Prove that a nonempty hypersurface in $\Bbb A^n_k$ has at least one non-singular point.

Question

Let $k$ be an algebraically closed field. Prove that a nonempty hypersurface in $\Bbb A^n_k$ has at least one non-singular point

My work is that i'm using the Hilbert’s Nullstellensatz , but i have no idea what to do next.

Answer 1

This is proved in Hartshorne's book Algebraic Geometry GTM-52, Theorem I.5.3. If $Y \subseteq \mathbb{A}^n_k$ is an affine variety it follows the set of singular points $Y_{sing} \subseteq Y$ is a proper closed subset of $Y$. In partucular it holds for a hypersurface $H:=V(f)$.

If the ideal $I$ defining $Y$ is generated by the polynomials $f_1,..,f_l$, the proof in HH.I.5.3 shows that the ideal of $Y_{sing}$ is defined using the jacobian matrix of the polynomials $f_j$. This is used to prove that $Y_{sing} \subsetneq Y$ is a proper subset. Let me cite the proof for hypersurfaces: If $Y:=V(f) \subseteq \mathbb{A}^n_k$ and $char(k)=0$ and $Y_{sing}=Y$ it follows

$$\partial f/\partial_{x_i} \in (f)$$

but $deg(\partial f/\partial_{x_i}) =deg(f)-1$ hence $\partial f/\partial_{x_i}=0$ - a contradiction since $deg(f) \geq 1$ and $char(k)=0$.