Hypersurface vs. Hyperslice

Hyperslices and Hypersurfaces in Affine Space.

Let $K$ be an algebraically closed field and let ${\mathbb{A}}^n$ be the $n$-dimensional affine space over $K$. The distinction between a general zero set of a single polynomial and a proper, non-empty one is formalized algebraically.

Definition (Affine Hyperslice). For any polynomial $f\in K[x_1,\dots ,x_n]$, the affine hyperslice defined by $f$ is the affine variety $V(f)\subseteq {\mathbb{A}}^n$.

Definition (Affine Hypersurface). An affine variety $X\subseteq {\mathbb{A}}^n$ is an affine hypersurface if $X=V(f)$ for some non-constant polynomial $f\in K[x_1,\dots ,x_n]$. This is equivalent to $f$ being neither a unit (a non-zero constant) nor the zero element in $K[x_1,\dots ,x_n]$.

Theorem (Hyperslices vs. Hypersurfaces in Affine Space). Let $f\in K[x_1,\dots ,x_n]$. The hyperslice $V(f)$ has the following properties:

1. $V(f)=\varnothing$ if and only if $f$ is a non-zero constant.
2. $V(f)={\mathbb{A}}^n$ if and only if $f=0$.

Proof. The identities $V(1)=\varnothing$ and $V(0)={\mathbb{A}}^n$ are immediate. Conversely, if $V(f)=\varnothing$, the Nullstellensatz implies that the ideal $⟨f⟩$ is $K[x_1,\dots ,x_n]$, so $1\in ⟨f⟩$, which means $f$ is a unit. If $V(f)={\mathbb{A}}^n$, then $f$ vanishes on all points; since $K$ is infinite, this implies $f=0$. $\square$

Hypersurfaces on an Irreducible Affine Variety.

Let $X\subseteq {\mathbb{A}}^n$ be an irreducible affine variety with coordinate ring $K[X]$.

Definition (Hyperslice on a Variety). For any regular function $f\in K[X]$, the hyperslice defined by $f$ is the set $V_X(f)=\{x\in X\mid f(x)=0\}$. This is a Zariski-closed subset of $X$.

Definition (Hypersurface on a Variety). A hyperslice $V_X(f)$ is a hypersurface in $X$ if $f$ is neither a unit nor the zero element in the coordinate ring $K[X]$.

Theorem (Hyperslices vs. Hypersurfaces on an Irreducible Affine Variety). Let $X$ be an irreducible affine variety and $f\in K[X]$. Then:

1. $V_X(f)=\varnothing$ if and only if $f$ is a unit in $K[X]$.
2. $V_X(f)=X$ if and only if $f=0$ in $K[X]$.

Proof. A regular function has no zeros on $X$ if and only if it is invertible in the ring of functions $K[X]$, making it a unit. Since $X$ is irreducible, $K[X]$ is an integral domain, so $f=0$ is the only function that vanishes everywhere on $X$. $\square$

Hypersurfaces in Projective Space.

Let $F$ be a homogeneous polynomial in the ring $K[x_0,\dots ,x_n]$.

Definition (Projective Hyperslice). For a homogeneous polynomial $F\in K[x_0,\dots ,x_n]$, the projective hyperslice defined by $F$ is the projective variety $V(F)\subseteq {\mathbb{P}}^n$.

Definition (Projective Hypersurface). A projective variety $X\subseteq {\mathbb{P}}^n$ is a projective hypersurface if $X=V(F)$ for some non-constant homogeneous polynomial $F$ (i.e., $\mathrm{deg}(F)\ge 1$).

Theorem (Hyperslices vs. Hypersurfaces in Projective Space). Let $F\in K[x_0,\dots ,x_n]$ be a homogeneous polynomial.

1. $V(F)=\varnothing$ if and only if $F$ is a non-zero constant.
2. $V(F)={\mathbb{P}}^n$ if and only if $F=0$.

Proof. By the projective Nullstellensatz, $V(F)=\varnothing$ if and only if the ideal $⟨F⟩$ contains all homogeneous polynomials of some sufficiently high degree. This occurs only if $F$ is a unit (a non-zero constant). The second statement is direct. For any non-constant $F$, $V(F)$ is non-empty. $\square$

Local Hypersurfaces in Quasiprojective Varieties.

Let $X$ be an irreducible quasiprojective variety.

Definition (Local Hyperslice). A subset $Y\subseteq X$ is a local hyperslice if for every point $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and a regular function $F\in K[U]$ such that $Y\cap U=V(F)$. This condition implies that $Y$ is a closed subvariety of $X$.

Definition (Local Hypersurface). A nonempty closed subvariety $Y\subseteq X$ is a local hypersurface if for every point $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and a regular function $F\in K[U]$ such that $F$ does not vanish on any irreducible component of $U$ and $Y\cap U=V(F)$.

Theorem (Local Hyperslices vs. Local Hypersurfaces). Let $Y\subseteq X$ be a local hyperslice. Let $\{U_i\}$ be an open cover of $X$ such that for each $i$, $Y\cap U_i=V(F_i)$ for some $F_i\in K[U_i]$. Then:

1. $Y=\varnothing$ if and only if for every $i$, $F_i$ is a unit in $K[U_i]$.
2. $Y=X$ if and only if for every $i$, $F_i=0$ in $K[U_i]$.

Proof. As being closed is a local property, $Y$ is closed in $X$. For (1), if $Y=\varnothing$, then $Y\cap U_i=\varnothing =V(F_i)$ for each $i$. Since each $U_i$ can be covered by affine open sets, this implies $F_i$ is locally a unit, and thus a unit in $K[U_i]$. The converse is immediate. For (2), if $Y=X$, then $Y\cap U_i=U_i=V(F_i)$, which implies $F_i=0$ in $K[U_i]$ for all $i$. The converse is immediate. $\square$