Commutative Algebra (IX): The Induced Map on the Spectrum

21 Mar 2019

• commutative algebra

In the previous post, we defined the prime spectrum of a ring as a topological space. In this post, we complete the definition of $\Spec$ as a contravariant functor from the category of (commutative unital) rings to the category of topological spaces.

The induced map

Let $A$ and $B$ be rings with spectra $X=\Spec(A)$ and $Y=\Spec(B)$. For this post, fix a ring homomorphism $\phi:A\to B$.

As in the coordinate ring case, $\phi$ induces a map $\phi^* :Y\to X$, defined by $\mf q\mapsto\phi^{-1}(\mf q)$. This is a well-defined map by the following result.

Proposition: If $\mf q\subseteq B$ is a prime ideal, then $\phi^{-1}(\mf q)$ is also a prime ideal.

Proof: For $f,g\in A$, we have

$$\begin{aligned} fg\in\phi^{-1}(\mf q)&\implies\phi(fg)=\phi(f)\phi(g)\in\mf q\\ &\implies\phi(f)\in\mf q\text{ or }\phi(g)\in\mf q\\ &\implies f\in\phi^{-1}(\mf q)\text{ or }g\in \phi^{-1}(\mf q).\ \qed \end{aligned}$$

This is also why we define the spectrum as the set of prime ideals rather than maximal ideals: the inverse image of a maximal ideal might not be maximal.

Proposition: The induced map of the identity $\id_A:A\to A$ is $(\id_A)^* =\id_X$.

Proof: Note that $(\id_A)^{-1}(\mf p)=\mf p$. $\qed$

Proposition: Let $f:A\to B$ and $g:B\to C$ be ring homomorphisms, which give the induced maps $f^* :Y\to X$, $g^* :Z\to Y$, and $(gf)^* :Z\to X$, where $Z=\Spec(C)$. Then

$$(gf)^* =f^* g^* .$$

Proof: This follows from $f^{-1}(g^{-1}(\mc p))=(gf)^{-1}(\mc p)$. $\qed$

Proposition: Let $\mf a\subseteq A$ be an ideal. Then $(\phi^* )^{-1}(V_A(\mf a))=V_B(\phi(\mf a))$.

Proof: For any $\mf q\in Y$, ie. $\mf q\subseteq B$ a prime ideal, we have

$$\begin{aligned} \mf q\in(\phi^* )^{-1}(V_A(\mf a))&\iff\phi^* (\mf q)\in V_A(\mf a)\\ &\iff\mf a\subseteq\phi^{-1}(\mf q)\\ &\iff\phi(\mf a)\subseteq\mf q\\ &\iff\mf q\in V_B(\phi(\mf a)).\ \qed \end{aligned}$$

Corollary: $\phi^* $ is a continuous map.

Proof: By the above result, the preimage of all closed sets in $X$ under $\phi^* $ are closed in $Y$. $\qed$

Combining all the results above, we obtain:

Theorem: The mapping $\Spec$ sending $A\mapsto\Spec(A)$ and $f\mapsto\Spec(f)\coloneqq f^* $ defines a contravariant functor from the category of (commutative unital) rings to the category of topological spaces. $\qed$

Direct image under the induced map

In the previous section, we worked out the inverse image of closed sets under $\phi^* $. For the direct image, we first prove some intermediate results.

Proposition: $\phi^* (Y)$ is dense in $X$ if and only if $\ker\phi\subseteq\nil(A)$.

Proof: ($\Leftarrow$) Note that for all $f\in A$, if $X_f\neq\emptyset$ then

$$\begin{aligned} &f\not\in\mf p\text{ for some }\mf p\subseteq A\text{ prime}\\ &\implies f\not\in\nil(A)\\ &\implies f^n\not\in\nil(A)\text{ for all }n\geq1\\ &\implies\phi(f)^n\neq0\text{ for all }n\geq1\\ &\implies\phi(f)\not\in\nil(B)\\ &\implies\phi(f)\not\in\mf q\text{ for some }\mf q\subseteq B\text{ prime}\\ &\implies f\not\in\phi^{-1}(\mf q)=\phi^* (\mf q)\\ &\implies\phi^* (\mf q)\in X_f. \end{aligned}$$

Hence $\phi^* (Y)$ intersects every nonempty basic open set $X_f\subseteq X$, so $\phi^* (Y)$ is dense in $X$.

($\Rightarrow$) We prove the contrapositive. If $\ker\phi\not\subseteq\nil(A)$, take any $f\in\ker\phi\backslash\nil(A)$. Then

$$\begin{aligned} &0=\phi(f)\in\mf q\text{ for all }\mf q\in Y\\ &\implies f\in\phi^{-1}(\mf q)=\phi^* (\mf q)\text{ for all }\mf q\in Y\\ &\implies\phi^* (Y)\subseteq V(f)\\ &\implies\phi^* (Y)\cap X_f=\emptyset. \end{aligned}$$

Now $f\not\in\nil(A)$, so $X_f$ is a nonempty open set in $X$. Hence $\phi^* (Y)$ is not dense in $X$. $\qed$

Proposition: Suppose that $\phi$ is surjective. Then $\phi^* $ is a homeomorphism of $Y$ onto $V_A(\ker\phi)\subseteq X$.

Proof: Note that $B\cong A/\ker\phi$. By the lattice isomorphism theorem, there is a correspondence

$$\begin{aligned} Y=\left\{\begin{gathered}\text{prime ideals}\\\text{in }B\end{gathered}\right\}&\leftrightarrow\left\{\begin{gathered}\text{prime ideals in }A\\\text{containing }\ker\phi\end{gathered}\right\}=V_A(\ker\phi)\\ \mf q\quad&\mapsto\quad\phi^* (\mf q)\\ \phi(\mf p)\quad&\gets\quad\mf p, \end{aligned}$$

so $\phi^* :Y\to V_A(\ker\phi)$ is a continuous bijection.

Moreover, this correspondence is inclusion-preserving, so for any ideal $\mf b\subseteq B$, the ideals containing $\mf b$ correspond exactly to the ideals containing $\phi^* (\mf b)$. In other words,

$$\phi^* (V_B(\mf b))=V_A(\phi^* (\mf b)),$$

so $\phi^* $ maps every closed set in $Y$ to a closed set in $V_A(\ker\phi)$. Hence $\phi^* $ is a homeomorphism. $\qed$

Corollary: $\Spec(A)$ and $\Spec(A/\nil(A))$ are naturally homeomorphic.

Theorem: For any ideal $\mf b\subseteq B$, we have

$$\overline{\phi^* (V_B(\mf b))}=V_A(\phi^* (\mf b)).$$

Proof: Note that the composition $A\overset\phi\to B\twoheadrightarrow B/\mf b$ has kernel $\phi^* (\mf b)$, so we have the induced map $\tilde\phi$ in the following commutative diagram, which is injective:

$$\begin{matrix} A&\overset\phi\longrightarrow&B\\ \bigg\downarrow&&\bigg\downarrow\\ A/\phi^* (\mf b)&\underset{\tilde\phi}\longrightarrow&B/\mf b \end{matrix}$$

Now applying the $\Spec$ functor gives the following commutative diagram:

$$\def\ua#1{\bigg\uparrow\raisebox{.5ex}{\mathrlap{$\scriptstyle{#1}$}}} \begin{matrix} X&\overset{\phi^* }\longleftarrow&Y\\ \ua{\subseteq}&&\ua{\subseteq}\\ V_A(\phi^* (\mf b))&&V_B(\mf b)\\ \ua{* }&&\ua{* }\\ \Spec(A/\phi^* (\mf b))&\overset{\tilde\phi^* }\longleftarrow&\Spec(B/\mf b) \end{matrix}$$

Now by the previous two results, the starred maps are homeomorphisms, and the image of $\tilde\phi^* $ is dense in the codomain (since $\ker\tilde\phi=\{0\}$). Hence $\phi^* $ maps $V_B(\mf b)$ into $V_A(\phi^* (\mf b))$ with dense image, as desired. $\qed$