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\begin{document}

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\section*{prime ideal}

\hypertarget{prime_ideals}{}\section*{{Prime ideals}}\label{prime_ideals}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

\emph{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 ring|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 \emph{not} a prime number, and the [[zero ideal]] of $\mathbb{Z}$ is not a maximal ideal either; however, it \emph{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.)

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

\begin{defn}
\label{rig}\hypertarget{rig}{}
Let $R$ be a [[rig]] (assumed unital and associative as usual), and let $P$ be a [[two-sided ideal]] in $R$. Then $P$ is \textbf{prime} if $P$ is [[proper ideal|proper]] and $x$ or $y$ belongs to $P$ whenever $x a y$ does for all $a$:

\begin{displaymath}
\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 .
\end{displaymath}
Also, $P$ is \textbf{completely prime} if $P$ is proper and $x$ or $y$ belongs to $P$ whenever $x y$ does:

\begin{displaymath}
\forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .
\end{displaymath}
\end{defn}
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:

\begin{defn}
\label{commrig}\hypertarget{commrig}{}
Let $R$ be a [[commutative ring|commutative]] rig, and let $P$ be an [[ideal]] in $R$. Then $P$ is \textbf{prime} if $P$ is [[proper ideal|proper]] and $x$ or $y$ belongs to $P$ whenever $x y$ does:

\begin{displaymath}
\forall\, x \in R,\; \forall\, y \in R,\; x y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .
\end{displaymath}
\end{defn}
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:

\begin{defn}
\label{lattice}\hypertarget{lattice}{}
Let $R$ be a [[lattice]], and let $P$ be an [[ideal]] in $R$. Then $P$ is \textbf{prime} if $P$ is [[proper ideal|proper]] and $x$ or $y$ belongs to $P$ whenever their meet does:

\begin{displaymath}
\forall\, x \in R,\; \forall\, y \in R,\; x \wedge y \in P \;\Rightarrow\; x \in P \;\vee\; y \in P .
\end{displaymath}
\end{defn}
There is an infinitary version of this that is called `complete' but is \emph{not} equivalent to the notion in Definition \ref{rig} of a completely prime ideal from the noncommutative theory:

\begin{defn}
\label{complattice}\hypertarget{complattice}{}
Let $R$ be a [[complete lattice]], and let $P$ be an [[ideal]] in $R$. Then $P$ is \textbf{completely prime} if some element belongs to $P$ whenever the meet of a subset of $R$ does:

\begin{displaymath}
\forall\, X \subseteq R,\; \bigwedge X \in P \;\Rightarrow\; \exists\, x \in R,\; x \in X \;\wedge\; x \in P .
\end{displaymath}
\end{defn}
With a little sublety, this makes sense even when [[meets]] (and [[joins]]) don't always exist:

\begin{defn}
\label{preorder}\hypertarget{preorder}{}
Let $R$ be a [[preorder]], and let $P$ be an ideal in $R$. Then $P$ is \textbf{prime} if $P$ is [[proper ideal|proper]] and $x$ or $y$ belongs to $P$ whenever every $z$ that precedes both $x$ and $y$ does:

\begin{displaymath}
\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 .
\end{displaymath}
Also, $P$ is \textbf{completely prime} if $P$ if some element belongs to $P$ whenever every $z$ that precedes every element of a subset of $R$ does:

\begin{displaymath}
\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 .
\end{displaymath}
\end{defn}
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 \ref{rig}.

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 [[lattice of ideals|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:

\begin{defn}
\label{universal}\hypertarget{universal}{}
Let $R$ be a rig or a proset, and let $P$ be an ideal in $R$. Then $P$ is \textbf{prime} if $P$ is [[proper ideal|proper]] and $I$ or $J$ is contained in $P$ whenever $I J$ is:

\begin{displaymath}
\forall\, I \in Idl(R),\; \forall\, J \in Idl(R),\; I J \subseteq P \;\Rightarrow\; I \subseteq P \;\vee\; J \subseteq P .
\end{displaymath}
\end{defn}
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 ideal|proper]]. This can be justified by removing [[bias]]. We state here the unbiased version of \ref{universal}:

\begin{defn}
\label{unbiased}\hypertarget{unbiased}{}
Let $R$ be a rig or a proset, and let $P$ be an ideal in $R$. Then $P$ is \textbf{prime} if some ideal in the list is contained in $P$ whenever a product of a [[finite list]] of ideals is contained in $P$:

\begin{displaymath}
\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 .
\end{displaymath}
\end{defn}
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.

\begin{defn}
\label{constructive}\hypertarget{constructive}{}
Let $R$ be a [[algebra with apartness|rig with apartness]], and let $M$ be a two-sided [[anti-ideal]] in $R$. Then $M$ is \textbf{prime} if $M$ is [[proper antiideal|proper]] (that is [[inhabited subset|inhabited]]) and $x a y$ belongs to $M$ for some $a$ whenever $x$ and $y$ do:

\begin{displaymath}
\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 .
\end{displaymath}
Also, $M$ is \textbf{completely prime} if $M$ is proper and $x y$ belongs to $M$ whenever $x$ and $y$ do:

\begin{displaymath}
\forall\, x \in R,\; \forall\, y \in R,\; x \in M \;\Rightarrow\; y \in M \;\Rightarrow\; x y \in M .
\end{displaymath}
\end{defn}
If we ignore the requirement that $M$ be an anti-ideal, then we say that $M$ is an \textbf{m-system} if it is inhabited and satisfies the binary condition of a prime ideal and \textbf{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).

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

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 \emph{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.)

[[!redirects prime ideal]] [[!redirects prime ideals]]

[[!redirects completely prime ideal]] [[!redirects completely prime ideals]]



\end{document}