Exercise 10.9

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $m$ is prime to the characteristic of $F$, show that the hypersurface defined by $a_0x_0^m+a_1x_1^m+ \cdots + a_n x_n^m$ has no singular points.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\medskip

Note: The sentence is not true if some coefficient $a_i$ is zero. To give an counterexample, the projective curve given by $f(x_0,x_1,x_2) = x_1^2 -x_2^2$ is the union of two lines, and the intersection point $a =[1,0,0]$ of these two lines is singular : $\partial f/ \partial x_0(a) = \partial f/ \partial x_1(a) = \partial f/ \partial x_2(a) = 0$. We must assume that $a_i 
e 0$ for every index $i$ (see the hint p. 371).

\begin{proof} Let $V$ be the projective hypersurface defined by $f(x_0,\ldots,x_n) = a_0x_0^m+a_1x_1^m+ \cdots + a_n x_n^m$.

If $m=1$, $V$ is an hyperplane, without singularity since $\frac{\partial f}{\partial x_i} (a) = a_i 
e 0$ for some index $i$.

We assume now that $m>1$. If $a = [u_0,\dots,u_n] \in V$ is a singular point, $$\frac{\partial f}{\partial x_i}(a) = ma_i u_i^{m-1} = 0\qquad (i=1,\ldots,n).$$

Since $m$ is prime with the characteristic, $m 
e 0$ in $F$, and $a_i 
e 0$, thus $u_i = 0$ for all indices $i$. Then $[u_0,\ldots,u_n]$ is not a projective point. This prove that $V$ has no singular point.
\end{proof}

Exercise 10.9

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Exercise 10.9

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