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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 $S$.

For a nonzero ideal $I$ in a Dedekind domain, $v_P(I)$ is the exponent of the nonzero prime ideal $P$ in the unique factorization of ideals in a Dedekind domain. Then $I\subseteq J$ if and only if $v_P(I)\ge v_P(J)$ for every $P$, and $v_P(I+J)=\min (v_P(I),v_P(J))$.

The prime radical is the intersection of all prime ideals of a noncommutative ring. For a nonzero ring satisfying the ascending chain condition on two-sided ideals, there exist prime ideals of a noncommutative ring $P_1,\dots ,P_r$ with $P_1\cdots P_r=0$. Indeed, a maximal counterexample among two-sided ideals could not itself be a prime ideal of a noncommutative ring, and two larger witness ideals would contradict its maximality. Every prime ideal of a noncommutative ring contains one $P_i$, so the intersection equals ${\bigcap }_i{P}_i$, and its $r$th power is zero. This nilpotence assertion fails without a chain condition.

Let $A,B$ be finite-dimensional commutative algebras, let $I\subset A$ and $J\subset B$ be ideals, and let $f:I\to J$ be a surjective algebra homomorphism. The ideals need not have identities. Then $f$ gives a bijection between the primitive idempotents of $A$ lying in $I$ with nonzero image and the primitive idempotents of $J$.

Decompose $A$ into its commutative Artinian local rings $A_i$, with identities $e_i$. Each $I\cap A_i$ is either all of $A_i$, in which case $e_i\in I$, or a proper ideal and therefore a nilpotent ideal. The sum of the latter ideals is nilpotent, so its image contributes no idempotents. Each nonzero $f(e_i)$ has local corner algebra $f(e_i)J=f(A_i)$, a quotient of $A_i$, so is primitive. These idempotents are orthogonal, and the remaining part of $J$ is nilpotent; hence they are exactly its primitive idempotents. This statement does not require the entire kernel of $f$ to be nilpotent.

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

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.

A right ideal $I$ in a unital associative algebra $A$ is a vector subspace with $IA\subseteq I$, equivalently a submodule of the right regular module. It need not be a left ideal. Its finite generation refers to right multiplication by elements of $A$.

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.

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

A topological ring is a ring equipped with a topology making addition, additive inversion and multiplication continuous functions. An adic topology is obtained from powers of an ideal.

Every ideal of a Dedekind domain is generated by at most two elements. Given a nonzero ideal $I$ and any prescribed nonzero $a\in I$, the Chinese remainder theorem supplies $b\in I$ with $v_P((b))=v_P(I)$ at every prime ideal dividing $(a)$. The prime-ideal valuation in a Dedekind domain then gives $I=(a,b)$.

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.

Every nonzero proper ideal of a Dedekind domain has a unique factorization, up to reordering, as a finite product of positive powers of distinct nonzero prime ideals. The unit ideal is the empty product. This is distinct from unique factorization of elements.

In $k[u,v]$, the ideal $I=(u^2,uv)$ has radical of an ideal $(u)$, a prime ideal, and hence a unique minimal prime ideal. Nevertheless $uv\in I$, $u\notin I$, and $v^j\notin I$ for all $j\ge 1$, so $I$ is not a primary ideal. The same example works after adjoining inverses of $1+u$ and $1+v$; thus it also gives an example in a two-variable Laurent polynomial ring.

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 ${\sum }_i{g}_i{f}_i=1$.