Exercise 10.10

Proof. Let $H_f(F)$ an affine hypersurface defined by $f(x_1,\dots ,x_n)$, with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=d$, and $a=(u_1,\dots ,u_n)\in F$.

$\text{∙}$

Suppose that the corresponding point $\bar{a}=[1,u_1,\dots ,u_n]\in \bar{F}$ is singular, and let $$\bar{f}(y_0,\dots ,y_n)=y_0^{d}f\left(\frac{y_1}{y_0},\dots ,\frac{y_i}{y_0},\dots ,\frac{y_n}{y_0}\right)$$

be the homogeneous polynomial defining $\bar{F}$. Then the chain rule gives

$$\frac{\partial \bar{f}}{\partial y_i}(x_0,\dots ,x_n)=x_0^{d-1}\frac{\mathit{\partial f}}{\partial x_i}\left(\frac{x_1}{x_0},\dots ,\frac{x_i}{x_0},\dots ,\frac{x_n}{x_0}\right).$$

Since $\bar{a}$ is singular,

$$0=\frac{\partial \bar{f}}{\partial y_i}(\bar{a})=\frac{\partial \bar{f}}{\partial y_i}(1,u_1,\dots ,u_n)=\frac{\mathit{\partial f}}{\partial x_i}(u_1,\dots ,u_n)=\frac{\mathit{\partial f}}{\partial x_i}(a).$$

This proves that $a$ is a common zero of $\mathit{\partial f}∕\partial x_i$ for $i=1,2,\dots ,n$

$\text{∙}$

Conversely, suppose that $\mathit{\partial f}∕\partial x_i(a)=0$ for $i=1,\dots ,n$. Then $$\frac{\partial \bar{f}}{\partial y_i}(\bar{a})=\frac{\partial \bar{f}}{\partial y_i}(1,u_1,\dots ,u_n)=\frac{\mathit{\partial f}}{\partial x_i}(u_1,\dots ,u_n)=0,$$

which proves that $\bar{a}$ is singular.