# Every real variety contains non-singular points

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 $n-d$.

• I think you have to be a little careful about using Nullstellensatz over a field that's not algebraically closed. For instance, if f = x^2 + y^2, then x vanishes on every real point where f does, but x isn't in the ideal generated by (f). Now you may not think of V = V(f) as a "variety of dimension n-1," and I can see why. But this shows someo of the subtleties.
– JSE
Feb 19, 2017 at 4:53
• See Prop. 3.3.10 and then Prop. 3.3.14 of the book Real Algebraic Geometry by Bochnak, Coste, and Roy; also see Remark 3.3.15 for a warning about failure of density in the classical topology for this algebraic notion of smoothness. I suggest that you first read the Introduction to appreciate some of the subtleties of real algebraic geometry (since your argument for hypersurfaces is not quite convincing since it seems to be mixing up what is a definition and what is a theorem, so that it seems you might not be aware of the pitfalls for passing between geometry and algebra over $\mathbf{R}$). Feb 19, 2017 at 6:00
• Thank you nfdc23. I indeed already looked for this in the Bochnak-Coste-Roy book. While the claim is stated in Prop 3.3.14, the proof refers to Prop 3.3.2 which in turn refers to two other sources. One of these is in French and the other I didn't easily find. I'd appreciate it if you could point out what is the issue with my argument. I skipped several details in the above sketch, but I can't see which step is problematic. Feb 19, 2017 at 7:22
• That "dimension $n-1$" implies $I(V)$ is principal is not explained (the analogue for schemes or over an algebraically closed field does not formally imply it). Your comment to JSE about complexification of $V$ (not a standard operation: one can make definitions, but they have real subtleties) and "real part" of an ideal suggest that the delicate nature of the passage between algebra and geometry in real algebraic geometry is something you may not be fully aware of; see the Intro of B-C-R. Also, problems in real algebraic geometry are almost never solved by importing results over $\mathbf{C}$. Feb 19, 2017 at 17:55
• I don't meant that your argument is definitely false, but rather than it seems to be making a leap from "dimension $n-1$" to asserting $I(V)$ is principal; I have never studied real algebraic geometry in a serious way, so I have no idea if that principality claim is true, or if true then whether it is hard to prove. The complexification as you now define it is reduced and so generically smooth, with "smooth locus" complementary to the zero locus of an ideal over $\mathbf{R}$ (i.e., comes from a Zariski-dense open over $\mathbf{R}$), there's no easy reason that should have $\mathbf{R}$-points. Feb 19, 2017 at 20:03

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.

• "... leaving us with the smooth points of $V_{\mathbb{C}}$ in $\mathbb{R}^{n}$." -- Could someone explain this please? Because I can only conclude that smooth points of $V_{\mathbb{C}}$ in $\mathbb{C}^{n}$ are Zariski dense in $V_{\mathbb{C}}$, but I don't see why there has to be any smooth point of $V_{\mathbb{C}}$ in $\mathbb{R}^{n}$ at all... Mar 28, 2023 at 14:16
• The point is not that the singular points are ignored, but that they are dealt with by induction. Thus, if there aren't any smooth points of $V_{\mathbb C}$ in $\mathbb R^n$, then everything has been dealt with by induction. Mar 28, 2023 at 14:28
• Thanks very much. Now it's the turn of the last hurdle: "From this it is easy to see that $V$ is a smooth manifold at this point $P$". For $V$ to be smooth at $P$ I think we want $dim_{R}(V)=n-rk_{P}(J_{R})$, where $J$ is the Jacobian of any choice of generators for the ideal $I_{R}(V)$ of real polynomials which are zero over $V$. I think that $rk_{P}(J_{R}(V))=rk_{P}(J_{C}(V_{C}))$, hence we use that $P$ is smooth for $V_{C}$ to reduce to proving $dim_{R}(V)=dim_{C}(V_{C})$. Is this true because $V_{C}$ is defined over $R$ and $V$ is its restriction to $R^{n}$? No need for induction? Mar 28, 2023 at 15:43
• I now note that my question in the comment above is related to mathoverflow.net/q/431559/501777 . Is my reasoning correct? Mar 28, 2023 at 15:48
• Something doesn't add up. The answer I linked in my comment above, proves that $dim_{R}(V)=dim_{C}(V_{C})$ as follows: "Now use the fact that if an irreducible real variety (in our case $V_{C}$) contains a smooth real point then the dimension of the set of real points (in our case $dim_{R}(V)$) is the same as the (algebraic) dimension of the variety (in our case $dim_{C}(V_{C})$)." So I think that they're actually using the relation between the Jacobians which I want to prove, to show equality between the dimensions. Therefore we cannot use their argument to finish our proof; I'm lost. Mar 28, 2023 at 15:59

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.

• The OP is speaking in terms of $I(V)$, so seems to be using an ideal determined by the given "variety" $V$ rather than the other way around (i.e., not picking an ideal $J$ that cuts out $V$ and then insisting to work with $J$), so for example this makes $I(V)$ always radical as when one proves "generic smoothness" for reduced schemes of finite type over perfect fields $k$ (but without trying to make a direct geometric theory out of $k$-points alone, as happens in real algebraic geometry). Since $I(V)$ and "variety" haven't been defined in the question, I am just guessing at the OP's intent. Feb 19, 2017 at 15:39
• It seems that I should apologize for not clearly defining my concepts. Since I'm an outsider, it seems that I don't use the standard definitions. Thank you nfdc23 for clarifying this. I did mean that when the ideal $I(V)$ is generated by a single polynomial $f$ then I can take $\nabla f$ (since this would be the Jacobian in this case). So taking $x^2$ or $f^2_1+\cdots+f^2_n$ would not work. Feb 19, 2017 at 19:16