*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; $0$ is a prime element of $\mathbb{Z}$ but not a prime number.)

Let $R$ be a rig (assumed unital and associative as usual), and let $P$ be a two-sided ideal in $R$. Then $P$ is **prime** if $P$ is proper and $x$ or $y$ belongs to $P$ whenever $x a y$ does for all $a$:

$\forall\, x \in R,\; \forall\, y \in R,\; (\forall\, a \in R,\; x a y \in P) \;\Rightarrow\; x \in P \;\vee\; y \in P .$

Also, $P$ is **completely prime** if $P$ is proper and $x$ or $y$ belongs to $P$ whenever $x y$ does:

$\forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .$

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

Let $R$ be a commutative rig, and let $P$ be an ideal in $R$. Then $P$ is **prime** if $P$ is proper and $x$ or $y$ belongs to $P$ whenever $x y$ does:

$\forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .$

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:

Let $R$ be a lattice, and let $P$ be an ideal in $R$. Then $P$ is **prime** if $P$ is proper and $x$ or $y$ belongs to $P$ whenever their meet does:

$\forall\, x \in R,\; \forall\, y \in R,\; x \wedge y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .$

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:

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

$\forall\, X \subseteq R,\; \bigwedge X \in P \;\Rightarrow\; \exists\, x \in R,\; x \in X \;\wedge\; x \in P .$

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

Let $R$ be a preorder, and let $P$ be an ideal in $R$. Then $P$ is **prime** if $P$ is proper and $x$ or $y$ belongs to $P$ whenever every $z$ that precedes both $x$ and $y$ does:

$\forall\, x \in R,\; \forall\, y \in R,\; (\forall\, z \in R,\; z \leq x \;\Rightarrow\; z \in P) \;\Rightarrow\; (\forall\, z \in R,\; z \leq y \;\Rightarrow\; z \in P) \;\Rightarrow\; x \in P \;\vee\; y \in P .$

Also, $P$ is **completely prime** if $P$ if some element belongs to $P$ whenever every $z$ that precedes every element of a subset of $R$ does:

$\forall\, X \subseteq R,\; (\forall\, z \in R,\; (\forall\, x \in R,\; x \in X \;\Rightarrow\; z \leq x) \;\Rightarrow\; z \in P) \;\Rightarrow\; \exists\, x \in R,\; x \in X \;\wedge\; x \in P .$

Note that a prime filter in a proset $R$ is a prime ideal in the opposite order $R^op$, and a completely prime filter in $R$ is a completely prime ideal in $R^op$. 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? $Idl(R)$ a quantale (cf. Day convolution). Namely, if $I, J$ are ideals, then their product $I J$ is the ideal generated by all products $x y$ with $x \in I, y \in J$ in the case of rigs, or generated by all meets $x \wedge y$ in the case of lattices, or generated by all $z$ satisfying $z \leq x$ and $z \leq y$ in the case of general prosets. (Note that for a lattice or other proset, $I J$ is equal to the intersection $I \cap J$.)

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

Let $R$ be a rig or a proset, and let $P$ be an ideal in $R$. Then $P$ is **prime** if $P$ is proper and $I$ or $J$ is contained in $P$ whenever $I J$ is:

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

Because $I J = I \cap J$ 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 :

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

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

As is typical, $n = 1$ is trivial, $n \gt 2$ can be proved from $n = 2$ by induction, and $n = 0$ is the mysterious preliminary clause, in this case that $P$ is proper. Note that completely prime ideals in a proset arise by generalizing from finite $n$ 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 $a_0 x_0 a_1 x_1 \cdots x_{n-2} a_{n-1} x_{n-1} a_n$; $a_0$ and $a_n$ can be ignored when $n \gt 0$, which is why only $a_1$ appears in the biased definition. The definition of completely prime ideals in order theory is already unbiased, since $X$ 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.

Let $R$ be a rig with apartness?, and let $M$ be a two-sided anti-ideal in $R$. Then $M$ is **prime** if $M$ is proper? (that is inhabited) and $x a y$ belongs to $M$ for some $a$ whenever $x$ and $y$ do:

$\forall\, x \in R,\; \forall\, y \in R,\; x \in M \;\Rightarrow\; y \in M \;\Rightarrow\; \exists\, a \in R,\; x a y \in M .$

Also, $M$ is **completely prime** if $M$ is proper and $x y$ belongs to $M$ whenever $x$ and $y$ do:

$\forall\, x \in R,\; \forall\, y \in R,\; x \in M \;\Rightarrow\; y \in M \;\Rightarrow\; x y \in M .$

If we ignore the requirement that $M$ be an anti-ideal, then we say that $M$ is an **m-system** if it is inhabited and satisfies the binary condition of a prime ideal and **multiplicatively closed** if it owns $1$ and satisfies the binary condition of a completely prime ideal. (A proper anti-ideal necessarily owns $1$, 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).

In a matrix ring? $M_n(k)$ over a field $k$, the zero ideal is prime (really because a matrix ring is a simple ring, where the zero ideal is a maximal ideal), but (for $n \gt 1$) 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 22:49:04. See the history of this page for a list of all contributions to it.