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I'd like some help to prove that a morphism of schemes $f:X\to Y$ if étale.

Here are the characters: $X=\textrm{Spec}\,k[x,x^{-1}]$, $Y=\textrm{Spec}\,k[t]$ and $f$ is induced by $t\mapsto x^2$. [We may assume $k=\overline k$]

I tried to show $f$ is étale in two different manners.

  1. By showing that it is flat and unramified. It is certainly flat, because it is dominant over a nonsingular curve (and it is dominant because $f^\sharp$ is injective). I am not able to show it is unramified, because I cannot figure this morphism concretely. What I can see is just that under $f^\sharp$ a polynomial $p(t)=\sum_{i\geq 0}a_it^i$ goes to $a_0+a_1x^2+a_2t^4+\dots$. Can you please help to me to actually see in the most concrete way what is the image of a point $x\in X$ under $f$?
  2. By showing that $f$ is smooth of relative dimension $0$. It is certainly flat of relative dimension $0$, but I still need to show that the geometric fibers of $f$ are nonsingular and pure of dimension $0$. So for every closed point $y\in Y$ we have $X_y=\textrm{Spec}\,(k[x,x^{-1}]\otimes_{k[t]}k)=X_{\overline y}$ and over the generic fiber $$ X_{\overline{\eta}}=\textrm{Spec}\,k[x,x^{-1}]\otimes_{k(t)}\overline{k(t)}. $$ Can you help me understand what "is" this $X_{\overline y}$?

Thank you all.

Jack
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    I suck at algebraic geometry, but I would think about this as follows. $X$ can be viewed as a plane curve $x_1x_2=1$, and $Y$ is just the affine line. The morfism is $(x_1,x_2)\mapsto x_1^2$. The Jacobian behaves badly at the points where $x_1=0$, but thankfully there aren't any on $X$. The fibers would be: empty above $t=0$, two distinct points above $t\neq0$, and above the generic point we get the spectrum of $k[x,x^{-1}]\otimes_{k[t]}k(t)$. As a $k[t]=k[x^2]$-module the ring $k[x,x^{-1}]$ is the union of free modules $V_i$ of rank two generated by $x^{-2i}$ and $x^{-2i+1}$. – Jyrki Lahtonen Jun 27 '13 at 08:12
  • cont'd It seems to me that this implies that $$k[x,x^{-1}]\otimes_{k[t]}k(t)$$ is a two-dimensional vector space over $k(t)$, but I don't trust myself 100% here. Doesn't that do it? – Jyrki Lahtonen Jun 27 '13 at 08:12
  • In 2. there seems to be a confusion between generic fiber and geometric fiber. I don't see why $\overline {k(t)}$ is particularly relevant here. – Georges Elencwajg Jun 27 '13 at 14:49
  • To check smoothness, doesn't one have to check regularity of all geometric fibers? (in particular the generic one) – Jack Jun 27 '13 at 14:56

2 Answers2

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0 You must assume $char.k\neq 2$, else the morphism $f$ is not étale at all.

1 The image of the generic point $(0)\in \textrm{Spec}\,k[x,x^{-1}]$ of $X$ is the generic point $(0)\in \textrm{Spec}\,k[t]$ of $Y$.
The other points of $X$ are closed and the image of such a closed point point $(x-a) \in \textrm{Spec}\,k[x,x^{-1}] \:(a\neq 0)$ of $X$ is the closed point $(t-a^2)\in \textrm{Spec}\,k[t]$ of $Y$.

2 The generic fiber of $f$ is the fiber of the generic point $\eta=(0)=\textrm{Spec}\,k[t]$.
That fiber consists of the generic point $\xi=(0)\in\ \textrm{Spec}\,k[x,x^{-1}]$.
At the generic point $\eta$ of $Y$ the dual morphism of local rings is the field extension $f^{\star}_\eta:\mathcal O_{Y,\eta}=k(t)\hookrightarrow k(x)=\mathcal O_{X,\xi}:t\mapsto x^2$.
This field extension is (as it should) étale of degree 2 or, in a more traditional terminology , separable of degree 2.

Georges Elencwajg
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  • Dear Georges, thank you! In 1, how can you represent a point of $X$ as $(x-a)$? shouldn't we have two "coordinates"? in a previous comment the morphism was described as $(x_1,x_2)\mapsto x_1^2$ ($=(t-x_1^2)$). Also, is it enough to check the generic fiber? – Jack Jun 27 '13 at 14:58
  • Dear Jack, I wrote my answer in the terminology of schemes, since you wrote the question in that language. A point of the scheme $X$ has no coordinates but rather is a prime ideal in the ring $k[x,x^{-1}]$. For example $(x-a)$ is the prime ideal consisting of all multiples of $x-a$ in the ring $k[x,x^{-1}]$: as an example $(x-a)\cdot(3x^{-2}+4x^7)$ is such a multiple. Notice tht this ideal $(x-a)$ equals the ideal $(1-ax^{-1})$ since $x$ is invertible. – Georges Elencwajg Jun 27 '13 at 15:05
  • As for checking the generic fiber, you just localize the comorphism $f^\ast:k[t]\to k[x,x^{-1}]:t\mapsto x^2$ at the prime ideals $(0)\subset k[t]$ of $Y$ and $(0)\subset k[x,x^{-1}]$ of $X$. – Georges Elencwajg Jun 27 '13 at 15:13
  • I wanted to make a list numbered 0.,1., 2. but when I tried to write it as above, the software replaced it by 1., 2., 3. How can I override that undesired transformation? – Georges Elencwajg Jun 27 '13 at 15:16
  • I have no idea, but it is clear! Thank you for your neat answers, @Georges. – Jack Jun 27 '13 at 21:29
  • You are welcome, Jack. – Georges Elencwajg Jun 27 '13 at 21:34
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Showing that it is unramified seems easier to me. Indeed, we have to show that for any $P\in X$ and $Q=f(P)$, you have that $f^\sharp(\mathfrak m_Q)\cdot\mathcal O_{X,P}=\mathfrak m_P$. Easily enough, $Y$ is just the line, so it is one-dimensional and smooth, and we know that $\mathfrak m_Q=(\pi)$ is a principal ideal, generated by a uniformizing parameter of the form $\pi=t-a$. Then, $f^\sharp(\pi)=x^2-a$. Since we are over an algebraically closed field, $f^\sharp(\pi)=(x-a_1)(x-a_2)$ for certain $a_1,a_2\in k$. Note that $a_1=a_2$ if and only if $a=0$ (this is where I assume sheepishly that the characteristic is not 2). We know for sure that $f^\sharp(\mathfrak m_Q)\subseteq\mathfrak m_P$ and $\mathfrak m_P=(x-b)$ for some $b\ne 0$, so we have $(x-a_1)(x-a_2)=\rho\cdot(x-b)$ for some $\rho\in\mathcal O_{X,P}$. Since $\mathcal O_{X,P}$ is the localization of the UFD $k[x]$, it is again a UFD. Since $x-a_i$ is either irreducible or a unit and $x-b$ is irreducible, we may assume without loss of generality that $b=a_1$ and $\mathfrak m_P=(x-a_1)$. Note that this already implies $a\ne 0$. Since $x-a_2$ is a unit in $\mathcal O_{X,P}$, we are done.

Jesko Hüttenhain
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  • Dear Jesko, thank you! Let me see if I understand correctly your last argument (maybe it's a bit different in fact): $x-a_1$ and $x-a_2$ are not both units because otherwise $x-b$ would not be a uniformizer. So if $x-a_1$ is irreducible then $x-a_2$ must be a unit by UF. Hence $m_P \subset f^\sharp(m_Q)\mathcal O_{X,P}$. Is it ok? – Jack Jun 27 '13 at 14:32
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    $x-a_1$ and $x-a_2$ are not both units because otherwise, $f^\sharp(\pi)$ would be a unit, which is impossible because $f^\sharp$ maps $\mathfrak m_Q$ into $\mathfrak m_P$, and $\mathfrak m_P$ contains no units. If $x-a_1$ is irreducible, then it must be equal to the factor $x-b$. Hence $b=a_1\ne a_2$, hence $x-a_2$ is a unit. I also added the part where I missed the fact that everything goes south in characteristic two. – Jesko Hüttenhain Jun 27 '13 at 19:35