---- > [!definition] Definition. ([[primary ideal]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]] and $I \subset R$ an [[ideal]]. > > 1. $I$ is [[prime ideal|prime]] iff $\frac{R}{I} \neq 0$ and [[zero-divisor|zero-divisors]]$\left( \frac{R}{I} \right)=\{ 0 \}$; > 2. $I$ is [[radical of an ideal|radical]] iff $\text{Nil}\left( \frac{R}{I} \right)=(0)$; > 3. $I$ is **primary** if $\frac{R}{I} \neq 0$ and [[zero-divisor|zero-divisors]]$\left(\frac{R}{I} \right)=\text{Nil}\left(\frac{R}{I} \right)$. > > Note that $I$ is [[prime ideal|prime]] $\iff$ $I$ is primary and [[radical of an ideal|radical]]. > If $I$ is primary, then $\sqrt{ I } \in \text{Spec }R$.[^1] The converse holds if in fact $\sqrt{ I } \in \text{mSpec }R$ is [[maximal ideal|maximal]].[^2] For $\mathfrak{p} \in \text{Spec }R$, a **$\mathfrak{p}$-primary ideal** is a primary ideal $I$ such that $\sqrt{ I }=\mathfrak{p}$. > A **primary decomposition** of $I$ is an expression $I=\mathfrak{q}_{1} \cap \dots \cap \mathfrak{q}_{n}$, $\mathfrak{q}_{i}$ primary. Called **minimal** if the [[prime ideal|prime ideals]] $\sqrt{ \mathfrak{q}_{1} }, \dots, \sqrt{ \mathfrak{q}_{n} }$ are distinct and $I \subsetneq \bigcap_{j \neq i} \mathfrak{q}_{i}$ for all $1 \leq j \leq n$. Any primary decomposition can be made minimal, since finite intersections[^3] preserve primary-ness. > [!justification] Making Primary Decompositions Minimal. > - [ ] todo ^justification > [!basicexample] Example. (The place to start, but not representative of the whole story) > Let $R=\mathbb{Z}$. Then $(0)$ is a [[prime ideal]] (hence radical + primary). Let $0 \neq x \in \mathbb{Z}$. Then, > > 1. $\langle x \rangle$ is [[prime ideal|prime]] if and only if $|x|$ is a [[prime number]] > 2. $\langle x \rangle$ is [[radical of an ideal|radical]] if and only if $\langle x \rangle=\sqrt{ \langle x \rangle }=\text{(product of prime factors of }x)$, if and only if $x$ is square-free > 3. $\langle x \rangle$ is *primary* if and only if $x=p^{n}$, $n \geq 1$, for some prime number $p$. I.e., $\langle x \rangle=\langle p \rangle^{n}$. > > For example: $\langle 6 \rangle$ is is radical ($6=3 \cdot 2$ is square-free) but not prime nor primary. $\langle 9 \rangle$ is primary, but not radical. > > Thus, the *primary decomposition* of $\langle x \rangle$ is exactly what one would hope: the ideals $\langle p _{i}^{k_{i}}\rangle$ for $p_{1}^{k_{1}} \cdots p_{m}^{k_{m}}$ the decomposition of $|x|$ [[Fundamental Theorem of Arithmetic|decomposition]] of $x$ into a product of primes numbers. For example: $\langle 90 \rangle=\langle 2 \rangle \cap \langle 3^{2} \rangle \cap \langle 5 \rangle$ is a primary decomposition. However, in general rings the concept only has partial uniqueness properties (see [[uniqueness properties of primary decomposition]]). > > In $\mathbb{Z}$, an ideal is primary if and only if it is a power $\mathfrak{p}^{n}$ of a [[prime ideal]] $\mathfrak{p}$. [[primary decomposition in Dedekind domains behaves like prime factorization of integers|This holds more generally in]] [[Dedekind domain|Dedekind domains]]. But in general rings, the powers of prime ideals are not the same as the primary ideals. > [!basicexample] Example. (Not every primary ideal is a power of a prime ideal) > > Let $R=k[X,Y]$ be the two-indeterminate [[polynomial 4|polynomial ring]] over $k$ a [[field]]. [[prime ideal|Recall]] $\text{Spec }k[X,Y]=\underbrace{ \{ \langle X-a, Y-b \rangle : a, b \in k \} }_{ \text{mSpec }k[X,Y] } \cup \underbrace{ \{ \langle f \rangle : f \text{ irreducible} \} }_{ \text{principal ideals} } \cup \underbrace{ \{ (0) \} }_{ k[X,Y] \text{ an }I.D. }$ > > Put $\mathfrak{q}:=\langle X, Y^{2} \rangle$. *Claim:* $\mathfrak{q}$ is primary, but not a power of any $\mathfrak{p} \in \text{Spec }k[X,Y]$. > > Indeed, $\mathfrak{q}$ is primary because $R / \mathfrak{q}=\frac{k[Y]}{\langle Y^{2} \rangle}$, so every [[zero-divisor]] is a multiple of $Y$ and is hence [[nilpotent element of a ring|nilpotent]], and so $\mathfrak{q}$ is primary. Moreover, using properties in [[radical of an ideal|radical ideal]], have $\sqrt{ \mathfrak{q} }=\langle X,Y \rangle$: > > $\begin{align} > \sqrt{ \mathfrak{q} } &= \sqrt{ \langle X \rangle + \langle Y \rangle^{2} } \\ > &= \sqrt{\underbrace{ \sqrt{ \langle X \rangle } }_{ =\langle X \rangle } + \underbrace{ \sqrt{ \langle Y \rangle^{2} } }_{ =\langle Y \rangle } } \\ > &= \sqrt{ \underbrace{ \langle X \rangle + \langle Y \rangle }_{ \langle X,Y \rangle } } \\ > &= \langle X, Y \rangle > \end{align}$ > where the last equality follows since $\mathfrak{m}=\langle X,Y \rangle$ is a [[maximal ideal]]. Thus, if $\mathfrak{q}=\mathfrak{p}^{n}$ for a [[prime ideal]] $\mathfrak{p}$, then $\sqrt{ \mathfrak{q} }=\sqrt{ \mathfrak{p}^{n} }=\mathfrak{p}$, i.e., $\mathfrak{p}=\mathfrak{m}$. This cannot happen, as $\mathfrak{m}^{2} \subsetneq \mathfrak{q} \subsetneq \mathfrak{m}$: $\mathfrak{q}$ cannot be a power of $\mathfrak{m}$. > [!basicproperties] > For Noetherian local rings: [[primary sandwiches for Noetherian local rings]] ^properties ---- #### [^1]: Indeed, $\frac{R}{\sqrt{ I }}$ is an [[integral domain]], since for $a \in R$, $\begin{align} a+\sqrt{ I } \in \text{zero-divisors}\left( \frac{R}{\sqrt{ I }} \right)& \iff \ex b \in R - \sqrt{ I } \text{ s.t. } ab \in \sqrt{ I } \\ & \iff \ex m \text{ s.t. } a ^{m} b^{m} \in I \\ & \iff a ^{m} \in \text{zero-divisors}(R / I) \cancel{\text{ or } b^{m} \in I}^{\text{assumed }b \notin \sqrt{ I } }\\ &\iff a ^{mn} \in I \text{ or } b \in \text{Nil}_{R / I} \\ & \iff a \in \sqrt{ I } \\ & \iff a + \sqrt{ I }= 0_{ R / \sqrt{ I }} \\ \end{align}$ [^2]: TODO ~~Assume $\sqrt{ I }=\mathfrak{m}$ is [[maximal ideal|maximal]]. WTS $I$ is $\mathfrak{m}$-primary, i.e., that ($a + I \neq 0$) $a+I \in \text{Nil }R/I \iff a+I \in \text{zero-divisors}\left( \frac{R}{I} \right)$. Equivalently, that ($a \notin I$) $a \in \sqrt{ I } \iff \ex b \in R-I \text{ s.t. } ab \in I.$ One direction is clear: If $a^{n} \in I$ with $n$ minimal, then $n \geq 2$ because $a \notin I$ and we can put $b=a^{n-1}$. For the other direction, assume there exists $b \notin I$ with $ab \in I$. ~~ Since $I \subset \sqrt{ I }$, we have $ab \in \sqrt{ I }$; since $\sqrt{ I }$ is assumed [[prime ideal|prime]] either $a \in \sqrt{ I }$ or $b \in \sqrt{ I }$. [^3]: TODO Since $\sqrt{ I }=\mathfrak{m}$ is [[maximal ideal|maximal]], $\frac{R}{\mathfrak{m}}$ is a [[field]], hence $ab+\mathfrak{m}=0$ $\mathfrak{m}+a=\mathfrak{m}$ if $a$ is not a [[unit]]. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```