2

I know that if $A$ is a finite ring, then $\pi_1(\text{Spec} \,A)=\prod_{\mathfrak{p}\in \text{Spec} A} \hat{\mathbb{Z}}$. (see for example Some basic examples of étale fundamental groups). In particular, if $A=\mathbb{F}_{p^n}[x]/(x^2)$, then $\pi_1(\text{Spec} A)= Gal(\bar{k}/k)$.

My question is whether or not this extends to non-finite fields, i.e. whether or not $\pi_1(\text{Spec}\,k[x]/(x^2))= Gal(\bar{k}/k)$ for any field $k$. I know by functoriality of $\pi$ and the fact that the identity on $k$ can be written as $k\rightarrow k[x]/(x^2) \rightarrow k$ that $\pi_1(\text{Spec} \,k[x]/(x^2))$ surjects onto $\pi_1(\text{Spec}\, k)=Gal(\bar{k}/k)$, but I don't understand the kernel of this map.

Can someone point me the right way or even show me the solution, that would be great!

Wish you all a great day!

Alejandro Tolcachier
  • 1,964
  • 2
  • 12
  • 28
curious math guy
  • 2,008
  • Why do you write $\pi_1(\operatorname{Spec} A) = Gal(\overline{k}/k)^2$? Isn't $Gal(\overline{k}/k)\cong \hat{\Bbb Z}$ here, and $(x)\subset \Bbb F_q[x]/(x^2)$ the only prime ideal? Or have I made a very silly mistake? – KReiser Jan 28 '20 at 21:47
  • No, it is me who has made a mistake, apologies. I've eddited my question. – curious math guy Jan 28 '20 at 22:04
  • 3
    Don't we have more generally that $\pi_1(X)=\pi_1(X_{red})$ ? See here : https://stacks.math.columbia.edu/tag/0BQB – Roland Jan 28 '20 at 23:49
  • @Roland that looks like an answer to me - would you care to post it as such? – KReiser Jan 29 '20 at 06:37

1 Answers1

3

This is a community wiki answer recording the solution from the comments, in order to remove the question from the unanswered queue:

Don't we have more generally that $\pi_1(X)=\pi_1(X_{red})$? See here: stacks.math.columbia.edu/tag/0BQB – Roland

KReiser
  • 74,746