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Let $Z\newcommand{\df}{:=}\df\newcommand{\C}{\mathbb C}\C$ and $T\df\C^\times$. Then, the coordinate ring of $Z$ is $\C[z]$ and that of $T$ is $\C[t,t^{-1}]$. Consider another copy of $T$ with coordinate ring $\C[u,u^{-1}]$. The map $$\begin{align*}\C[z]\otimes_\C\C[t,t^{-1}]&\longrightarrow\C[u,u^{-1}]\\z^a\otimes t^b&\longmapsto u^{a+b}\end{align*}$$ should be a surjective ring homomorphism, inducing a closed immersion $\kappa:T\hookrightarrow Z\times T$. However, the usual (open!) inclusion $\omega: T\hookrightarrow Z$ factors as $\omega=\DeclareMathOperator{\pr}{pr}\pr_Z\circ\kappa$ where $\pr_Z:Z\times T\twoheadrightarrow Z$ is the canonical projection. But this would imply that $\kappa(T)=\pr_Z^{-1}(\omega(T))$ is both closed and open and $Z\times T$ is connected.

What am I doing wrong?

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    $\begingroup$ I would suggest drawing some pictures. $\endgroup$ – David E Speyer Aug 29 '12 at 13:29
  • $\begingroup$ That's what I did, it's where the confusion came from. $\endgroup$ – user38830 Aug 29 '12 at 13:46
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    $\begingroup$ However - I drew the wrong picture. I embedded $T$ as $T\times\{1\}$, but it has to be embedded diagonally - with that in mind, everything becomes clear. $\endgroup$ – user38830 Aug 29 '12 at 14:44
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The commutativity of maps $\omega = \textrm{pr}_{Z} \circ \kappa$ only implies that $\textrm{pr}_{Z}^{-1}(\omega(T)) \supset \kappa(T)$. The equality holds iff $\textrm{pr}_{Z}$ is injective, which is not the case.

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    $\begingroup$ Ahh! That is, of course, true. I now also realized why this is not confusing (for anyone still confused): $\kappa$ embeds $T$ diagonally, which is why it becomes closed in $Z\times T$. $\endgroup$ – user38830 Aug 29 '12 at 14:43

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