Associated Primes and Primary Decomposition Examples

Let $\bar{x}=x+I$. Then $f\bar{x}=0$ exactly when $fx\in (x^2,xy)=x(x,y)$, that is, exactly when $f\in (x,y)$. Hence ${\mathrm{Ann}}_R(\bar{x})=(I:x)=(x,y)\mathrm{.}$

Let $\bar{y}=y+I$. Then $f\bar{y}=0$ exactly when $fy\in (x^2,xy)=x(x,y)$. Because $x$ and $y$ are relatively prime in $k[x,y]$, this happens exactly when $x$ divides $f$. Thus ${\mathrm{Ann}}_R(\bar{y})=(I:y)=(x)\mathrm{.}$ Both ideals are prime.

Every class in $R\mathrm{/}I$ has a unique representative $g(y)+cx$ with $g(y)\in k[y]$ and $c\in k$, because $x^2=xy=0$ modulo $I$. Let $r=g(y)+cx\mathrm{\notin }I$. If $g=0$, then $c\mathrm{\ne }0$ and $(I:r)=(x,y)$ by step 1.1. Assume $g\mathrm{\ne }0$. For $a(y)+dx\in R\mathrm{/}I$, multiplication gives $(a(y)+dx)(g(y)+cx)=a(y)g(y)+(a(0)c+dg(0))x\mathrm{.}$ If this vanishes, then $a(y)g(y)=0$ in the domain $k[y]$, so $a=0$. Thus $(I:r)=(x)$ when $g(0)=0$, while $(I:r)=I$ when $g(0)\mathrm{\ne }0$. The ideal $I$ is not prime because $x^2\in I$ but $x\mathrm{\notin }I$. Hence the only prime colon ideals are $(x)$ and $(x,y)$.

By [L1] and step 3.1, the associated primes of $R\mathrm{/}I$ are exactly $(x)$ and $(x,y)$.

In $A$, every quadratic monomial vanishes, so ${\mathfrak{m}}^2=(\bar{x},\bar{y}{)}^2=0.$ Consequently $(\mathfrak{m}\mathrm{/}J{)}^2=0$ in the quotient ring $A\mathrm{/}J$.

The quotient $A\mathrm{/}J$ is local with maximal ideal $\mathfrak{m}\mathrm{/}J$. Any zero divisor in $A\mathrm{/}J$ is a nonunit, hence lies in the maximal ideal $\mathfrak{m}\mathrm{/}J$. By step 1.1 every element of $\mathfrak{m}\mathrm{/}J$ is square-zero, so every zero divisor on $A\mathrm{/}J$ acts nilpotently.

Fact [L1] now shows that $J$ is primary, and its radical is $\mathfrak{m}$ by assumption. Hence $J$ is $\mathfrak{m}$-primary.

The field $k$ is Noetherian because its only ideals are $0$ and $k$, so [L3] makes $R=k[x,y]$ Noetherian. As an $R$-module, $R$ is finitely generated by $1$. The inclusion $(x^2,xy)\subseteq (x)\cap (x,y{)}^2$ is immediate. Conversely, every element of $(x)\cap (x,y{)}^2$ has the form $xf$ with $f\in (x,y)$, hence lies in $(x^2,xy)$. So $(x^2,xy)=(x)\cap (x,y{)}^2$.

Since a field is an integral domain, [L4] makes $k[y]\cong R\mathrm{/}(x)$ an integral domain, so [L5] makes $(x)$ prime. If multiplication by a class $\bar{a}$ on the domain $R\mathrm{/}(x)$ has nontrivial kernel, then $\bar{a}=0$; hence that multiplication is zero and therefore nilpotent. Also ${\mathrm{Ann}}_R(R\mathrm{/}(x))=(x)$, whose radical is $(x)$. Thus [L6] shows that $(x)$ is $(x)$-primary.

Put $Q=(x,y{)}^2$. Every class in $R\mathrm{/}Q$ has the form $c+\alpha \bar{x}+\beta \bar{y}$. If $c\mathrm{\ne }0$, this class is a unit, with inverse $c^{-1}-c^{-2}(\alpha \bar{x}+\beta \bar{y})$ because $(\bar{x},\bar{y}{)}^2=0$. Therefore every zero divisor lies in $(\bar{x},\bar{y})$ and is square-zero, so every zero divisor on $R\mathrm{/}Q$ acts nilpotently. Moreover ${\mathrm{Ann}}_R(R\mathrm{/}Q)=Q$ and $\sqrt{Q}=(x,y)$: every element of $(x,y)$ has square in $Q$, while $f^n\in Q$ forces the constant term of $f$ to vanish. Finally, $R\mathrm{/}(x,y)\cong k$ is a domain, so [L5] makes $(x,y)$ prime. Thus [L6] shows that $(x,y{)}^2$ is $(x,y)$-primary.

The two prime radicals $(x)$ and $(x,y)$ are distinct, with $(x)⊊(x,y)$. The decomposition from step 1.1 is irredundant: $y^2\in (x,y{)}^2\setminus (x)$, so the component $(x)$ is not redundant, while $x\in (x)\setminus (x,y{)}^2$, so the component $(x,y{)}^2$ is not redundant. By [L7], the displayed primary decomposition is minimal, and its $(x)$-primary component is isolated because $(x)$ is the smaller of the two radicals.

At the prime $(x)$, the element $y$ becomes a unit. Since $y^2\in (x,y{)}^2$, fact [L1] and steps 1.1–2.1 give $((x,y{)}^2{)}_{(x)}=R_{(x)}$, while $(x{)}_{(x)}$ survives. Hence $(x^2,xy{)}_{(x)}=(x{)}_{(x)}\mathrm{.}$ Since step 2.1 proves that $(x)$ is the isolated component of a minimal primary decomposition, fact [L2] says it contracts back to $(x)$.

At the maximal ideal $(x,y)$, neither prime radical meets the denominator set, so [L1] and steps 1.1–2.1 preserve both primary components. Thus the whole decomposition survives in $R_{(x,y)}$.

This computation shows concretely how localization removes exactly the components whose radicals meet the denominator set.

The ideal $(x)$ is prime because $R\mathrm{/}(x)\cong k[y]$. The quotients $R\mathrm{/}(x^2,y)\cong k[x]\mathrm{/}(x^2)$ and $R\mathrm{/}(x,y{)}^2$ are local rings whose maximal ideals are square-zero, so every zero divisor is nilpotent. Hence $(x^2,y)$ and $(x,y{)}^2$ are $(x,y)$-primary, while $(x)$ is $(x)$-primary. Thus the displayed intersection is a primary decomposition.

Since $(x,y{)}^2=(x^2,xy,y^2)\subseteq (x^2,y)$, the factor $(x^2,y)$ is redundant, and the repeated copy of $(x,y{)}^2$ is redundant as well. Fact [L1] therefore cleans the four-term intersection down to $I=(x)\cap (x,y{)}^2\mathrm{.}$

The two surviving radicals are $(x)$ and $(x,y)$, already distinct. If one first combines the two equal-radical components $(x,y{)}^2$ and $(x,y{)}^2$, [L2] replaces them by their intersection, which is again $(x,y{)}^2$. Hence both cleanup orders lead to the same two-term presentation.

Finally, $(x)\cap (x,y{)}^2=(x)\cap (x^2,xy,y^2)=(x^2,xy),$ because an element in the intersection has the form $xf$ with $f\in (x,y)$. The decomposition is irredundant: $x\in (x)\setminus (x,y{)}^2$ and $y^2\in (x,y{)}^2\setminus (x)$. Together with the distinct radicals, this proves minimality.

This example isolates the two routine cleanup moves in a primary decomposition: deletion of redundant components and combination of equal radicals.

The inclusion $(x^2,xy)\subseteq (x)\cap (x,y{)}^2$ is immediate, and every element of the right side has the form $xf$ with $f\in (x,y)$, so $(x^2,xy)=(x)\cap (x,y{)}^2$. Likewise $(x^2,xy)\subseteq (x)\cap (x^2,y)$, and if $xf\in (x^2,y)$ then $xf=x^{2}a+yb$ for some $a,b$. The right-hand side lies in $(x)$, so $yb\in (x)$; thus $b=xc$ and $xf=x(xa+yc)\in (x^2,xy)$. Hence $(x^2,xy)=(x)\cap (x^2,y)$.

The field $k$ is Noetherian because its only ideals are $0$ and $k$, so [L3] makes $R=k[x,y]$ Noetherian. The ideal $(x)$ is prime because $R\mathrm{/}(x)\cong k[y]$. The quotients $R\mathrm{/}(x,y{)}^2$ and $R\mathrm{/}(x^2,y)\cong k[x]\mathrm{/}(x^2)$ are local rings with square-zero maximal ideals, so $(x,y{)}^2$ and $(x^2,y)$ are $(x,y)$-primary. Both decompositions are irredundant: $x\in (x)$ lies in neither $(x,y{)}^2$ nor $(x^2,y)$, while $y^2\in (x,y{)}^2\setminus (x)$ and $y\in (x^2,y)\setminus (x)$. Hence both displayed decompositions are minimal primary decompositions with prime radicals $(x)$ and $(x,y)$.

Fact [L1] now predicts exactly the common radical set from step 2.1. The second component differs: one decomposition uses $(x,y{)}^2$, the other uses $(x^2,y)$.

Localizing at the minimal prime $(x)$ kills the $(x,y)$-primary component in either decomposition, so [L2] recovers the same isolated component $(x)$ from both. The difference therefore lies only in the embedded component.

This is the standard warning that first uniqueness does not imply componentwise uniqueness of embedded pieces.

The only element of the zero module is $0$, and its annihilator is the whole ring $R$, which is not a prime ideal of itself. Therefore [L1] gives ${\mathrm{Ass}}_R(0)=\mathrm{\varnothing }$.

Every localization of the zero module is again the zero module, so [L2] gives ${\mathrm{Supp}}_R(0)=\mathrm{\varnothing }$.

The inclusion $0\subseteq 0$ is not proper, so [L3] shows that it is not a candidate for the primary-submodule definition. This is the exact boundary that prevents a vacuous use of primaryity at the zero module.

Thus the zero module sits outside associated-prime and primaryity claims in exactly the stated way.