I come back today with an interesting and basic question: "Is there any variety that has singularities at all points?" Imagine that it will be extremely complicated when working with these varieties, if they exists. Fortunately, it is never the case.
Well, we come back to the question, what is a singular point? We already know that if $f\in k[x_1,...,x_n]$ is a non-constant polynomial, then all points in $X=Z(f)$ such that they vanish on all partial derivatives of $f$, then they are called singular points. If follows that if $P\in Z(f)$, then $P$ is singular iff the matrix $(\frac{df}{dx_1}(P),...\frac{df}{dx_n}(P))$ is of rank 0.
However, what is the case if our variety is given by finitely many polynomials $f_1,...,f_r\in k[x_1,...,x_n]$? One can extend the earlier observation to the Jacobian matrix at a point $P$ as follows $\{\frac{df_i}{dx_j}(P)\}$, and we get stuck of defining the singularities, because it is unclear that in which case the rank of the Jacobian matrix is suitable to define the singular point.
However, if we look closer to the case $r=1$, we can see that the singularity is a local concept. If $P$ is not singular point, then the rank of the Jacobian matrix is (at least) 1. We can see that $X$ is the zero of the non-constant polynomial, and hence, due to the Krull's principal ideal, the dimension of any irreducible component of $X$ is $n-1$. If $P$ is a point on $X$, then $P$ belongs to some irreducible component, and hence, $codim\{P\}= n-1$, where $codim\{P\}$ is the co-dimension of $P$. We recall that $codim\{P\}$ is the height of the maximal ideal corresponding to $P$. In the case $P$ is singular, we can see that the rank of the Jacobian matrix is 0, which is less than $n-codim\{P\}$. This observation seems to be suitable for the general case.
Definition 1. Let $X\subset \mathbb{A}^n$ be an affine variety, and $I(x)=(f_1,...,f_r), f_i\in k[x_1,...,x_n]$. Let $P\in X$ be a point, then $P$ is called a smooth point if the rank of the Jacobian matrix (denoted $J_P$) defined above is at least $n-codim\{P\}$.
By change of coordinate, we can assume that $P$ is the origin. We see later that when $P$ is non-singular, the Jacobian matrix has the rank equal to $n-codim\{P\}$ actually. This follows from the observation via the kernel of $J_P$. We can easily see that $\ker J_P$ is actually the tangent space of $X$ at $P$ (denoted $T_P(X)$, which is defined by $V(f_{\text{linear part}}|f\in I(X))$, where $f_{\text{linear part}}$ is the linear part of all polynomials $f\in I(X)$. One can prove (based on the dimension of the tangent cone of $X$) that the dimension (as $k$-vector space) of the tangent space at $P$ is equal to $codim\{P\}$ iff $\dim_k(T_P(X))\le codim(P)$ iff $P$ is non-singular.
Also, the $T_P(X)\cong M_P/M_P^2$ as $k$-vector space, and hence, we can realize that $P$ is non-singular iff $codim\{P\}=dim_k(M_P/M_P^2)$, which is very similar to the case $X$ is an irreducible curve. In this case, $\dim_k(M_P/M_P^2)=1$, and $M_P$ is a DVR. When $P$ is singular, then the dimension of $\dim_k(M_P/M_P^2)$ is obvious larger than 1, and it must be 2, due to the isomorphism with the tangent space, whose dimension does not exceed 2.
Back to the general case, we can see that $P$ is non-singular iff $n-codim\{P\}=rank J_P$. This gives us the following
Proposition 2. Let $X\subset \mathbb{A}^n$ be an affine variety, then the set of singular points of $X$ is closed.
Proof. It easy follows from the Jacobian matrix, due to the rank, we can establish the set of non-singular points are not vanishing points of some determinants, which is defined by the polynomial equations, and hence, open in $X$. That implies the complement, i.e. the set of singular points is closed. (Q.E.D)
If the set of non-singular points is non-empty, then they must be dense in $X$, due to the previous proposition. If $X=V(f)$, where $f\in k[x_1,...,x_n]$ is an irreducible polynomial, Assume the contrary that every point of $X$ is singular, then $\frac{df}{dx_i}\in I(X)=(f)$. It cannot be the case if the characteristic of $k$ is zero, because in this case the degree of the partial derivatives of $f$ is strictly smaller than $f$, and they are non-zero. Let us consider the case $char(k)=p>0$. In this case, all partial derivatives vanish implies that $f$ contains monomial of the form $ax_1^{pa_1}...x_n^{pa_n}$, and hence, it is a power $p$ of another polynomial, which is the contradiction to the irreducibility of $f$. Hence, the set of non-singular points of $X$ is dense.
It can be generalized for arbitrary variety $X$, whose proof can be found in Harshone, Chapter I, Section 5.
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