----- > [!proposition] Proposition. ([[incomparability]]) > Let $A \subset_{\iota} B$ be an [[integral algebra|integral extension]] of ([[commutative ring|commutative]]) [[ring|rings]], and let $\mathfrak{q}, \mathfrak{q}'$ be [[prime ideal|prime ideals]] of $B$ such that $\mathfrak{q} \cap A=\mathfrak{q}' \cap A$[^1] and $\mathfrak{q} \subset \mathfrak{q}'$. Then $\mathfrak{q}=\mathfrak{q}'$. > Thus, there is nothing to be gained by comparing via inclusion [[ideal|ideals]] in a given fiber of $\iota^{*}:\text{Spec }B \to \text{Spec }A$: hence *incomparability*. ^proposition > [!NOTE] Remark. > > When one thinks of a prototypical (say) topological map $f:X \to Y$, they think of a fiber $f ^{-1}(\{ q \})$ as a collection of unrelated, *incomparable* points in $X$. > > $\iota^{*}:\text{Spec }B \to \text{Spec } A$ is nominally atypical, however, in the sense that the fiber $(\iota^{*})^{-1}(\mathfrak{q})$ is a collection of *subsets* of $X$. Incomparability assures us that we can still think of its elements as 'a bunch of unrelated points', in the sense that there are no proper inclusions ('subpoints') among these subsets. Summary: - Put $\mathfrak{p}=\mathfrak{q} \cap A=\mathfrak{q'}\cap A$, and consider the localization $B_{\mathfrak{p}}=(A - \mathfrak{p})^{-1}B$. - It will be enough to show $\mathfrak{q}B_{\mathfrak{p}}=\mathfrak{q'}B_{\mathfrak{p}}$, where $\mathfrak{q}B_{\mathfrak{p}}$, $\mathfrak{q}'B_{\mathfrak{p}}$ denote the extensions of of $\mathfrak{q}$, $\mathfrak{q}'$ under the localization map $B \to B_{\mathfrak{p}}$ are prime ideals of $B_{\mathfrak{p}}$. - That $\mathfrak{q}B_{\mathfrak{p}} \subset \mathfrak{q}'B_{\mathfrak{p}}$ follows from the assumption $\mathfrak{q} \subset \mathfrak{q'}$. Can then show $\mathfrak{q}B_{\mathfrak{p}}$ and $\mathfrak{q}'B_{\mathfrak{p}}$ are both in $\text{mspec }B_{\mathfrak{p}}$, meaning this inclusion cannot be proper - To proceed, look at the map $A_{\mathfrak{p}} \to B_{\mathfrak{p}}$, an integral extension as a localization thereof. Enough to show contraction under this map to the unique maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$. So explain why that's the case. > [!proof]+ Proof. ([[incomparability]]) > Put $\mathfrak{p}=\mathfrak{q} \cap A= \mathfrak{q}' \cap A$ and $S=A- \mathfrak{p}$. Note that since $S$ is a [[multiplicative subset of a ring|multiplicative subset]] of $B$, the [[localization]] $S ^{-1}B$ makes sense. It will be suggestively denoted $B_{\mathfrak{p}}$, but importantly this is *not* the [[localization]] of a [[ring]] at a [[prime ideal]], e.g. $B_{\mathfrak{p}}$ is *not* a [[local ring]]. > > > Consider the [[localization|localization map]] $B \to B_{\mathfrak{p}}$. From [[extension and contraction under localization]], [[extension of an ideal|extension]] and [[contraction of an ideal|contraction]] along this map give a [[bijection]] $\{ \mathfrak{b} \in \text{Spec }B : \mathfrak{b} \cap S = \emptyset \} \leftrightarrow \text{Spec }B_{\mathfrak{p}}$. Since $\mathfrak{q}$ and $\mathfrak{q}'$ do not intersect $S$, this implies that $\underbrace{S ^{-1} \mathfrak{q}}_{=\mathfrak{q}B_{\mathfrak{p}}}$ and $\underbrace{S ^{-1} \mathfrak{q'}}_{=\mathfrak{q'}B_{\mathfrak{p}}}$ are [[prime ideal|prime ideals]] of $B_{\mathfrak{p}}$ satisfying $\mathfrak{q}=(S ^{-1} \mathfrak{q})^{c}$, $\mathfrak{q}'=(S ^{-1} \mathfrak{q}')^{c}$. So it will suffice to prove $\underbrace{S ^{-1} \mathfrak{q}}_{=\mathfrak{q}B_{\mathfrak{p}}}=\underbrace{S ^{-1} \mathfrak{q'}}_{=\mathfrak{q'}B_{\mathfrak{p}}}$. Since $\mathfrak{q} \subset \mathfrak{q}'$, it is clear that $S ^{-1}\mathfrak{q} \subset S ^{-1} \mathfrak{q}'$. With that, we will show the reverse inclusion by showing $\underbrace{S ^{-1} \mathfrak{q}}_{=\mathfrak{q}B_{\mathfrak{p}}}$ and $\underbrace{S ^{-1} \mathfrak{q'}}_{=\mathfrak{q'}B_{\mathfrak{p}}}$ are [[maximal ideal|maximal ideals]] of $B_{\mathfrak{p}}$, and therefore the containment cannot be proper. > > > To this end, first note $A_{\mathfrak{p}} \subset B_{\mathfrak{p}}$ is an [[integral algebra|integral extension]] since [[integral algebra|localization preserves integral extensions]]. Hence by [[integral extensions, units, and fields#^corollary|this corollary]], it will be equivalent for us to show that $\underbrace{S ^{-1} \mathfrak{q}}_{=\mathfrak{q}B_{\mathfrak{p}}}$ and $\underbrace{S ^{-1} \mathfrak{q'}}_{=\mathfrak{q'}B_{\mathfrak{p}}}$ contract to [[maximal ideal|maximal ideals]] in $A_{\mathfrak{p}}$. Now, $A_{\mathfrak{p}}=S ^{-1} A$ is a [[local ring]] whose unique [[maximal ideal]] is $\underbrace{\mathfrak{p}A_{\mathfrak{p}}}_{=S ^{-1} \mathfrak{p}}$. [[localization commutes with intersections, sums, and quotients|And]] > > $S ^{-1} \mathfrak{q} \cap S ^{-1} A=S ^{-1}(\underbrace{\mathfrak{q} \cap A}_{\mathfrak{p}})$; similarly $S ^{-1} \mathfrak{q'} \cap S ^{-1} A=S ^{-1}(\underbrace{\mathfrak{q}' \cap A}_{\mathfrak{p}})$. > ----- #### [^1]: In other words, $\mathfrak{q},\mathfrak{q'}$ contract to the same thing under the [[contraction of an ideal|contraction map]] $\iota^{*}:\text{Spec }B \to \text{Spec }A$ induced by the [[inclusion map|inclusion]] $\iota:A \to B$. That is, they belong to the same [[spec functor|fiber of]] $\iota^{*}=\text{Spec }\iota$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```