Can somebody point me good reference about the triviality of etale fundamental group for rational affine space over zero characteristic fields?
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1$\begingroup$ No, it's not simply connected. $\endgroup$– Dan PetersenCommented Sep 13, 2018 at 7:13
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$\begingroup$ I consider zero characteristic fields only. $\endgroup$– KonstantinCommented Sep 13, 2018 at 7:25
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5$\begingroup$ You consider $\Bbb{Q}$, which is very far from being simply connected. $\endgroup$– abxCommented Sep 13, 2018 at 7:56
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$\begingroup$ Simply connected means any covering of Spec has ramification. $\endgroup$– KonstantinCommented Sep 13, 2018 at 8:07
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8$\begingroup$ The point is that $\pi_1(\mathrm{Spec}(\mathbb Q)) = \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ sits inside $\pi_1(\mathrm{Spec}(\mathbb Q[x_1,\ldots,x_n]))$. Konstantin, you would be better off reading something like Milne's notes on étale cohomology rather than trying to ask questions here at this stage of your learning. $\endgroup$– Dan PetersenCommented Sep 13, 2018 at 8:43
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$\mathrm{Spec}(\mathbb{Q}[x_1,...,x_n])=\mathbb{A}^n_\mathbb{Q}$ is not étale simply connected as $\widehat{\pi}_1(\mathbb{A}^n_\mathbb{Q})=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, but $\mathrm{Spec}(\overline{\mathbb{Q}}[x_1,...,x_n])=\mathbb{A}^n_{\overline{\mathbb{Q}}}$ is étale simply connected, as is the case for any algebraically closed field of characteristic 0. One way to see this is that for X a quasi-projective scheme over a field k we have a short exact sequence $0\rightarrow\widehat{\pi}_1(X_{\overline{k}})\rightarrow\widehat{\pi}_1(X)\rightarrow\mathrm{Gal}(\overline{k}/k)\rightarrow 0$, as any finite étale cover of $\mathrm{Spec}(k)$ induces a finite étale cover of $X$ by extension of scalars, which induces the trivial finite étale cover of $X_{\overline{k}}$ by extension of scalars to $\overline{k}$. In the case $X=\mathbb{A}^n_\mathbb{Q}$ we have $\widehat{\pi}_1(X_{\overline{\mathbb{Q}}})=0$ by a simple Riemann-Hurwitz argument so it follows that $\widehat{\pi}_1(\mathbb{A}^n_\mathbb{Q})=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ by exactness.
