# When is the diagonal a closed immersion with open image?

Let $$A$$ be a commutative unital $$R$$-algebra and consider the multiplication map $$m:A\otimes_R A\rightarrow A$$ given by $$m(a\otimes b) = ab$$ with kernel $$I$$. The map $$m$$ is surjective, hence the image of $$f: \operatorname{Spec}A\rightarrow \operatorname{Spec}A\times\operatorname{Spec}A$$ is $$V(I)$$.

If we assume that $$V(I)$$ is also open, then is $$f$$ an open immersion?

As far as I know, an immersion with closed image is a closed immersion, but an immersion with open image maybe not an open immersion?

Thanks.

• If $X$ is a nonreduced scheme, $X_{red} \rightarrow X$ is a closed immersion with open image but isn’t an open immersion. Commented Feb 22, 2021 at 20:15
• @Mindlack It may be worth pointing out that if $X$ is connected then this is essentially the only obstruction. Commented Feb 22, 2021 at 21:01
• @Alex Youcis: what does connectedness have to do with it? If $f: Y \rightarrow X$ is a closed immersion with open image and $X$ reduced, then if $F \subset X$ is the image of $f$ (and thus open in $X$), $Y \rightarrow F$ is a surjective closed immersion with $F$ reduced so is an isomorphism – thus $f$ is an open immersion. Commented Feb 22, 2021 at 21:40
• @Mindlack I misread the problem as 'isomorphism', not 'open embedding'. Commented Feb 22, 2021 at 23:44

It's helpful to remember the sheaf-theoretic conditions on immersions: an open immersion $$f:Y\to X$$ must have that the induced maps $$\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$$ are isomorphisms for all $$y\in Y$$, while a closed immersion only implies that those maps are surjections. Assuming $$f$$ is a closed immersion and adding the condition $$f(Y)$$ open only implies that the kernel of $$\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$$ is in the nilradical of $$\mathcal{O}_{X,f(y)}$$, which is not enough to guarantee that we have an open immersion unless $$\mathcal{O}_{X,f(y)}$$ is reduced for all $$y\in Y$$.

Here is an explicit example in your specific case. Let $$R=k$$ be a field and let $$A=k[\epsilon]/(\epsilon^2)$$. Then $$A\otimes_R A\cong k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)$$, so both $$\operatorname{Spec} A$$ and $$\operatorname{Spec} A\otimes_R A$$ are single points, and if $$\operatorname{Spec} m$$ is an open immersion it must be an isomorphism. Hence $$m$$ should be an isomorphism, but one can immediately see that this is not the case as $$1\otimes\epsilon-\epsilon\otimes1$$ is in the kernel.

I believe the following holds (by an exercise in Atiyah-Macdonald (AM)): Let $$B:=A \otimes A$$ and let $$I:=ker(m)$$ be the kernel of the multiplication map. It follows $$V(I) \cup V(I)^c=S:=Spec(B)$$. If $$V(I)^c=V(J)$$ is closed it follows by AM Exercise I.15 that

$$V(I) \cup V(J)= V(I\cap J)=S$$

hence $$I \cap J \subseteq \mathfrak{p}$$ for all prime ideals $$\mathfrak{p}\subseteq B$$, hence $$I\cap J \subseteq nil(B)$$. Since $$V(I) \cap V(J)=V(I+J)=\emptyset$$ it follows $$I+J=(1)$$ is the unit ideal. Hence if $$B$$ is a domain it follows $$I\cap J =(0)$$ and $$I+J =(1)$$ hence there is by the Chinese remainder theorem an isomorphism

$$B:=A\otimes A \cong A\otimes A/I \oplus A\otimes A/J \cong A \oplus A\otimes A/J.$$

Hence

$$S:=Spec(A\otimes A) \cong Spec(A) \cup Spec(A\otimes A/J):=S_1 \cup S_2$$

is a disjoint union. It seems the diagonal map $$\Delta$$ is an open immersion in this case, since $$Spec(A)\cong S_1 \subseteq Spec(B)$$ is an open subscheme. The two schemes $$S_1:=Spec(A),S_2:=Spec(B/J)$$ are open and closed in $$S$$.

Question: "If we assume that $$V(I)$$ is an open subscheme, then is $$f$$ an open immersion?"

Answer: If the product $$S$$ is a reduced scheme it seems the following holds:

Lemma. If $$V(I)$$ is an open subscheme it follows the diagonal map $$\Delta$$ is an open immersion.

Proof. This holds since $$nil(B)=0$$ and hence by the above argument it follows the diagonal map $$\Delta$$ induces an isomorphism between $$Spec(A)$$ and the image $$\Delta(Spec(A))\cong S_1 \subseteq S$$.

Example. In general if $$S:=Spec(A)$$ is a reduced affine scheme and $$Z:=V(I)$$ is open and closed it follows $$V(I)^c=V(J)$$ and $$I \cap J =(0), I+J=(1)$$ hence by the Chinese remainder theorem we get a direct sum decomposition

$$A \cong A/I \oplus A/J$$

and a disjoint union of schemes

$$Spec(A) \cong Spec(A/I) \cup Spec(A/J):=V(I) \cup V(I)^c.$$

Conversely if $$A \cong A_1 \oplus A_2$$ there are ideals $$I_i\subseteq A$$ with $$A/I_i \cong A_i$$ and $$I_1 \cap I_2=(0), I_1+I_2=(1).$$ The subschemes $$Spec(A_i)\subseteq Spec(A)$$ are open and closed.