---- > [!definition] Definition. ([[prime ideal]]) > An [[ideal]] $I \neq \langle 1 \rangle$ of a [[commutative ring|commutative]] [[ring]] $R$ is called a **prime ideal** if $R / I$ is an [[integral domain]]. > > The set of all [[prime ideal|prime ideals]] of a [[ring]] $R$ is called its **spectrum** and denoted $\text{Spec }R$. > Endowing spectra with the [[Zariski topology on a ring spectrum|Zariski topology]], $\text{Spec }$is a [[contravariant functor]] $\mathsf{CRing} \to \mathsf{Top}$. It sends morphisms to their [[contraction of an ideal|contraction maps]]: $\text{Spec}(A \xrightarrow{\varphi}B):=\varphi ^{-1}(\cdot)$.[^1] $\text{Spec}$ may be upgraded to a [[covariant functor|functor]] $\mathsf{CRing}^{\text{op}} \to \mathsf{Aff}$: [[spec functor]]. > [!equivalence] > $I \neq \langle 1 \rangle$ is a prime ideal if and only if $ab \in I \implies (a \in I \text{ or } b \in I)$ for all $a,b \in R$. ^equivalence > [!basicproperties] > - [[prime and maximal ideals align for finite quotients]] (since [[a finite commutative ring is an integral domain iff it is a field]]) > - [[prime iff maximal for nonzero ideals in PID]] > - If $I_{1}$ and $I_{2}$ are two [[ideal|ideals]] of $R$, and $\mathfrak{p}$ is prime with $\mathfrak{p} \supset I_{1} \cap I_{2}$, then in fact either $\mathfrak{p} \supset I_{1}$ or $\mathfrak{p} \supset I_{2}$. Induction extends the result to hold for $\mathfrak{p} \supset I_{1} \cap \dots \cap I_{r}$. (The converse trivially holds.) > - If in fact $\bigcap_{i=1}^{r}I_{i}=\mathfrak{p}$ then ${I}_{i}=\mathfrak{p}$ for some $i$. ^properties If $I_{1}$ and $I_{2}$ are two ideals of $R$ and $\mathfrak{p}$ is prime with $I_{1}$ > [!basicexample] Example. ($\text{Spec }k[X], k[X,Y]$ for $k=\overline{k}$ ) > Let $k=\overline{k}$ be [[algebraically closed]]. **$\text{Spec }k[X]$.** $k[X]$ is a [[PID]]; by [[prime iff maximal iff irreducible for nonzero ideals in a PID]] its prime ideals are just its maximal ideals and zero, i.e., $\text{Spec }k[X]=\{ \langle X-a \rangle : a \in k\} \cup \{ (0) \}$. > **$\text{Spec }k[X,Y]$.** Much trickier. Can show $\text{Spec }k[X,Y]=\underbrace{ \{ \langle X-a, Y-b \rangle : a, b \in k \} }_{ \text{mSpec }k[X,Y] } \cup \underbrace{ \{ \langle f \rangle : f \text{ irreducible} \} }_{ \text{principal ideals} } \cup \underbrace{ \{ (0) \} }_{ k[X,Y] \text{ an }I.D. }$ In general, classifying $\text{Spec }k[T_{1},\dots.,T_{n}]$ is difficult. ^basic-example If $\mathfrak{p} \in \text{Spec } A$ then $\text{Spec }A/\mathfrak{p} \cong V(\mathfrak{p})$. Indeed, [[the correspondence theorem for rings]] gives a [[bijection]] $ \text{Spec } A / \mathfrak{p}\leftrightarrow \{ \mathfrak{q} \in \text{Spec } A : \mathfrak{q} \supset \mathfrak{p}\}=V(\mathfrak{p})$ which is in fact a [[homeomorphism]]. > [!proof]+ Proof of Equivalence. > Throughout, $a,b$ are arbitrary elements of $R$. $\begin{align} I \text{ is prime } \iff & \text{R/I} \text{ is an integral domain} \\ \iff & [(a+ I)(b+I) = I \implies (a+I) = I \text{ or } (b+I)=I] \\ \iff & [(ab + I) = I \implies (a+I) = I \text{ or } (b+I)=I] \\ \iff & [ab \in I \implies a \in I \text{ or } b \in I], \end{align}$ where the last equivalence uses the general fact regarding [[coset|cosets]] that for any [[group]] $(G, \cdot)$ and [[subgroup]] $H$, $g \cdot H=H \iff g \in H$. ^proof > [!proof] Proof of Basic Properties. > Suppose not; then we could find $a_{1} \in I_{1}$ and $a_{2} \in I_{2}$ such that $a_{1} \notin \mathfrak{p}$ and $a_{2} \notin \mathfrak{p}$. This is impossible: let $a_{1} \in I_{1}$ and $a_{2} \in I_{2}$ be arbitrary. $a_{1}a_{2} \in I_{1}$ since $a_{1} \in I_{1}$ and $I_{1}$ an [[ideal]]; similarly $a_{1}s_{2} \in I_{2}$. So $a_{1}a_{2} \in I_{1} \cap I_{2} \subset \mathfrak{p}$. Since $\mathfrak{p}$ [[prime ideal|prime]], this means either $a_{1} \in \mathfrak{p}$ or $a_{2} \in \mathfrak{p}$. > Then if we are given $\mathfrak{p} \supset I_{1} \cap\dots \cap I_{r}$, we can recurse/induct to argue that $\mathfrak{p} \supset I_{i}$ for some $i \in [r]$. > If $I_{1},I_{2}$ are [[ideal|ideals]] of $R$ such that $I_{1} \neq \mathfrak{p}$ and $I_{2} \neq \mathfrak{p}$, then we can obtain $a_{1} \in I_{1} \cap (R - \mathfrak{p})$ and $a_{2} \in I_{2} \cap (R -\mathfrak{p})$. Then $a_{1}a_{2} \in I_{1}$ and $a_{1}a_{2} \in I_{2}$ (because ideal) so $a_{1}a_{2} \in I_{1} \cap I_{2}$. But $R-\mathfrak{p} \subset R$ is obviously a [[multiplicative subset of a ring|multiplicative set]] so $a_{1}a_{2} \in R-\mathfrak{p}$. It follows that $I_{1} \cap I_{2} \neq \mathfrak{p}$. Contrapositively, we conclude that $I_{1} \cap I_{2}=\mathfrak{p}$ implies $I_{1}=\mathfrak{p}$ or $I_{2}=\mathfrak{p}$. ---- #### [^1]: Recall from [[contraction of an ideal#^properties]] that [[prime ideal|prime ideals]] contract to prime ideals. So a [[ring homomorphism]] $A \to B$ indeed gives a map $f:\text{Spec }B \to \text{Spec }A$ via $f(\mathfrak{q}):=\varphi ^{-1}(\mathfrak{q})$. To see $f$ is [[continuous]], let $V(I)$ be any [[closed set|closed]] subset of $\text{Spec }A$. Then $\begin{align} \mathfrak{h} \in f ^{-1}\big( V(I) \big)& \iff \varphi ^{-1}(\mathfrak{h}) \in V(I) \\ & \iff \varphi ^{-1}(\mathfrak{h}) \supset I \\ & \iff \varphi \varphi ^{-1} (\mathfrak{h}) \supset \varphi(I) \\ & \iff \mathfrak{h} \supset \langle \varphi(I) \rangle \\ & \iff \mathfrak{h} \in V(\langle \varphi(I) \rangle ) \end{align}$ from which it follows that the inverse image of a closed set under $f$ is closed. ---- #### 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 > ```