= Solution
Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, when
$$
\dim T_pX=d,
$$
where $T_pX$ is the <Zariski tangent space>; it is a <singular point of an algebraic variety>[singular point] when $\dim T_pX>d$. Equivalently, the local ring $\mathcal O_{X,p}$ is regular exactly at a nonsingular point.
For an irreducible affine variety $X\subseteq\mathbb A^N$ over a perfect field, the <Jacobian criterion> expresses the singular locus by the vanishing of the relevant Jacobian minors, so it is <Zariski-closed set>[Zariski closed]. At a point where the tangent-space dimension is minimal it equals $d$; equivalently, some $(N-d)$-rowed Jacobian minor is nonzero. Its nonvanishing locus is therefore a nonempty <Zariski-open set> contained in the smooth locus. Every nonempty open subset of an <irreducible topological space> is dense, proving the <density of the smooth locus>.
Assume the ground field has characteristic other than two. For
$$
Q=V(X_0^2+\cdots+X_{n-1}^2)\subseteq\mathbb P^n,
$$
the partial derivatives are $2X_0,\ldots,2X_{n-1},0$. They vanish simultaneously at the unique projective point
$$
\boxed{[0:\cdots:0:1]}.
$$
Thus this point is the singular locus of the projective quadric cone.
For varieties $X,Y$,
$$
T_{(x,y)}(X\times Y)\cong T_xX\oplus T_yY,
\qquad
\dim(X\times Y)=\dim X+\dim Y.
$$
If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. Hence
$$
\operatorname{Sing}(X\times Y)=Z\times Y
$$
and
$$
\boxed{\dim\operatorname{Sing}(X\times Y)=k+m}.
$$
This is the <singular locus of a product with a smooth variety>.
To construct the requested examples, put $r=n-k>0$. If $r\geq2$, let $Q_r\subseteq\mathbb P^{r+1}$ be an irreducible quadric cone of dimension $r$ with one singular vertex. If $r=1$, use instead the irreducible cuspidal cubic
$$
C=V(Y^2Z-X^3)\subseteq\mathbb P^2,
$$
whose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embed
$$
W\times\mathbb P^k
$$
in projective space by the <Segre embedding>. It is irreducible and has dimension $r+k=n$, while its singular locus is the vertex or cusp point times $\mathbb P^k$, which is nonempty and has dimension exactly $k$. This is the <projective variety with a prescribed-dimensional singular locus> construction.
Finally, suppose that the irreducible plane curve $C\subseteq\mathbb P^2$ were smooth of degree $d$. Since it is birational to a smooth projective curve of genus two, its <geometric genus> would be two. But the <genus of a smooth plane curve> is
$$
g(C)=\frac{(d-1)(d-2)}2,
$$
and no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.
Back to article page
