In An Invitation to Algebraic Geometry Karen Smith writes at the beginning of the book:
The zero set of a single polynomial in arbitrary dimension is called a hypersurface in $\mathbb C^n$. The quadratic cone is a typical example of a hypersurface.
I believe I have come across the term hypersurface before and I believe that it was used to mean that in $\mathbb R^n$ a hypersurface is an $n-1$ dimensional manifold.
As a consequence I expect the definition of hypersurface in the context of algebraic geometry to mean a variety of dimension $n-1$.
Consider for example $\mathbb R^2$ and the polynomial $p(x,y) = xy$. This is the two axes and is a variety of dimension $1$. (According to the perhaps naive definition given in the book so far.)
Since I was still skeptical I tried to find a different book giving the defnition of hypersurface and so I came across Hartshorne:
Proposition 1.13: A variety $Y$ in $\mathbb A^n$ has dimension $n-1$ if and only if it is the zero set $Z(f)$ of a single nonconstant irreducible polynomial in $k[x_1, \dots, x_n]$.
Assuming my understanding of the term hypersurface is correct this proposition adds the condition of irreducibility to the polynomial. As a consequence I would expect the zero sets of reducible polynomials to not be hypersurfaces.
So I tried to find the simplest example possible. For example, in $\mathbb R^2$ the polynomial $p(x,y) = x^2 - 4 $ is reducible.
But the graph (using this online grapher) of $x^2 - 4 = 0$ is, as expected, two vertical lines. Which, according to my current naive understanding of dimension of a variety, is indeed an $n-1=1$ dimensional variety.
Please could someone help me resolve my confusion?
a) An algebraic subset $X\subset\mathbb C^n$ has all its irreducible components of dimension $n-1$ if and only if it is the zero set $X=V(P)$ of some non-constant polynomial $P\in \mathbb C[T_1,...,T_n]$.
b) Beware that $X=\{(1,0)\}\cup V(T_1)\subset \mathbb C^2$ is an algebraic subset of dimension $1$ (=the maximum dimension of its irreducible components) but is not equal to any $V(P)$ because one its irreducible components, namely $\{(1,0)\}=V(T_1-1,T_2)$, has dimension $0\neq n-1=2-1$.