## Primary ideals

Last update: 23 December 2011

## Primary ideals

Let $R$ be a commutative ring.

• A prime ideal is an ideal $𝔭$ of $R$ such that $R/𝔭$ is an integral domain.
• A maximal ideal is an ideal $𝔪$ of $R$ such that $R/𝔪$ is a field.

Let $R$ be a ring.

• The nilradical of $R$ is the ideal $\mathrm{nil}\left(R\right)=\left\{nilpotent elements of\phantom{\rule{.5em}{0ex}}R\right\}$.
• The radical of an ideal $𝔞$ is the ideal $\sqrt{𝔞}$ of $R$ corresponding to the ideal $\mathrm{nil}\left(R/𝔞\right)$ in $R/𝔞$.

Let $R$ be a commutative ring. Let $𝔭$ be a prime ideal of $R$.

• A primary ideal is an ideal $𝔞$ of $R$ such that $A/𝔞\ne 0$ and every zero divisor in $A/𝔞$ is nilpotent.
• A $𝔭-$primary ideal is a primary ideal $𝔮$ such that $\sqrt{𝔮}=𝔭$.

Let $R$ be a commutative ring and let $𝔞$ be an ideal of $R$.

1. The ideal $𝔞$ is primary if and only if for every $a,b\in R$ such that $ab\in \sqrt{𝔞}$ either $a\in \sqrt{𝔞}$ or $b\in \sqrt{𝔞}$.
2. If $\sqrt{𝔞}$ is maximal then $𝔞$ is primary.
3. If $𝔞$ is primary then $\sqrt{𝔞}$ is the minimal prime ideal containing $𝔞$.

 Proof. a. If $xy\in \sqrt{𝔞}$ then $\left(xy{\right)}^{m}\in 𝔞$ for some $m\in {ℤ}_{>0}$. So ${x}^{n}\in 𝔞$ or ${y}^{n}\in 𝔞$ for some $n\in {ℤ}_{>0}$. So $x\in \sqrt{𝔞}$ or $y\in \sqrt{𝔞}$. b. If $\sqrt{𝔞}=𝔪$ is maximal then $\sqrt{0}=𝔪/𝔞$ in $R/𝔞.$ So $A/𝔞$ has only one prime ideal (since $\sqrt{0}=\bigcap 𝔭\right).$ So every element of $A/𝔞$ is a unit or is nilpotent. So every zero divisor of $A/𝔞$ is nilpotent. So $𝔞$ is primary. c. $\square$

Let $R$ be a commutative ring. An irreducible ideal is an ideal $𝔞$ of $R$ such that $if 𝔞 = 𝔟 ∩ 𝔠 then 𝔞 = 𝔟 or 𝔞 = 𝔠 .$

Let $R$ be a Noetherian ring.

1. Every ideal is a finite intersection of irreducible ideals.
2. Every irreducible ideal is primary.

 Proof. a. If there is an ideal which is not a finite intersection of irreducible then, by Zorn's lemma, the set of ideals which are not a finite intersection of irreducibles has a maximal element $𝔞$. Then $𝔞$ is reducible and $𝔞=𝔟\cap 𝔠$ with $𝔟$ and $𝔠$ strictly larger than $𝔞$. So $𝔟$ and $𝔠$ are finite intersections of irreducibles. So $𝔞$ is a finite intersection of irreducibles. b. To show $𝔞$ is primary, it is sufficient to show that $\left(0\right)$ is primary in $A/𝔞.$ Let us show that the $0$ ideal is primary in a Noetherian ring. Let $xy=0$ with $y\ne 0$ then $ann ( x ) ⊆ ann ( x2 ) ⊆ ⋯$ must stabilize and so $\mathrm{ann}\left({x}^{n}\right)=\mathrm{ann}\left({x}^{n+1}\right)$ for large enough $n$. Then $\left({x}^{n}\right)\cap \left(y\right)=0.$ Now, if $a\in \left(y\right)\cap \left({x}^{n}\right)$ then $0=ax=\left(b{x}^{n}\right)x=b{x}^{n+1}$ and so $b\in \mathrm{ann}\left({x}^{n+1}\right)=\mathrm{ann}\left({x}^{n}\right).$ Thus $a=b{x}^{n}=0.$ Since $\left({x}^{n}\right)\cap \left(y\right)=\left(0\right)$ is irreducible and $\left(y\right)\ne 0,\phantom{\rule{.5em}{0ex}}\left({x}^{n}\right)=0.$ So $\left(0\right)$ is primary. $\square$

Let $A$ be a commutative ring. Let $𝔞$ be an ideal of $A$.

• A primary decomposition of $𝔞$ is $𝔞 = ⋂ i=1 n 𝔮 i , with 𝔮 i primary ideals.$
• A minimal primary decomposition is a primary decomposition of $𝔞$ such that all ${𝔮}_{i}$ are primary ideals, all $\sqrt{{𝔮}_{i}}$ are distinct, and, for all $1\le i\le n,\phantom{\rule{.5em}{0ex}}{𝔮}_{i}⊉\bigcap _{j\ne i}{𝔮}_{j}.$
• A decomposable ideal is an ideal with a primary decomposition.
• $\left(𝔞;x\right)= {y\in A|yx\in 𝔞} .$

1. Any primary decomposition can be reduced to a minimal primary decomposition.
2. If $𝔞=\bigcap _{i}{𝔮}_{i}$ is a minimal primary decomposition of $𝔞$ then $𝔮 i 1 ≤ i ≤ n = ( 𝔞 ; x ) x ∈ A ∩ Spec ( A ) .$
3. The minimal primes in ${\sqrt{{𝔮}_{i}}|1\le i\le n}$ are the minimal primes in $V\left(𝔞\right).$

 Proof. a. b. c. If $𝔭$ is a prime ideal such that $𝔭\supseteq 𝔞=\bigcap _{i}{𝔮}_{i}$ then $𝔭 = 𝔭 ⊇ ⋂ 𝔮 i .$ (If $𝔭⊉\sqrt{{𝔮}_{i}}$ for all $i$ then there exists ${x}_{i}\in \sqrt{{𝔮}_{i}},{x}_{i}\notin 𝔭$ for each $1\le i\le n.$ So ${x}_{1}\cdots {x}_{n}\in \sqrt{{𝔮}_{1}}\cdots \sqrt{{𝔮}_{n}}\subseteq \bigcap \sqrt{{𝔮}_{i}}.$ But ${x}_{1}\cdots {x}_{n}\notin 𝔭,$ since $𝔭$ is prime. So $𝔭⊉\bigcap _{i}\sqrt{{𝔮}_{i}}.$) So $𝔭\supseteq \sqrt{{𝔮}_{i}}$ for some $i$. Thus, by part b., $𝔭$ contains a minimal prime. $\square$

## Notes and References

Where are these from?

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