# Algebraic variety as a union of nonsingular subvarieties

### Definitions

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

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.