Prime Ideal is Prime Element
Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $I$ be an ideal in $L$.
Then:
- $I$ is a prime ideal
- $I$ is a prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$
where:
- $\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$
- $\mathord \precsim := \mathord \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Proof
Sufficient Condition
Let $I$ be a prime ideal.
Let $x, y \in \map {\mathit {Ids} } L$ such that
- $x \wedge y \precsim I$
By definition of $\precsim$:
- $x \wedge y \subseteq I$
By Meet is Intersection in Set of Ideals:
- $x \cap y \subseteq I$
Aiming for a contradiction, suppose:
- $x \not \precsim I$ and $y \not \precsim I$
By definition of $\precsim$: $x \nsubseteq I$ and $y \nsubseteq I$
By definition of subset:
- $\exists a \in x: a \notin I$
and
- $\exists b \in y: b \notin I$
- $a \wedge b \preceq a$ and $a \wedge b \preceq b$
By definition lower section:
- $a \wedge b \in x$ and $a \wedge b \in y$
By definitions subset and intersection:
- $a \wedge b \in I$
By Characterization of Prime Ideal:
- $a \in I$ or $b \in I$
This contradicts $a \notin I$ and $b \notin I$
Hence the assumption:
- $x \not \precsim I$ and $y \not \precsim I$
was false.
$\Box$
Necessary Condition
Let $I$ be prime element in $\struct {\map {\mathit {Ids} } L, \precsim}$
Let $x, y \in S$ such that:
- $x \wedge y \in I$
By Lower Closure of Element is Ideal:
- $X := x^\preceq$ and $Y := y^\preceq$ are ideals in $L$
By Meet is Intersection in Set of Ideals:
- $X \cap Y = X \wedge Y$
We will prove that:
- $X \wedge Y \subseteq I$
Let $a \in X \wedge Y$.
By definition of intersection:
- $a \in X$ and $a \in Y$
By definition of lower closure of element:
- $a \preceq x$ and $a \preceq y$
By definition of infimum:
- $a \preceq x \wedge y$
Thus by definition of lower section:
- $a \in I$
$\Box$
By definition of $\precsim$:
- $X \wedge Y \precsim I$
By definition of prime element:
- $X \precsim I$ or $Y \precsim I$
By definition of $\precsim$:
- $X \subseteq I$ or $Y \subseteq I$
By definition of reflexivity:
- $x \preceq x$ and $y \preceq y$
By definition of lower closure of element:
- $x \in X$ and $y \in Y$
Thus by definition of subset:
- $x \in I$ or $y \in I$
Hence by Characterization of Prime Ideal:
- $I$ is prime ideal.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_7:19