in $k[x,y]$ gives different $(x,y)$-primary ideals as embedded components, with the same isolated component $(x)$.

An isolated prime of a minimal primary decomposition is a minimal member of the set of radicals of its primary components.

In a Noetherian ring, the primary component belonging to each isolated prime $\mathfrak{p}$ is uniquely determined byEmbedded primary components need not be unique.

A set $S\subseteq I$ generates an ideal $I$ when every element of $I$ is a finite sum of ring multiples of elements of $S$.

A principal ideal is generated by one element: $(a)=\{ra:r\in R\}$.

The product $IJ$ consists of finite sums of products $xy$ with $x\in I$ and $y\in J$.

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 set-theoretic intersection of any family of ideals is again an ideal.

Ideals $I$ and $J$ are comaximal when $I+J=R$.

A prime-power ideal has the form ${\mathfrak{p}}^m$ for a prime ideal $\mathfrak{p}$ and positive integer $m$.

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

Reduction modulo an ideal $I$ is the quotient homomorphism $R\to R/I$.

For a two-sided ideal $I$ of a ring $R$, the quotient $R/I$ has addition and multiplication induced from $R$.

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.

The radical of an ideal $I$ isAn ideal equal to its radical is a radical ideal.

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.

The nilradical of a commutative ring is its ideal of nilpotent elements, equivalently the radical of an ideal applied to its zero ideal.

In $R=k[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots )$, the nilradical is generated by all ${\bar{t}}_i$. Every element uses finitely many variables and is nilpotent, while each square-free product ${\bar{t}}_1\cdots {\bar{t}}_j$ is nonzero. Thus this nilradical is not a nilpotent ideal. A Noetherian ring cannot have this behavior: finitely many nilpotent generators give a nilpotent ideal.

For every commutative ring $R$,Every nilpotent element lies in every prime ideal. If $a$ is not nilpotent, an ideal maximal among those disjoint from $\{1,a,a^2,\dots \}$ is a prime ideal avoiding $a$, by the Zorn lemma and the maximality argument.