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Let $M$ be algebraic variety and let $I$ be the defining ideal of $M$, that is $$I(M) = \{ f \in K[X_1,...,X_n] \mid \forall x \in M : f(x)=0 \}$$ Let $f_1,...,f_m$ be the generators of $I(M)$. Let $$ J = \frac{\partial(f_1,...,f_m)}{\partial(X_1,...,X_n)}$$ be the Jacobian matrix.

A point $x \in M$ in an algebraic variety is called simple if $\mathrm{rk}J(x)=\mathrm{rk}(J)$ where rk is the rank of the $J$. The notation $J(x)$ is the matrix $J$ evaluated in $x \in M$. Clearly, $\mathrm{rk}J(x) \le \mathrm{rk}(J)$.

Let us denoted by $M^{\mathrm{reg}}$ the set of all simple points of $M$. A non-simple point is called regular. A nonsingular variety is a variety without non-simple points.

I want to prove the following:

Any algebraic variety $M$ is the union of finite number of nonintersecting nonsingular subvarieties. That is, $$M = \biguplus_{i=1}^q M_i$$

My attempt

If $M$ is nonsingular then we won. Therefore assume there are singular (non-simple) points in $M$.

Let $M = N_1 \cup ... \cup N_q$ be the decomposition of $M$ into irreducible components. Take one of the $M_i$ to be $M^{\mathrm{reg}}$ which is nonsingular. Now I need to add the rest of the singular points such they each is included in a subvariety in which it is simple. Therefore, I somehow need to take the singular points such that each is contained in a single irreducible component and is simple there (but not in $M$).

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Can you do it for the variety $xy=0$ in the plane? – Mariano Suárez-Alvarez Jan 11 '13 at 17:21
In that case you take the open subvariety $M^{\mathrm{reg}}$ defined by the algebra $K[x,y]_{I(0,0)}$ (by that I mean the localization of $K[x,y]$ by the ideal of functions that vanish at $(0,0)$) and then $M = M^{\mathrm{reg}} \uplus \{ (0,0) \}$. – LinAlgMan Jan 11 '13 at 17:26
Do you know that simple points form an nonempty open set? – user27126 Jan 11 '13 at 17:40
@Sanchez Yes. So if we take $N_i^{\mathrm{reg}} \subset N_i$ then it is dense ($N_i$ is irreducible). But does it imply that $N_i - N_i^{\mathrm{reg}}$ contains finitely many points? If it is true then I take the singleton subvarieties of these points. – LinAlgMan Jan 11 '13 at 17:46
If I understand your hint, let us denote $\mu = \mathrm{rk}(J)$ and $$ M_s= \{ x \in M \mid \mathrm{rk}J(x) = s \} \ .$$ Then $M_\mu = M^{\mathrm{reg}}$ and $$ M = \biguplus_{s=0}^{\mu} M_s \ . $$ It remains to show that $M^{(s)}$ are indeed subvarieties. – LinAlgMan Jan 11 '13 at 17:54
up vote 1 down vote accepted

To amplify Sanchez's comment: suppose the ground field is perfect (e.g. algebraically closed or of characteristic $0$). Then $M_0:=M^{reg}$ is open and dense in $M$. So $N_1:=M\setminus M_0$ is a closed subset of $M$ of dimension $\dim N_1 < \dim M$. Endow $N_1$ with the structure of reduced closed subvariety of $M$. Let $$M_1=N_1^{reg}, \quad N_2=N_1\setminus M_1, \quad M_2=N_2^{reg} \dots $$ This sequence stops because $\dim N_{n+1} < \dim N_n$. So $$ M = M_0 \cup M_1 \cup ... $$ (finite disjoint union of smooth subvarieties). If you want integral smooth subvarieties, you can replace each $M_i$ by the disjoint union of its connected components.

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Is there anything we can say about nonsingular points over imperfect fields? (As in dense/open?) I don't know where perfectness came in.. – user27126 Jan 12 '13 at 1:54
Dear @Sanchez: the set of nonsingular points is always open, but can be empty. The simplest example is the spec of a finite radicial extension of the ground field. However, the set of regular points is open and dense (if the variety is reduced). Regularity coincides with smoothness (=nonsingularity) over perfect fields. – user18119 Jan 12 '13 at 12:58
Thanks! {}{}{}{} – user27126 Jan 12 '13 at 19:08
Dear @user18119: your terminology is not the standard one. Regular and non-singular are synonyms and are absolute notions, not depending on a base scheme. Smoothness depends on some base scheme and means: regular and remaining regular after any base-change.See section 27.9 in the Stacks Project. See also section 12.2.3 in Vakil's Foundations of algebraic Geometry. – Georges Elencwajg Feb 17 '14 at 23:10

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