Smooth varieties are local complete intersection

Question

I read in Harris Book that any Smooth variety is local complete intersection but I don't know why. I wonder what can one say about singular points in curves that are local complete intersection. Remember that a point $P$ of a variety is said to be local complete intersection if there exists $U$ an affine open such that $V\cap U\subseteq \mathbb{A}^n$ is the zero locus of $n-d$ polinomials where $d=\dim_P(V)$.

Answer 1

Let $k$ be a field and $f \in k[x_1,\dots,x_n]$ be any non-zero polynomial. Then $\operatorname{Spec} \big( k[x_1,\dots,x_n]/f \big) \longrightarrow \operatorname{Spec}(k)$ is a global (hence local) complete intersection.

You see, this has nothing to do with non-singularness of $\operatorname{Spec}\big( k[x_1,\dots,x_n]/f \big)$.

You can of course take some $f$ with vanishing derivatives at some point to make $\operatorname{Spec}\big( k[x_1,\dots,x_n]/f \big)$ singular.

I believe what the author wants to say is that, if $X/k$ is smooth at $x$, then locally around $x$ it is standard smooth, i.e. of the form $\operatorname{Spec} \big( k[x_1,\dots,x_n]/f_1,\dots,f_r \big)$ with $\dim_x X = n-r$.