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2

$begingroup$

I am trying to prove an exercise in Hartshorne's. Let $k_0$ be a field of characteristic $p>0$, let $k=k_0(t)$, and let $Xsubseteq A_k^2$ be a curve defined by $y^2=x^p-t$. Show that every local ring on $X$ is a regular local ring but $X$ is not smooth over $k$.

For the first statement, a local ring on a point $q$ should be the localization of ring $mathcalO_X$, i.e. $(k[x,y]/y^2-x^p+t)_q$. And a maximal ideal correspondes to a irreducible polynomial in $(k[x,y]/y^2-x^p+t)$, thus, it is regular local ring.

For the second statement, notice that $0=d(y^2-x^p+t)=2dy-px^p-1dx=2dy$, so $Omega_X/k=0$. But $f$ has relative dimension 1, and we conclude that $f$ is not smooth.

I am not sure whether my argument is right. I hope someone can point out my mistake if there is any.

algebraic-geometry

share|cite|improve this question

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are making a mistake when you say $Omega_X/k=0$, please check the definition.
  $endgroup$
  – Mohan
  Mar 29 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mohan - Actually I don't know how to compute $Omega_X/k$ explictly, and I hope someone could help me. It only need to show that $dim_k Omega_X/k=0$, and then the conclusion still follows.
  $endgroup$
  – Yuyi Zhang
  Mar 29 at 15:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ y^2' = 2y $ .
  $endgroup$
  – Youngsu
  Mar 29 at 16:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for the ugly shape of the formula. I needed to put at least 15 characters.
  $endgroup$
  – Youngsu
  Mar 29 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you have a curve $X$ defined as $f(x,y)=0$ (valid more generally), the exact sequence for 1-forms is, with $R=k[x,y]$, $Rfto R/f dxoplus R/f dyto |Omega_X/kto 0$, where the left map is $fmapsto df=f_xdx+f_ydy$. In your case, $f_x=0, f_y=2y$ and so, $Omega_X/k=R/f dxoplus R/(f,2y) dy$.
  $endgroup$
  – Mohan
  Mar 29 at 23:11
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

I am trying to prove an exercise in Hartshorne's. Let $k_0$ be a field of characteristic $p>0$, let $k=k_0(t)$, and let $Xsubseteq A_k^2$ be a curve defined by $y^2=x^p-t$. Show that every local ring on $X$ is a regular local ring but $X$ is not smooth over $k$.

For the first statement, a local ring on a point $q$ should be the localization of ring $mathcalO_X$, i.e. $(k[x,y]/y^2-x^p+t)_q$. And a maximal ideal correspondes to a irreducible polynomial in $(k[x,y]/y^2-x^p+t)$, thus, it is regular local ring.

For the second statement, notice that $0=d(y^2-x^p+t)=2dy-px^p-1dx=2dy$, so $Omega_X/k=0$. But $f$ has relative dimension 1, and we conclude that $f$ is not smooth.

I am not sure whether my argument is right. I hope someone can point out my mistake if there is any.

algebraic-geometry

share|cite|improve this question

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are making a mistake when you say $Omega_X/k=0$, please check the definition.
  $endgroup$
  – Mohan
  Mar 29 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mohan - Actually I don't know how to compute $Omega_X/k$ explictly, and I hope someone could help me. It only need to show that $dim_k Omega_X/k=0$, and then the conclusion still follows.
  $endgroup$
  – Yuyi Zhang
  Mar 29 at 15:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $ y^2' = 2y $ .
  $endgroup$
  – Youngsu
  Mar 29 at 16:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for the ugly shape of the formula. I needed to put at least 15 characters.
  $endgroup$
  – Youngsu
  Mar 29 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you have a curve $X$ defined as $f(x,y)=0$ (valid more generally), the exact sequence for 1-forms is, with $R=k[x,y]$, $Rfto R/f dxoplus R/f dyto |Omega_X/kto 0$, where the left map is $fmapsto df=f_xdx+f_ydy$. In your case, $f_x=0, f_y=2y$ and so, $Omega_X/k=R/f dxoplus R/(f,2y) dy$.
  $endgroup$
  – Mohan
  Mar 29 at 23:11
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am trying to prove an exercise in Hartshorne's. Let $k_0$ be a field of characteristic $p>0$, let $k=k_0(t)$, and let $Xsubseteq A_k^2$ be a curve defined by $y^2=x^p-t$. Show that every local ring on $X$ is a regular local ring but $X$ is not smooth over $k$.

For the first statement, a local ring on a point $q$ should be the localization of ring $mathcalO_X$, i.e. $(k[x,y]/y^2-x^p+t)_q$. And a maximal ideal correspondes to a irreducible polynomial in $(k[x,y]/y^2-x^p+t)$, thus, it is regular local ring.

For the second statement, notice that $0=d(y^2-x^p+t)=2dy-px^p-1dx=2dy$, so $Omega_X/k=0$. But $f$ has relative dimension 1, and we conclude that $f$ is not smooth.

I am not sure whether my argument is right. I hope someone can point out my mistake if there is any.

algebraic-geometry

share|cite|improve this question

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

$endgroup$

share|cite|improve this question

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

share|cite|improve this question

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117

asked Mar 29 at 3:31

Yuyi ZhangYuyi Zhang

17117