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prime ideal

Prime ideals

Idea

Prime ideals are supposed to be a generalization of prime numbers from elements of the ring of integers to ideals in the sense of ‘ideal elements’ of an arbitrary ring (usually commutative, but also possibly something more general than a ring). It's not clear that they do so; maximal ideals may do a better job. (In particular, zero is not a prime number, and the zero ideal of \mathbb{Z} is not a maximal ideal either; however, it is a prime ideal.) Nevertheless, they have assumed an importance that dwarfs any question of original motivation; indeed, the general definition of prime element follows prime ideals rather than prime numbers. (An element of a ring is prime iff its principal ideal is prime; 00 is a prime element of \mathbb{Z} but not a prime number.)

Definitions

Definition

Let RR be a rig (assumed unital and associative as usual), and let PP be a two-sided ideal in RR. Then PP is prime if PP is proper and xx or yy belongs to PP whenever xayx a y does for all aa:

xR,yR,(aR,xayP)xPyP. \forall\, x \in R,\; \forall\, y \in R,\; (\forall\, a \in R,\; x a y \in P) \;\Rightarrow\; x \in P \;\vee\; y \in P .

Also, PP is completely prime if PP is proper and xx or yy belongs to PP whenever xyx y does:

xR,yR,xyPxPyP. \forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .

Note that every completely prime ideal is prime (using that RR is unital). The converse holds if the rig is commutative (using that PP is an ideal). So in commutative algebra one usually uses the (simpler) definition of completely prime ideal as the definition of prime ideal:

Definition

Let RR be a commutative rig, and let PP be an ideal in RR. Then PP is prime if PP is proper and xx or yy belongs to PP whenever xyx y does:

xR,yR,xyPxPyP. \forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .

Essentially the same definition applies in order theory, using the analogy (which is more than an analogy in the case of a distributive lattice) between multiplication in a rig and the meet in an order:

Definition

Let RR be a lattice, and let PP be an ideal in RR. Then PP is prime if PP is proper and xx or yy belongs to PP whenever their meet does:

xR,yR,xyPxPyP. \forall\, x \in R,\; \forall\, y \in R,\; x \wedge y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .

There is an infinitary version of this that is called ‘complete’ but is not equivalent to the notion in Definition of a completely prime ideal from the noncommutative theory:

Definition

Let RR be a complete lattice, and let PP be an ideal in RR. Then PP is completely prime if some element belongs to PP whenever the meet of a subset of RR does:

XR,XPxR,xXxP. \forall\, X \subseteq R,\; \bigwedge X \in P \;\Rightarrow\; \exists\, x \in R,\; x \in X \;\wedge\; x \in P .

With a little sublety, this makes sense even when meets (and joins) don’t always exist:

Definition

Let RR be a preorder, and let PP be an ideal in RR. Then PP is prime if PP is proper and xx or yy belongs to PP whenever every zz that precedes both xx and yy does:

xR,yR,(zR,zxzP)(zR,zyzP)xPyP. \forall\, x \in R,\; \forall\, y \in R,\; (\forall\, z \in R,\; z \leq x \;\Rightarrow\; z \in P) \;\Rightarrow\; (\forall\, z \in R,\; z \leq y \;\Rightarrow\; z \in P) \;\Rightarrow\; x \in P \;\vee\; y \in P .

Also, PP is completely prime if PP if some element belongs to PP whenever every zz that precedes every element of a subset of RR does:

XR,(zR,(xR,xXzx)zP)xR,xXxP. \forall\, X \subseteq R,\; (\forall\, z \in R,\; (\forall\, x \in R,\; x \in X \;\Rightarrow\; z \leq x) \;\Rightarrow\; z \in P) \;\Rightarrow\; \exists\, x \in R,\; x \in X \;\wedge\; x \in P .

Note that a prime filter in a proset RR is a prime ideal in the opposite order R opR^op, and a completely prime filter in RR is a completely prime ideal in R opR^op. Again, the meaning of ‘completely prime’ here is unrelated to its meaning in Definition .

All of these definitions may be justified by looking at the quantale of ideals. As discussed at ideals in a monoid, there is for two-sided ideals an operation of ideal multiplication, making the ideal lattice? Idl(R)Idl(R) a quantale (cf. Day convolution). Namely, if I,JI, J are ideals, then their product IJI J is the ideal generated by all products xyx y with xI,yJx \in I, y \in J in the case of rigs, or generated by all meets xyx \wedge y in the case of lattices, or generated by all zz satisfying zxz \leq x and zyz \leq y in the case of general prosets. (Note that for a lattice or other proset, IJI J is equal to the intersection IJI \cap J.)

With IJI J suitably defined, every definition above (except the ‘complete’ ones) can be subsumed below:

Definition

Let RR be a rig or a proset, and let PP be an ideal in RR. Then PP is prime if PP is proper and II or JJ is contained in PP whenever IJI J is:

IIdl(R),JIdl(R),IJPIPJP. \forall\, I \in Idl(R),\; \forall\, J \in Idl(R),\; I J \subseteq P \;\Rightarrow\; I \subseteq P \;\vee\; J \subseteq P .

Because IJ=IJI J = I \cap J in order theory, prime ideals there are the same as strongly irreducible ideals, so the completely prime ideals of that theory are really just completely strongly irreducible ideals (generalizing from a pair of ideals to an arbitrary set of ideals). In contrast, the completely prime ideals of noncommutative ring theory are not of much interest; they are a naïve definition that works in the commutative case but not so well in the noncommutative case.

Finally, note that most of these definitions have an extra clause that a prime ideal must be proper. This can be justified by removing bias. We state here the unbiased version of :

Definition

Let RR be a rig or a proset, and let PP be an ideal in RR. Then PP is prime if some ideal in the list is contained in PP whenever a product of a finite list of ideals is contained in PP:

n,IIdl(R) n, k[n]I kPk[n],I kP. \forall\, n \in \mathbb{N},\; \forall\, I \in Idl(R)^n,\; \prod_{k\in[n]} I_k \subseteq P \;\Rightarrow\; \exists\, k \in [n],\; I_k \subseteq P .

As is typical, n=1n = 1 is trivial, n>2n \gt 2 can be proved from n=2n = 2 by induction, and n=0n = 0 is the mysterious preliminary clause, in this case that PP is proper. Note that completely prime ideals in a proset arise by generalizing from finite nn to arbitrary cardinality.

Unbiased versions of the other definitions are fairly straightforward. (For prime ideals in a noncommutative rig, the unbiased definition involves a product of the form a 0x 0a 1x 1x n2a n1x n1a na_0 x_0 a_1 x_1 \cdots x_{n-2} a_{n-1} x_{n-1} a_n; a 0a_0 and a na_n can be ignored when n>0n \gt 0, which is why only a 1a_1 appears in the biased definition. The definition of completely prime ideals in order theory is already unbiased, since XX could always be the empty set.)

Sometimes it's more fruitful to consider the complement of a prime ideal. This is known in constructive mathematics as an anti-ideal, and this becomes a necessary perspective there, as many common examples fail to satisfy the above definitions constructively. (To support anti-ideals, a rig must be equipped with a tight apartness relation, which is vacuous in classical mathematics.) However, the concept is useful even classically.

Definition

Let RR be a rig with apartness?, and let MM be a two-sided anti-ideal in RR. Then MM is prime if MM is proper? (that is inhabited) and xayx a y belongs to MM for some aa whenever xx and yy do:

xR,yR,xMyMaR,xayM. \forall\, x \in R,\; \forall\, y \in R,\; x \in M \;\Rightarrow\; y \in M \;\Rightarrow\; \exists\, a \in R,\; x a y \in M .

Also, MM is completely prime if MM is proper and xyx y belongs to MM whenever xx and yy do:

xR,yR,xMyMxyM. \forall\, x \in R,\; \forall\, y \in R,\; x \in M \;\Rightarrow\; y \in M \;\Rightarrow\; x y \in M .

If we ignore the requirement that MM be an anti-ideal, then we say that MM is an m-system if it is inhabited and satisfies the binary condition of a prime ideal and multiplicatively closed if it owns 11 and satisfies the binary condition of a completely prime ideal. (A proper anti-ideal necessarily owns 11, but an m-system might not, even though by definition it must be inhabited.) Thus, an ideal in a commutative ring is (classically) prime iff its complement is multiplicatively closed (and analogously for noncommutative rings).

Examples

In a matrix ring? M n(k)M_n(k) over a field kk, the zero ideal is prime (really because a matrix ring is a simple ring, where the zero ideal is a maximal ideal), but (for n>1n \gt 1) not completely prime.

In the ring of integers (or the rig of natural numbers), the prime ideals are precisely the principal ideals of the prime numbers together with the zero ideal. (This is the motivating example, despite not lining up perfectly.)

Last revised on September 4, 2018 at 18:49:04. See the history of this page for a list of all contributions to it.