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I have asked a question here

Applying Chevalley's Theorem to Elimination of quantifiers

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And now I'm having some trouble showing that $(im \pi) \cap \bar{k}^m =Y = im{\pi}^{cl}$

What I have tried : It's suffice to show that there exists a closed point in $\pi^{-1}(c)$, where $c$ is any closed point of $\mathbb{A}^{m}$，so I want to use the fact if $X$ is a quasicompact scheme, then every nonempty closed subset of $X$ contains a closed point of $X$(vakil's FOAG, 5.1.E). So it remains to show $X$ is quasicompact, which could be done by noticing that intersection of an quasicompact open set and quasicompact closed set is again quasicompact.

Does this approach work? But I think it could be done directly without ivoking the fact about closed point in quasicompact scheme, which I can't figure out, could you help me? Thanks in advance.

Mugenen
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1 Answers1

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Your proposed solution using quasicompactness works. Alternatively, one may note that there is a closed point by the nullstellensatz: $X$ is the projection from $\Bbb A^{m+n+q}$ (with coordinates $(w_1,\cdots,w_m,x_1,\cdots,x_n,t_1,\cdots,t_q)$) of the variety cut out by the ideal $(f_1,\cdots,f_p,g_1t_1-1,g_2t_2-1,\cdots,g_qt_q-1)$, which has a closed point of the form $(w_1-a_1,\cdots,x_1-b_1,\cdots,t_1-c_1,\cdots)$, and it's clear to see that the image of this closed point under the map $\Bbb A^{m+n+q}\to \Bbb A^{m+n}\to \Bbb A^m$ is again a closed point.

KReiser
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