Every real variety contains non-singular points

Question

I am looking for a relatively "elementary" proof that every variety in ${\mathbb R}^n$ contains at least one non-singular point.

So far I only have such a proof for the case of hypersurfaces. Hopefully this proof would also explain what I mean by an "elementary" proof. Consider a variety $V \subset {\mathbb R}^n$ of dimension $n-1$, and note that the ideal $I(V)$ is generated by a single polynomial $f\in {\mathbb R}[x_1,\ldots,x_n]$. A point $p\in V$ is singular if the gradient $\nabla f$ vanishes on $p$. Without loss of generality, assume that $f$ depends on $x_1$, so $f_1 = \frac{\partial f}{\partial x_1}$ is not identically zero. If $f_1$ vanishes on $V$ then $f_1\in I(V)$, which is a contradiction to $I(V)$ being generated by $f$. Thus, there exists a point $p\in V$ such that $f_1(p)\neq 0$. By definition, $p$ is not a singular point of $V$.

I could not get a similar idea to work for lower-dimensional varieties. In case it helps, let me also add the definition of a singular point in this case. Consider a variety $V \subset {\mathbb R}^n$ of dimension $d$ and assume that $I(V)$ is generated by $f_1,\ldots,f_k$. Then a point $p\in V$ is singular if the rank of the Jacobian matrix of $f_1,\ldots,f_k$ at $p$ is smaller than $n-d$.

Answer 1

If you're willing to view a variety as a set of points rather than as a scheme, then it is fairly easy to show that every real algebraic variety $V$ in ${\bf R}^n$ is equal as a set to the finite union of smooth manifolds (of various dimensions, and typically not closed). Namely, one can view $V$ as the restriction of some complex variety $V_{{\bf C}} \subset {\bf C}^n$ to ${\bf R}^n$. By decomposing into irreducible components, we may assume without loss of generality that $V_{{\bf C}}$ is irreducible. If $V_{{\bf C}}$ is not invariant with respect to complex conjugation $(z_1,\dots,z_n) \mapsto (\overline{z_1}, \dots, \overline{z_n})$, then one may intersect this variety with its complex conjugate, dropping the dimension of this variety without affecting $V$ as a set. Thus, by an induction on dimension, one may assume that $V_{{\bf C}}$ is invariant with respect to complex conjugation (or equivalently, is definable over the reals). We remove the (complex) singular points of $V_{{\bf C}}$, since these can be handled by the induction on dimension, leaving us with the smooth points of $V_{{\bf C}}$ in ${\bf R}^n$. At such a point, the complex tangent space is invariant with respect to complex conjugation and is thus the complexification of a real space of the same dimension. From this it is easy to see that $V$ is a smooth manifold at this point, and the claim follows.

Answer 2

It appears, from the computation you provide, you want to say a point is singular if all the partial derivatives vanish at that point. What about $V(x^2)$ in $\mathbb{R}$? The variety is the point $0$ and the derivative of $x^2$ vanishes at $0$? So there is no "smooth point".

Comment: you claim to have a proof for all hyper-surfaces (which has already been shown to have flaws). However all real varieties $V(f_1,...,f_n)=V(f_1^2+\cdots +f_n^2)$ are hyper-surfaces. So if you do construct a valid proof for what you need for hyper-surfaces, you will have proved the result you want in general.

Lastly, even over $\mathbb{C}$ there are examples of varieties some, like myself, would say are entirely singular. For example, the GIT quotient of $\mathrm{SL}(2,\mathbb{C})$ acting on itself by conjugation is $\mathbb{C}$ and so smooth. But the generic stabilizer of the action is positive dimensional. I would say (I am sure there would be disagreement here) the entire variety is singular with its singular locus smooth.