“How weird was it that so many bits and pieces, all diverse, could make someting whole. Something with potential. Perfect.”
―Sarah Dessen
Remark12.45.
We refine our decomposition theory once again, and introduce primary decompositions of ideals. One of the fundamental classical results in commutative algebra is the fact that every ideal in any noetherian ring has a primary decomposition. This can be thought of as a generalization of the Fundamental Theorem of Arithmetic:
Theorem12.46.Fundamental Theorem of Arithmetic.
Every nonzero integer \(n \in \mathbb{Z}\) can be written as a product of primes: there are distinct prime integers \(p_1, \ldots, p_n\) and integers \(a_1, \ldots, a_n \geqslant 1\) such that
\begin{equation*}
n = p_1^{a_1} \cdots p_n^{a_n}.
\end{equation*}
Moreover, such a a product is unique up to sign and the order of the factors.
Remark12.47.
We will soon discover that such a product a primary decomposition, perhaps after some light rewriting. But before we get to the and the of primary decomposition, it is worth discussing the . If we wanted to extend the Fundamental Theorem of Arithmetic to other rings, our first attempt might involve irreducible elements. Unfortunately, we don’t have to go far to find rings where we write elements as a unique product of irreducibles up to multiplying by a unit.
are two ways to write \(6\) as a product of irreducible elements. In fact, we cannot obtain \(2\) or \(3\) by multiplying \(1 + \sqrt{-5}\) or \(1 - \sqrt{-5}\) by a unit.
Definition12.49.Primary Ideal.
We say that an ideal is primary if
\begin{equation*}
xy\in I \implies x\in I \text{ or } y\in \sqrt{I}.
\end{equation*}
We say that an ideal is \(P\)-primary, where \(P\) is prime, if \(I\) is primary and \(\sqrt{I}=P\text{.}\)
Lemma12.50.Radical of Primary Ideal is Prime.
The radical of a primary ideal is prime.
Proof.
Suppose that \(Q\) is primary and \(xy \in \sqrt{Q}\text{.}\) Then \(x^n y^n\in Q\) for some \(n\text{.}\) If \(y\notin \sqrt{Q}\text{,}\) then \(y^n\notin \sqrt{Q}\text{.}\) Since \(Q\) is primary, we must have \(x^n\in Q\text{,}\) so \(x\in \sqrt{Q}\text{.}\)
Example12.51.Prime Ideals are Primary.
Any prime ideal is also primary.
Example12.52.Principal Primary Ideals in UFDs.
If \(R\) is a UFD, we claim that a principal ideal is primary if and only if it is generated by a power of a prime element.
Solution.
Indeed, if \(a=f^n\text{,}\) with \(f\) irreducible, then
\begin{equation*}
xy\in (f^n) \Longleftrightarrow f^n | xy \Longleftrightarrow f^n | x \text{ or } f| y \Longleftrightarrow x\in (f^n) \text{ or } y\in \sqrt{(f^n)} = (f).
\end{equation*}
Conversely, if \(a\) is not a prime power, then \(a=gh\text{,}\) for some \(g\text{,}\)\(h\) nonunits with no common factor, then take \(gh\in (a)\) but \(g\notin (a)\) and \(h\notin \sqrt{(a)}\text{.}\)
Example12.53.Primary Ideals in \(\Z\).
The nonzero primary ideals in \(\mathbb{Z}\) are of the form \((p^n)\) for some prime \(p\) and some \(n \geqslant 1\text{.}\) This example is a bit misleading, as it suggests that primary ideals are the same as powers of primes. We will soon see that it not the case.
Example12.54.\((y^2,yz,z^2)\) Primary in \(R=k[x,y,z]\).
In \(R=k[x,y,z]\text{,}\) the ideal \(I=(y^2,yz,z^2)\) is primary.
Solution.
Give \(R\) the grading with weights \(\deg(y)=\deg(z)=1\) and \(\deg(x)=0\text{.}\) If \(g\notin \sqrt{I}=(y,z)\text{,}\) then \(g\) has a degree zero term. If \(f\notin I\text{,}\) then \(f\) has a term of degree zero or one. The product \(fg\) has a term of degree zero or one, so is not in \(I\text{.}\)
Example12.55.\((x^2,xy)\) Not Primary in \(R=k[x,y,z]\).
In \(R=k[x,y,z]\text{,}\) the ideal \(Q =(x^2,xy)\) is not primary, even though \(\sqrt{Q}=(x)\) is prime.
Solution.
The offending product is \(xy\text{:}\) we have \(x \notin Q\) and \(y \notin \sqrt{Q}\text{.}\)
Remark12.56.
One thing that can be confusing about primary ideals is that the definition is not symmetric. For \(Q\) to be a primary ideal, given a product \(xy \in Q\text{,}\) the definition says that if \(x \notin Q\text{,}\) then \(y \in \sqrt{Q}\text{,}\) and it also says that if \(y \notin Q\text{,}\) then \(x \in \sqrt{Q}\text{.}\) In , we found that \(x \notin Q\) and \(y \notin \sqrt{Q}\text{,}\) so \(Q\) is not primary. Notice that if we switch the roles of \(x\) and \(y\text{,}\) we do have \(x \in \sqrt{Q}\text{,}\) but that is not sufficient to make \(Q\) a primary ideal.
Theorem12.57.Characterization of Primary Ideals.
If \(R\) is noetherian, the following are equivalent: (a) \(Q\) is primary. (b) Every zerodivisor in \(R/Q\) is nilpotent on \(R/Q\text{.}\) (c) \(\Ass(R/Q)\) is a singleton. (d) \(Q\) has exactly one minimal prime, and no embedded primes. (e) \(\sqrt{Q} = P\) is prime and for all \(r, w \in R\) with \(w \notin P\text{,}\)\(rw \in Q \Rightarrow r \in Q\text{.}\) (f) \(\sqrt{Q}= P\) is prime, and \(Q R_{P} \cap R = Q\text{.}\)
Proof.
\((1) \iff (2)\text{:}\) Saying \(y\) is a zerodivisor modulo \(Q\) if there is some \(x\notin Q\) with \(xy\in Q\text{.}\) So the condition that every zerodivisor on \(R/Q\) must be nilpotent is equivalent to
This is exactly the condition that \(Q\) is a primary ideal.
\((2) \iff (3)\text{:}\) First, note that the associated primes of \(R/Q\) are the associated primes of the ideal \(Q\text{,}\) while the minimal primes of \(R/Q\) are the minimal primes of \(Q\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Zerodivisors Associated Primes|Theorem]],
\begin{equation*}
\bigcup_{P \in \Ass(Q)} P \, = \, \mathcal{Z}(R/Q).
\end{equation*}
Every minimal prime of \(Q\) is associated to \(Q\text{,}\) so
\begin{equation*}
\bigcap_{P \in \Min(Q)} P = \bigcap_{P \in \Ass(Q)} P.
\end{equation*}
Finally, every nilpotent element is always a zerodivisor. Putting all these together, we always have the following:
\begin{equation*}
\bigcup_{P \in \Ass(Q)} P \, = \, \mathcal{Z}(R/Q) \, \supseteq \{ r \in R \mid r + Q \in \mathcal{N}(R/Q) \} \, = \bigcap_{P \in \Min(Q)} P = \bigcap_{P \in \Ass(Q)} P.
\end{equation*}
On the one hand, (2) says that the set of zerodivisors on \(R/Q\) coincides with the elements in the nilradical of \(R/Q\text{;}\) thus (2) is the statement that we have equality throughout. So (2) holds if and only if
\begin{equation*}
\bigcup_{P \in \Ass(Q)} P \, = \bigcap_{P \in \Ass(Q)} P.
\end{equation*}
The rest of the proof is elementary set theory: the intersection and union of a collection of sets agree if and only if there is only one set. More precisely, we have equality above if and only if there is only one associated prime.
\((3) \iff (4)\) is clear, given that every ideal has a minimal prime and minimal primes are always associated, so having a single associated prime means having only one minimal prime and no embedded primes.
\((1) \iff (5)\text{:}\) Given the observation that the radical of a primary ideal is prime, this is just a rewording of the definition.
\((5) \iff (6)\text{:}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Lemma - Localization Ideals|Lemma]], we have the following characterization:
\begin{equation*}
Q R_P \cap R = \lbrace r \in R \mid sr \in Q \textrm{ for some } s \notin P \rbrace.
\end{equation*}
Thus the second condition in \((5)\) is equivalent to \(Q R_Q \cap R = Q\text{.}\)
Lemma12.58.Radical Max Implies Primary.
Let \(R\) be a noetherian ring and \(I\) be an ideal. If \(\sqrt{I}=\fm\) a maximal ideal, then \(I\) is a primary ideal.
Proof.
By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Support and Associated Primes|Theorem]], \(\Ass(R/I)\) is nonempty and contained in \(\Supp(R/I)=V(I)=\{\fm\}\text{,}\) so \(\Ass(R/I) = \{ \fm \}\text{,}\) and hence by [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Characterization of Primary Ideals|Proposition]] \(I\) is primary.
Remark12.59.
Note that the assumption that \(\fm\) is maximal was necessary here. Indeed, having a prime radical does not guarantee an ideal is primary, as we saw [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Exercise - \((x2,xy)\) Not Primary in \(R=kx,y,z\)|here]]. Moreover, even the powers of a prime ideal may fail to be primary.
Example12.60.Powers of Prime not Primary.
Fix a field \(k\) and an integer \(n \geqslant 2\text{,}\) and let \(R = k[x,y,z]/(xy-z^n)\text{.}\) Consider the prime ideal \(P = (x,z)\) in \(R\text{.}\) On the one hand, \(xy = z^n \in P^n\text{,}\) while \(x \notin P^n\) and \(y \notin \sqrt{P^n} = P\text{.}\) Therefore, \(P^n\) is not primary, even though its radical is the prime \(P\text{.}\)
Lemma12.61.Contraction of Primary Ideals is Primary.
Let \(f:R\to S\) be a ring homomorphism. If \(Q\) is a primary ideal in \(S\text{,}\) then \(Q \cap R\) is a primary ideal in \(R\text{.}\)
Proof.
if \(xy \in Q \cap R\text{,}\) and \(x \notin Q \cap R\text{,}\) then \(f(x) \notin Q\text{,}\) so \(f(y^n) = f(y)^n \in Q\) for some \(n\text{.}\) Therefore, \(y^n \in Q \cap R\text{,}\) and \(Q \cap R\) is indeed primary.
The canonical map \(R \to R/I_1 \oplus R/I_2\) sending \(r \mapsto (r+I_1, r+I_2)\) has kernel \(I_1 \cap I_2\text{,}\) so there is an inclusion \(R/(I_1 \cap I_2) \subseteq R/I_1 \oplus R/I_2\text{.}\) Hence, by and ,
By , \(\Ass(I_1 \cap \cdots \cap I_t)\) is nonempty, and thus \(\Ass(I_1 \cap \cdots \cap I_t) = \{ P \}\text{.}\) Then \(I_1 \cap \cdots \cap I_t\) is \(P\)-primary by the characterization of primary in (3) above.
Definition12.64.Primary Decomposition.
A primary decomposition of an ideal \(I\) is an expression of the form
\begin{equation*}
I = Q_1 \cap \cdots \cap Q_t
\end{equation*}
with each \(Q_i\) primary. A primary decomposition is irredundant if
\begin{equation*}
\sqrt{Q_i} \neq \sqrt{Q_j} \textrm{ for } i\neq j \quad \textrm{ and } \quad Q_i \not\supseteq \displaystyle\bigcap_{j\neq i} Q_j \textrm{ for all } i.
\end{equation*}
Remark12.65.
By Corollary 12.63, the intersection of \(P\)-primary ideals is \(P\)-primary. Thus we can turn any primary decomposition into an irredundant one by combining the terms with the same radical, then removing redundant terms.
Example12.66.Primary decomposition in \(\mathbb{Z}\).
Given a decomposition of \(n \in \mathbb{Z}\) as a product of distinct primes, say \(n = p_1^{a_1} \cdots p_k^{a_k}\text{,}\) then the primary decomposition of the ideal \((n)\) is \((n) = (p_1^{a_1}) \cap \cdots \cap (p_k^{a_k})\text{.}\) However, this example can be deceiving, in that it suggests that primary ideals are just powers of primes; as we saw in Example 12.60, powers of primes may fail to be primary! Moreover, an ideal might be primary but not a power of a prime.
Remark12.67.
The existence of primary decompositions was first shown by Emanuel Lasker (yes, the chess champion!) for polynomial rings and power series rings in 1905, and then extended to any noetherian ring (which were not called that yet at the time) by Emmy Noether in 1921.
Theorem12.68.Lasker, 1905, Noether, 1921.
Every ideal in a noetherian ring admits a primary decomposition.
Proof.
We will say that an ideal is irreducible if it cannot be written as a proper intersection of larger ideals. If \(R\) is noetherian, we claim that any ideal of \(R\) can be expressed as a finite intersection of irreducible ideals. If the set of ideals that are not a finite intersection of irreducibles were nonempty, then by noetherianity there would be an ideal maximal with the property of not being an intersection of irreducible ideals. Such a maximal element must be an intersection of two larger ideals, each of which are finite intersections of irreducibles, giving a contradiction.
We claim that every irreducible ideal is primary. If we show this, any decomposition into an intersection of irreducible ideals will be a primary decomposition. To prove the contrapositive, suppose that \(Q\) is not primary, and take \(xy\in Q\) with \(x\notin Q\text{,}\)\(y\notin \sqrt{Q}\text{.}\) The ascending chain of ideals
stabilizes for some \(n\text{,}\) since \(R\) is moetherian. Note that this means that for any element \(f \in R\text{,}\) we have \(y^{n+1} f\in Q \implies y^n f\in Q\text{.}\) Using this, we will show that
\begin{equation*}
b y^{n+1} = ay - qy \in Q \implies b \in (Q : y^{n+1}) = (Q : y^n).
\end{equation*}
By definition, this means that \(by^n \in Q\text{,}\) and thus \(a = q+by^n \in Q\text{.}\) This shows that \(Q\) is not irreducible, concluding the proof.
Example12.69.Irredundant Decompositions for \(I=(x^2,xy)\).
Let \(R=k[x,y]\text{,}\) where \(k\) is a field, and \(I=(x^2,xy)\text{.}\) We can write
These are two different irredundant primary decompositions of \(I\text{.}\)
Solution.
To check this, we just need to see that each of the ideals \((x^2,xy,y^2)\) and \((x^2,y)\) are primary. Observe that each has radical \(\fm=(x,y)\text{,}\) which is maximal, so these ideals are both primary. In fact, our ideal \(I\) has infinitely many minimal primary decompositions: given any \(n \geqslant 1\text{,}\)
\begin{equation*}
I = (x) \cap (x^2, xy, y^n)
\end{equation*}
is an irredundant primary decomposition. One thing all of these have in common is the radicals of the primary components: they are always \((x)\) and \((x,y)\text{.}\)
Theorem12.70.1st Uniqueness Theorem for Primary Decompositions.
Suppose \(I\) is an ideal in a noetherian ring \(R\text{.}\) Given any irredundant primary decomposition of \(I\text{,}\) say
\begin{equation*}
I = Q_1 \cap \cdots \cap Q_t,
\end{equation*}
We just need to show that in an irredundant decomposition as above, every \(P_j := \sqrt{Q_j}\) is indeed an associated prime of \(I\text{.}\) So fix \(j\text{,}\) and let
\begin{equation*}
I_j = \displaystyle\bigcap_{i\neq j} Q_i \supseteq I.
\end{equation*}
Since the decomposition is irredundant, the module \(I_j/I\) is nonzero. By , \(I_j/I\) has an associated prime, say \(\mathfrak{a}\text{.}\) Fix \(x_j\in R\) such that \(\mathfrak{a}\) is the annihilator of \(x_j+I\) in \(I_j/I\) for some \(x_j \in I_j\text{.}\) Since
we conclude that \(Q_j \subseteq \ann(x_j+I) = \mathfrak{a}\text{.}\) Since \(P_j\) is the unique minimal prime of \(Q_j\) and \(\mathfrak{a}\) is a prime containing \(Q_j\text{,}\) we must have \(P_j \subseteq \mathfrak{a}\text{.}\) On the other hand, for any \(r\in \mathfrak{a}\text{,}\) we have \(rx_j\in I \subseteq Q_j\text{,}\) and since \(x_j\notin Q_j\text{,}\) we must have \(r \in \sqrt{Q_j} = P_j\) by the definition of primary ideal. Thus \(\mathfrak{a} \subseteq P_j\text{,}\) so we can now conclude that \(\mathfrak{a} = P_j\text{.}\) This shows that \(P_j\) is an associated prime of \(I_j/I\) for all \(j\text{.}\) But this is a submodule of \(R/I\text{,}\) and thus \(P_j\) is associated to \(R/I\text{.}\)
Theorem12.71.2nd Uniqueness Theorem for Primary Decompositions.
If \(I\) is an ideal in a noetherian ring \(R\text{,}\) then the minimal components in any irredundant primary decomposition of \(I\) are unique. More precisely, if
\begin{equation*}
I = Q_1 \cap \cdots \cap Q_t
\end{equation*}
is an irredundant primary decomposition, and \(\sqrt{Q_i} \in \Min(I)\text{,}\) then \(Q_i\) is given by the formula
\begin{equation*}
Q_i = I R_{\sqrt{Q_i}} \cap R,
\end{equation*}
which does not depend on our choice of irredundant decomposition.
Proof.
Let \(Q\) be a primary ideal, and let \(P\) be any prime. If \(Q \subseteq P\text{,}\) then \(\sqrt{Q} \subseteq P\text{.}\) Since the associated primes of an ideal localize well, by , \(Q_P\) will still have a unique associated prime. Thus, the localization \(Q_P\) is either: - the unit ideal, if \(Q \not\subseteq P\text{,}\) or - a primary ideal, if \(Q \subseteq P\text{.}\) Finite intersections commute with localization, by , so for any prime \(P\text{,}\)
is a primary decomposition, although not necessarily irredundant. Fix a minimal prime \(P=P_i\) of \(I\text{,}\) and let \(Q = Q_i\text{.}\) When we localize at \(P\text{,}\) all the other components become the unit ideal, since their radicals are not contained in \(P\text{,}\) and thus \(I_{P} = Q_P\text{.}\) We can then contract to \(R\) to get \(I_{P} \cap R = (Q_i)_{P_i} \cap R = Q_i\text{,}\) since \(Q_i\) is \(P_i\)-primary and we can then apply (6).
Remark12.72.
If \(R\) is not noetherian, we may or may not have a primary decomposition for a given ideal. It is true that if an ideal \(I\) in a general ring has a primary decomposition, then the primes occurring are the same in any irredundant decomposition. However, they are not the associated primes of \(I\) in general; rather, they are the primes that occur as radicals of annihilators of elements.
Example12.73.Decomposition of Radical Ideal is Unique.
If \(R\) is noetherian, and \(I\) is a radical ideal, then we have seen that \(I\) coincides with the intersection of its minimal primes \(P_i\text{,}\) meaning \(I=P_1 \cap \cdots \cap P_t\text{.}\) This is the only primary decomposition of a radical ideal.
For a more concrete example, take the ideal \(I = \left( xy, xz, yz \right)\) in \(k[x,y,z]\text{.}\) This ideal is radical, so we just need to find its minimal primes. And indeed, one can check that \(\left( xy, xz, yz \right) = \left( x, y \right) \cap \left( x, z \right) \cap \left( y, z \right)\text{.}\) More generally, the radical monomial ideals are precisely those that are squarefree, and the primary components of a monomial ideal are also monomial.
Example12.74.Decomposition of \((6)\) in \(\Z[\sqrt{-5}]\) Unique.
Let’s get back to our motivating example in \(\mathbb{Z}[\sqrt{-5}]\text{,}\) where some elements can be written as products of irreducible elements in more than one way. For example, we saw that
So \((6) = (2) \cap (3)\text{,}\) but while \((2)\) is primary, \((3)\) is not. In fact, \((3)\) has two distinct minimal primes, and the following is a minimal primary decomposition for \((6)\text{:}\)
In fact, all of these come components are minimal, and so this primary decomposition is unique. Primary decomposition saves the day!
Definition12.75.Symbolic Power.
If \(P\) is a prime ideal in a ring \(R\text{,}\) the \(n\th\)symbolic power of \(P\) is \(P^{(n)} := P^n R_{P} \cap R\text{.}\)
Proposition12.76.Characterizations of Symbolic Powers.
Let \(R\) be noetherian, and \(P\) a prime ideal of \(R\text{.}\) (a) \(P^{(n)}=\{ r\in R \ \mid \ rs\in P^n \textrm{ for some } s\notin P \}\text{.}\) (b) \(P^{(n)}\) is the unique smallest \(P\)-primary ideal containing \(P^n\text{.}\) (c) \(P^{(n)}\) is the \(P\)-primary component in any minimal primary decomposition of \(P^n\text{.}\)
Proof.
The first characterization follows from the definition, and the fact that expanding and contraction to/from a localization is equivalent to saturating with respect to the multiplicative set, which we proved in .
We know that \(P^{(n)}\) is \(P\)-primary from one of the characterizations of primary we gave in . Any \(P\)-primary ideal satisfies \(\fq R_{P} \cap R = \fq\text{,}\) and if \(\fq\supseteq P^{n}\text{,}\) then \(P^{(n)} = P^n R_{P} \cap R \subseteq \fq R_{P} \cap R = \fq\text{.}\) Thus, \(P^{(n)}\) is the unique smallest \(P\)-primary ideal containing \(P^n\text{.}\)
The last characterization follows from the second uniqueness theorem, .
Example12.77..
In \(R=k[x,y,z]\text{,}\) the prime \(P=(y,z)\) satisfies \(P^{(n)}=P^n\) for all \(n\text{.}\) This follows along the same lines as .
In \(R=k[x,y,z]=(xy-z^n)\text{,}\) where \(n \geqslant 2\text{,}\) we have seen in Example that the square of \(P=(y,z)\) is not primary, and therefore \(P^{(2)}\neq P^2\text{.}\) Indeed, \(xy=z^n \in P^2\text{,}\) and \(x\notin P\text{,}\) so \(y\in P^{(2)}\) but \(y \notin P^2\text{.}\)
Let \(X=X_{3\times 3}\) be a \(3\times 3\) matrix of indeterminates, and \(k[X]\) be a polynomial ring over a field~\(k\text{.}\) Let \(P=I_2(X)\) be the ideal generated by \(2\times 2\) minors of \(X\text{.}\) Write \(\Delta_{\substack{i|k \\j |l}}\) for the determinant of the submatrix with rows \(i,j\) and columns \(k,l\text{.}\) We find
Note that in the second row, we subtracted the Laplace expansion of the determinant of the matrix with row 3 replaced by another copy of row 1. That is, we subtracted zero.
Remark12.78.
While we will not discuss symbolic powers in detail, they are ubiquitous in commutative algebra. They show up as tools to prove various important theorems of different flavors, and they are also interesting objects in their own right. In particular, symbolic powers can be interpreted from a geometric perspective, via the Zariski-Nagata Theorem. Roughly, this theorem says that when we consider symbolic powers of prime ideals over \(\mathbb{C}[x_1, \ldots, x_d]\text{,}\) the polynomials in \(P^{(n)}\) are precisely the polynomials that vanish on the variety corresponding to \(P\text{.}\) This result can be made sense of more generally, for any radical ideal in \(\mathbb{C}[x_1, \ldots, x_d]\) over any perfect field \(k\text{,}\) and even when \(k = \mathbb{Z}\text{.}\)
