What is the dimension of a hypersurface? We say a circle is one dimensional and a sphere is two dimensional.
More generally,  we say the hypersurface $f(x_1,\cdots,x_n)＝0$ in $\mathbb{R}^n$ is $n-1$ dimensional.
But these hypersurfaces may not be linear subspaces. So I wonder how we define the dimension of a general hypersurface？ Can we find a basis of them? And what is the relation between this definiton with the usual dimension definition of linear spaces?
Many thanks for your hints.
 A: Note that your definition is not standard, and has some problems. For example, if $f(x_1, x_2) = x_1^2 + x_2^2 + 3$, then the set of zeros of $f$ has no points; we don't call that $1$-dimensional even though it's a hypersurface in $\mathbb{R}^2$. Another problematic example would be the zero locus of $(x^2+y^2)z$ in $\mathbb{R}^3$, which is the union of a plane (the points where $z = 0$) and a line (the points where $x$ and $y$ are zero).
To make matters even worse, we could describe what is basically the same set as being cut out by one polynomial in both $R^n$ and $R^m$ for $m \neq n$, so the dimension wouldn't depend on the geometric properties of the set, but on the ambient space. For example, $f(x, y) = x$ cuts out a line in $\mathbb{R}^2$, but $f(x, y, z) = x^2 + z^2$ cuts out the "same" line in $\mathbb{R}^3$. We don't want to say it's 1-dimensional or 2-dimensional depending on where you think of it.
Still, there's definitely coherent notions of dimension for more general shapes than vector spaces. Here's one, which encompasses the examples you're thinking of: a subset $X$ of $\mathbb{R}^n$ is called an $m$-dimensional manifold if there exists $m$ such that for every point $P$ in $X$, there exists $\epsilon$ such that the set $Y$ consisting of the intersection of a ball of radius $\epsilon$ about $P$ and $X$ is "homeomorphic" to $\mathbb{R}^n$. 
Here, "homeomorphic" means that there is a continuous map from $Y$ to $\mathbb{R}^n$ which is invertible, and whose inverse is also continuous. 
