Primary ideals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 23 December 2011

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 $nil (R) = {nilpotent elements of R}$ .
The radical of an ideal $𝔞$ is the ideal $\sqrt{𝔞}$ of $R$ corresponding to the ideal $nil (R / 𝔞)$ 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 $A / 𝔞 \neq 0$ 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 $R$ .

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

		Proof.
	a. If $x y \in \sqrt{𝔞}$ then $(x y)^{m} \in 𝔞$ 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 $\sqrt{0} = ⋂ 𝔭) .$ 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. $□$

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

Let $R$ be a Noetherian ring.

		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 $(0)$ is primary in $A / 𝔞 .$ Let us show that the $0$ ideal is primary in a Noetherian ring. Let $x y = 0$ with $y \neq 0$ then $ann (x) \subseteq ann (x^{2}) \subseteq \dots$ must stabilize and so $ann (x^{n}) = ann (x^{n + 1})$ for large enough $n$ . Then $(x^{n}) \cap (y) = 0 .$ Now, if $a \in (y) \cap (x^{n})$ then $0 = a x = (b x^{n}) x = b x^{n + 1}$ and so $b \in ann (x^{n + 1}) = ann (x^{n}) .$ Thus $a = b x^{n} = 0 .$ Since $(x^{n}) \cap (y) = (0)$ is irreducible and $(y) \neq 0, (x^{n}) = 0 .$ So $(0)$ is primary. $□$

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 $1 \leq i \leq n, 𝔮_{i} ⊉ ⋂_{j \neq i} 𝔮_{j} .$
A decomposable ideal is an ideal with a primary decomposition.
$(𝔞; x) = {y \in A| y x \in 𝔞} .$

Any primary decomposition can be reduced to a minimal primary decomposition.
If $𝔞 = ⋂_{i} 𝔮_{i}$ is a minimal primary decomposition of $𝔞$ then ${\sqrt{𝔮_{i}}| 1 \leq i \leq n} = {\sqrt{(𝔞; x)}| x \in A} \cap Spec (A) .$
The minimal primes in ${\sqrt{𝔮_{i}}| 1 \leq i \leq n}$ are the minimal primes in $V (𝔞) .$

		Proof.
	a. b. c. If $𝔭$ is a prime ideal such that $𝔭 \supseteq 𝔞 = ⋂_{i} 𝔮_{i}$ then $𝔭 = \sqrt{𝔭} \supseteq ⋂ \sqrt{𝔮_{i}} .$ (If $𝔭 ⊉ \sqrt{𝔮_{i}}$ for all $i$ then there exists $x_{i} \in \sqrt{𝔮_{i}}, x_{i} \notin 𝔭$ for each $1 \leq i \leq n .$ So $x_{1} \dots x_{n} \in \sqrt{𝔮_{1}} \dots \sqrt{𝔮_{n}} \subseteq ⋂ \sqrt{𝔮_{i}} .$ But $x_{1} \dots x_{n} \notin 𝔭,$ since $𝔭$ is prime. So $𝔭 ⊉ ⋂_{i} \sqrt{𝔮_{i}} .$ ) So $𝔭 \supseteq \sqrt{𝔮_{i}}$ for some $i$ . Thus, by part b., $𝔭$ contains a minimal prime. $□$

Notes and References

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