---- The following corollary of the [[weak Nullstellensatz|weak]] and [[strong Nullstellensatz|strong Nullstellensätze]] relates geometry and algebra. Let $k \subset \Omega$ be a [[field]] [[field extension|extension]], where $\Omega$ is [[algebraically closed]]. > [!theorem] Theorem. ([[Hilbert's geometry-algebra correspondence]]) > Let $n \geq 0$. Then > **1.** We have an inclusion-reversing [[bijection]] $\{ k\text{-algebraic subsets of } \Omega^{n} \} \leftrightarrow \{ \text{radical ideals of }k[T_{1},\dots,T_{n}] \}$ > given by $X \mapsto I(X)$ and $V(\mathfrak{a}) \leftarrow^{|}_{|} \mathfrak{a}$. > > > **2.** This [[bijection]] restricts to a [[bijection]] $\{ \text{irreducible }k\text{-algebraic subsets} \} \leftrightarrow \text{Spec }k[T_{1},\dots,T_{n}]$ > **3.** Since this [[bijection]] reverses inclusion and every [[maximal ideal]] is [[radical of an ideal|radical]], we get a further restricted bijection $\{ \text{minimal nonempty }k\text{-algebraic subsets }\} \leftrightarrow \text{mSpec }k[T_{1},\dots,T_{n}].$This map gives rise to an embedding $k^{n} \hookrightarrow \text{mSpec}(k[T_{1},\dots,T_{n}])$. *It is a [[bijection]] when $k=\Omega$ is [[algebraically closed]].* Specifically, the embedding associates to $\boldsymbol x=(x_{1},\dots,x_{n})$ the [[maximal ideal]] $\mathfrak{m}_{\boldsymbol x}=\langle T_{1}-x_{1},\dots,T_{n}-x_{n} \rangle$. > > [!proposition] Summary for $k=\overline{k}$ [[algebraically closed]]. > For $k=\overline{k}=\Omega$ [[algebraically closed]], get: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#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-HE82U7Y02Bbaw0TogA > \begin{tikzcd} > \{\text{algebraic subsets of }k^n\} \arrow[r] \arrow[d, "\cup" description, no head, dotted] & {\{\text{radical ideals of }k[T_1,\dots, T_n]\}} \arrow[l] \arrow[d, "\cup" description, no head, dotted] \\ > \{\text{irreducible algebraic subsets}\} \arrow[r] \arrow[d, "\cup" description, no head, dotted] & {\{\text{prime ideals of }k[T_1,\dots, T_n]\}} \arrow[l] \arrow[d, "\cup" description, no head, dotted] \\ > k^n \arrow[r] & {\{\text{maximal ideals of }k[T_1,\dots, T_n]\}} \arrow[l] > \end{tikzcd} > \end{document} > ``` > > > > > > | Type of [[algebraic set]] | Type of [[ideal]] | Type of [[coordinate ring]] | > | ----------------------------------------- | -------------------------------- | --------------------------------------------- | > | [[algebraic set\|(affine) algebraic set]] | [[radical of an ideal\|radical]] | [[nilpotent element of a ring\|reduced ring]] | > | [[affine variety\|affine (sub)variety]] | [[prime ideal\|prime]] | [[integral domain]] | > | point of $k^{n}$ | [[maximal ideal\|maximal]] | [[field]] | > > [!proof]+ Proof. ([[Hilbert's geometry-algebra correspondence]]) > From the properties in [[algebraic set]] we saw that $I(X)$ is [[radical of an ideal|radical]] for every $k$-algebraic set $X \subset \Omega^{n}$, and that $X=V(I(X))$. > > On the other hand, if $\mathfrak{a}$ is a [[radical of an ideal|radical ideal]] of $[T_{1},\dots,T_{n}]$, then by the [[strong Nullstellensatz]] $I(V(\mathfrak{a}))=\sqrt{ \mathfrak{a} }=\mathfrak{a}$. This proves the result. > > The [[bijection]] is inclusion-reversing because, as we've seen, $I(\cdot)$ and $V(\cdot)$ are inclusion-reversing. > > *An algebraic set $X \subset \Omega^{n}$ is irreducible if and only if the radical ideal $I(X)$ of $k[T_{1},\dots,T_{n}]$ is prime.* > > Write $X=V(\mathfrak{a})$, $\mathfrak{a} \subset \Omega[T_{1},\dots,T_{n}]$ an [[ideal]]; suppose $X$ is [[irreducible algebraic set|irreducible]]. WTS $I(X)$ is [[prime ideal|prime]]. So take $f,g \in \Omega(T_{1},\dots,T_{n})$ such that $fg \in I(X)$. Then for every $\boldsymbol x \in X$, $f(\boldsymbol x)=0$ or $g(\boldsymbol x)=0$ (recall that $\Omega$ is a [[field]], hence an [[integral domain]]). That is, $X\subset V(\langle f \rangle) \cup V(\langle g \rangle)$. Hence $V(f)$ and $V(g)$ [[cover]] $X$, in the sense that $X=\big(\underbrace{ X \cap V(\langle f \rangle }_{ V(\mathfrak{a}+\langle f \rangle ) } ) \big) \cup \big( \underbrace{ X \cap V(\langle g \rangle ) }_{ V(\mathfrak{a}+\langle g \rangle ) } \big),$ > a union of algebraic sets. So (WLOG) $X=V(\mathfrak{a}+\langle f \rangle)$. Then $X \subset V(\langle f \rangle)$; taking $I(\cdot)$ gives $I(X) \supset I(V(\langle f \rangle )) \supset \sqrt{ \langle f \rangle } \ni f$[^1] . Thus $f \in I(X)$, so $I(X)$ is prime. > > Conversely assume $I(X)$ is prime. Take $X=X_{1} \cup X_{2}$, a union of algebraic subsets of $\Omega^{n}$. Then $I(X)=I(X_{1} \cup X_{2})=I(X_{1}) \cap I(X_{2})$. By the properties in [[prime ideal]] (this uses that $I(X)$ is prime), it follows that either $I(X)=I(X_{1})$ or $I(X)=I(X_{2})$. WLOG $I(X)=I(X_{1})$. Then $\underbrace{ V(I(X)) }_{ =X }=\underbrace{ V(I(X_{1})) }_{ =X_{1} }$. So $X=X_{1}$. Hence $X$ is irreducible so long as it is nonempty. Indeed, $X \neq \emptyset$ by the [[weak Nullstellensatz]], as $1 \notin I(X)$ (prime ideals are proper ideals by definition). > > > *Minimal nonempty $k$-algebraic subsets $\leftrightarrow$ maximal ideals of $k[T_{1},\dots,T_{n}]$.* > > There is actually not much to show, but we do invoke the [[strong Nullstellensatz]] (but even that I think could be sidestepped with a more elementary, albeit longer, argument). > > Write $\mathcal{A}$ for the set of minimal nonempty $k$-algebraic subsets of $\Omega^{n}$. As was said already in the theorem statement, because the [[bijection]] reverses inclusion, it restricts to a bijection between minimal nonempty $k$-algebraic subsets and maximal ideals of $k[T_{1},\dots,T_{n}]$. We now prove: > 1. Every $\{ \boldsymbol x \}$, $\boldsymbol x \in k^{n}$, is minimal $k$-algebraic, so we get an embedding $k^{n} \hookrightarrow \text{mSpec }k[T_{1},\dots,T_{n}]$. > 2. Specifically, the claim is that the restricted map $I(\cdot) |_{k^{n}}: k^{n} \hookrightarrow \text{mSpec }k[T_{1},\dots,T_{n}]$ is given by $\boldsymbol x \mapsto \mathfrak{m}_{x}$, where $\mathfrak{m}_{\boldsymbol x}=\langle T_{1}-x_{1},\dots,T_{n}-x_{n} \rangle$. > 3. If $k=\Omega$ is [[algebraically closed]], then we have a [[bijection]] between $k^{n}$ and $\text{mSpec }k[T_{1},\dots,T_{n}]$. > > > > **1.** Put $\mathcal{C}:=\{ \{ \boldsymbol x \}: \boldsymbol x \in k^{n} \} \subset \Omega^{n}$. Want to show $\mathcal{C} \subset \mathcal{A}$. Indeed, $\{ \boldsymbol x \}=V(\mathfrak{m}_{\boldsymbol x})$ for $\mathfrak{m}_{\boldsymbol x}=\langle T_{1}-x_{1},\dots,T_{n}-x_{n} \rangle$. $\mathfrak{m}_{\boldsymbol x}$ is [[maximal ideal|maximal]] because $\frac{k[T_{1},\dots,T_{n}]}{\mathfrak{m}_{\boldsymbol x}} \cong k$ is a [[field]]. > > **2.** Applying $I(\cdot)$, we see that $I(\{ \boldsymbol x \})=I(V(\mathfrak{m}_{\boldsymbol x}))=\mathfrak{m}_{\boldsymbol x}$ by the [[strong Nullstellensatz]]. > > **3.** So we have an injective map $\boldsymbol x \xmapsto{{I(\{ \cdot \})}}\mathfrak{m}_{x}$, but in general the inclusion $\mathcal{C} \subset \mathcal{A}$ is proper so it is not a bijection. But if $k=\Omega$, then $\mathcal{C}=\{\{ \boldsymbol x\} : \boldsymbol x \in \Omega^{n} \}$ and so the only minimal nonempty $k$-algebraic sets must be the singletons (if every singleton is $k$-algebraic, then every minimal nonempty $k$-algebraic set will contain, and hence have to be, a singleton). > ---- #### [^1]: By the [[strong Nullstellensatz]] we have $I(V(\langle f \rangle))=\sqrt{ \langle f \rangle }$, but here we only need the easy-to-prove inclusion. ----- #### 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 > ```