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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 $f(x \cdot y)=f(xy)=yx\neq x^2 = x\cdot f(y)$.

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1 Answer 1

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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.

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