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David Lampert
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Let $S$ be the local ring of nodal curve, $R=\varinjlim_{\text{Frob}} S \to S$$R$ = inverse limit $Frob: S \to S$. For example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
  • $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.

Here's a complete local example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
  • $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.

In each example $(y/x)$ is integral over $R$.

Let $S$ be the local ring of nodal curve, $R=\varinjlim_{\text{Frob}} S \to S$. For example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
  • $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.

Here's a complete local example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
  • $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.

In each example $(y/x)$ is integral over $R$.

Let $S$ be the local ring of nodal curve, $R$ = inverse limit $Frob: S \to S$. For example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
  • $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.

Here's a complete local example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
  • $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.

In each example $(y/x)$ is integral over $R$.

Let S=(local$S$ be the local ring of) nodal curve, R=inverse limit Frobenius S->S$R=\varinjlim_{\text{Frob}} S \to S$. For example, k=perfect field, f(x,y)=yp+1-xp+1(1+x), R=k[x1/p,y1/p]/(f1/p). Here's:

  • $k$ a perfect field,
  • $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
  • $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.

Here's a complete local example: k=perfect field, f(x,y)=yp-xpy-xp+1, R=k[[x1/p,y1/p]]/(f1/p). In

  • $k$ a perfect field,
  • $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
  • $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.

In each example (y/x)$(y/x)$ is integral over R$R$.

Let S=(local ring of) nodal curve, R=inverse limit Frobenius S->S. For example, k=perfect field, f(x,y)=yp+1-xp+1(1+x), R=k[x1/p,y1/p]/(f1/p). Here's a complete local example: k=perfect field, f(x,y)=yp-xpy-xp+1, R=k[[x1/p,y1/p]]/(f1/p). In each example (y/x) is integral over R.

Let $S$ be the local ring of nodal curve, $R=\varinjlim_{\text{Frob}} S \to S$. For example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p+1}-x^{p+1}(1+x)$,
  • $R=k[x^{1/p^{\infty}},y^{1/p^{\infty}}]/(f^{1/p^{\infty}})$.

Here's a complete local example:

  • $k$ a perfect field,
  • $f(x,y)=y^{p}-x^{p}y-x^{p+1}$,
  • $R=k[[x^{1/p^{\infty}},y^{1/p^{\infty}}]]/(f^{1/p^{\infty}})$.

In each example $(y/x)$ is integral over $R$.

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David Lampert
  • 2.7k
  • 1
  • 14
  • 12

Let S=(local ring of) nodal curve, R=inverse limit Frobenius S->S. For example, k=perfect field, f(x,y)=yp+1-xp+1(1+x), R=k[x1/p,y1/p]/(f1/p). Here's a complete local example: k=perfect field, f(x,y)=yp-xpy-xp+1, R=k[[x1/p,y1/p]]/(f1/p). In each example (y/x) is integral over R.