# If $V=V_1 \cup V_2$ is a connected reducible algebraic set, then we have a strict inclusion of coordinate rings.

I want to show that if $$V=V_1\cup V_2$$ is a connected reducible affine algebraic set, where $$V_1,V_2$$ are proper closed subsets (necessarily with nonempty intersection since $$V$$ is connected), we have a strict inclusion of coordinate rings $$k[V] \subset k[V_1] \times k[V_2]$$.

Is this argument correct?

The inclusion follows from the Chinese Remainder Theorem: let $$R=k[X_1,\ldots,X_n]$$, then

$$k[V] \simeq R/(\mathcal{I}(V))=R/(\mathcal{I}(V_1)\cap \mathcal{I}(V_2)) \hookrightarrow R/(\mathcal{I}(V_1)) \times R/(\mathcal{I}(V_2)) \simeq k[V_1] \times k[V_2]$$.

But why is the inclusion strict? I guess I'd want to appeal to a converse of the Chinese Remainder Theorem (CRT), see here: Converse to Chinese Remainder Theorem

The Converse to CRT says that if $$R/(\mathcal{I}(V_1)\cap \mathcal{I}(V_2)) \simeq R/(\mathcal{I}(V_1)) \times R/(\mathcal{I}(V_2))$$ as $$R$$-modules, then we may deduce that $$\mathcal{I}(V_1)+\mathcal{I}(V_2)=R$$, which implies $$V_1$$ and $$V_2$$ are disjoint, a contradiction. Therefore, we can at least say that $$R/(\mathcal{I}(V_1)\cap \mathcal{I}(V_2)) \not\simeq R/(\mathcal{I}(V_1)) \times R/(\mathcal{I}(V_2))$$ as $$R$$-modules.

However, what we want to show is that they are not isomorphic as rings, or as $$k$$-algebras. So I am done, if the following result holds:

Lemma A ring isomorphism $$R/I \simeq R/J \times R/K$$ must be $$R$$-linear (when viewing both sides as $$R$$-modules in the obvious way).

The problem is, I don't believe this last Lemma really holds: if $$K=R=\mathbb{R}[x,y]$$ and $$I=J=0$$, then there is a ring isomorphism $$f: R/I \rightarrow R/J \times R/K$$, that is, $$f: \mathbb{R}[x,y] \rightarrow \mathbb{R}[x,y]$$, given by $$x \mapsto y, y \mapsto x$$. This does not respect the respective $$R$$-module structures, since $$f(x \cdot y)=f(xy)=yx\neq x^2 = x\cdot f(y)$$.

The key is to look at $$W=V_1\cap V_2$$. For any function $$f\in k[V]$$, it's restriction to $$V_1$$ and then $$W$$ must match it's restriction to $$V_2$$ and then to $$W$$. But there is no such conditions on functions in $$k[V_1]\times k[V_2]$$, and any pair of functions $$(g,h)$$ which do not match on $$W$$ cannot be in the image of the inclusion of $$k[V]$$. So certainly $$(0,1)$$ is not in the image of the inclusion, and thus the inclusion is strict.