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A point on $F=0$ where all first partial derivatives vanish. For a reduced hypersurface this is the Jacobian criterion; a nondegenerate quadratic leading part defines an ordinary double point.

A monomial $x^a{y}^b$ is invariant precisely when $a-b$ is divisible by $n$: pulling back along the generator multiplies it by ${\zeta }^{a-b}$. Since distinct monomials are linearly independent, an invariant polynomial is a linear combination of exactly these invariant monomials. If $a\ge b$, write $a-b=kn$ and obtainIf $b\ge a$, the analogous expression uses $y^n$ and $xy$. Thus the polynomial invariant ring is generated bywhich satisfy $uv=w^n$.

To check that this is the only relation, reduce each monomial in $\mathbb{C}[U,V,W]$ using $UV=W^n$ until it has one of the formsTheir images are, respectively, $x^c{y}^c$, $x^{nk+c}{y}^c$ and $x^c{y}^{nk+c}$. These are pairwise distinct monomials in $x,y$, so no nontrivial linear combination of the reduced forms maps to zero. ThereforeThis identifies the determinant-one cyclic quotient of the affine plane with the affine hypersurface $UV=W^n$, and the quotient map is $(x,y)\mapsto (x^n,y^n,xy)$.

Apply the Jacobian criterion to $f=UV-W^n$. Its gradient isOver $\mathbb{C}$, with the printed $n\ge 3$, this vanishes only at $U=V=W=0$. Every other point is smooth. This point is exactly the image of the origin, and its inverse image is only the origin, since $x^n=y^n=0$ forces $x=y=0$. In fact the origin itself is singular: its tangent space has dimension three while the hypersurface has dimension two. Hence the quotient is smooth away from the image of the origin, with that point its unique singularity. Primality is not needed for this calculation; the same argument works for every integer $n\ge 2$.

Parametrize the line as $\ell (t)=x+t(x^{\prime }-x)$ and set $F(t)=f(\ell (t))$, a polynomial of degree at most three. The Jacobian criterion at a singular point of a hypersurface gives vanishing of its equation and all first derivatives. ConsequentlyThus $t^2$ and $(t-1{)}^2$ both divide $F$. They are coprime in every characteristic, so their degree-four product divides a polynomial of degree at most three. It follows that $F=0$, provingThis is the secant line through singular points of a cubic hypersurface argument. If the given defining polynomial has repeated factors, use its square-free part for the reduced variety's Jacobian criterion: vanishing of that polynomial and its gradient at a singular point implies vanishing of the original polynomial's gradient there as well, by the product rule. The same restriction argument therefore still applies.

First suppose $\mathrm{char}k\ne 2$. The quadratic form has rank $n>2$, so it is irreducible: a reducible homogeneous quadratic would be a product of two linear forms, whose associated symmetric matrix has rank at most two. Thusis a domain of dimension $n-1$. Its gradient is $(2t_1,\dots ,2t_n)$, so the Jacobian criterion shows that the only singular point is the origin. Its codimension is $n-1\ge 2$, giving regularity at all codimension-one points, the $(R_1)$ condition.

At every localization the polynomial ambient ring is regular and hence a Cohen-Macaulay ring. The nonzero quadratic equation is a nonzero divisor; quotienting by it decreases both depth and dimension by one. Hence the hypersurface local rings remain Cohen-Macaulay, so their depth is their dimension and in particular at least $\min (2,\dim )$. This verifies $(S_2)$. The Serre criterion for normality now yields $\boxed{X\text{ is normal}}$, establishing normality of a quadratic cone with all its hypotheses checked.

In characteristic two, $t_1^2+\cdots +t_n^2=(t_1+\cdots +t_n{)}^2$. The reduced zero variety is therefore the hyperplane $t_1+\cdots +t_n=0$, isomorphic to affine $(n-1)$-space, and its polynomial coordinate ring is integrally closed. Thus the reduced variety is normal in this case too. If one instead keeps the nonreduced hypersurface scheme defined by the squared equation, its coordinate ring has a nonzero square-zero class and is not normal. The variety interpretation in the question requires the reduced zero set.

We first prove the local algebra fact behind regularity in codimension one. Let $(A,\mathfrak{m})$ be a one-dimensional normal Noetherian local domain. Choose $0\ne a\in \mathfrak{m}$. Its radical is $\mathfrak{m}$, so some ${\mathfrak{m}}^n$ lies in $(a)$; take minimal $n$. If $n=1$, then $\mathfrak{m}=(a)$. Otherwise choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$ and put $z=b/a$ in the fraction field. We have $z\notin A$ but $z\mathfrak{m}\subseteq A$.

If $z\mathfrak{m}\subseteq \mathfrak{m}$, express multiplication by $z$ on a finite generating set of $\mathfrak{m}$. The determinant trick gives a monic polynomial in $z$ annihilating $\mathfrak{m}$; as this ideal is nonzero in a domain, the polynomial itself vanishes. Normality would then give $z\in A$, a contradiction. Hence there is $u\in \mathfrak{m}$ with $zu$ a unit of $A$. Every $v\in \mathfrak{m}$ satisfies $v=u(zv)/(zu)$, proving $\mathfrak{m}=(u)$. The maximal ideal therefore has one generator in dimension one, so $A$ is a regular local ring, indeed a discrete valuation ring. This proves one-dimensional normal local rings are discrete valuation rings directly.

Now a codimension-one point of the normal variety has such a local ring, and a codimension-zero local ring is a field. Both are regular. Since the algebraically closed ground field is perfect, regularity and smoothness agree for this finite-type variety, and the Jacobian criterion makes the singular locus closed. It contains no codimension-zero or codimension-one point. Therefore every irreducible component of the singular locus has codimension at least two, givingFor $d\le 1$ this means the singular locus is empty. In particular the singular set itself is the required closed subset, not merely a set contained in some larger one.

Over ${\overline{\mathbb{F}}}_3$, take the projective cubicIn characteristic three its partial derivatives are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the Jacobian criterion proves smoothness everywhere, including infinity.

It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in ${\overline{\mathbb{F}}}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. Gauss lemma for polynomials then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the smooth plane cubic in characteristic three. In particular, choosing a Fermat cubic here would fail, because its cubic partial derivatives vanish in characteristic three.

Take the projective hypersurfaceAs a polynomial in $X_0$ over $\mathbb{C}[X_1,X_2,X_3]$, it is primitive because $X_1$ and $X_2^2$ are coprime. It is linear and irreducible over the fraction field, so Gauss lemma for polynomials proves irreducibility. Thus it is an irreducible surface.

Its gradient is $(X_1,X_0,-2X_2,0)$. The simultaneous zero set on $S$ is exactlyThe Jacobian criterion proves every other point is smooth. At that vertex the affine equation in the chart $X_3=1$ has no linear term, so the tangent space has dimension three whereas the surface has dimension two; the vertex is genuinely singular. This is a projective quadric cone with one singular vertex.

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

On an affine chart $\mathrm{Spec}A\subseteq V$, the module of Kähler differentials ${\Omega }_{A/k}$ is generated by symbols $da$ subject to $k$-linearity and $d(ab)=adb+bda$. It represents $k$-derivations. These modules commute with localization, so their associated quasi-coherent sheaves glue to the Kähler differential sheaf ${\Omega }_{V/k}^1$. It is coherent: if $A=k[x_1,\dots ,x_N]/I$ and $I=(f_1,\dots ,f_s)$, then it has the finite presentation of a moduleFor a closed point $P$, put $R={\mathcal{O}}_{V,P}$ with maximal ideal $\mathfrak{m}$ and residue field $k$. The Zariski tangent space is ${\mathrm{Hom}}_k(\mathfrak{m}/{\mathfrak{m}}^2,k)$, equivalently ${\mathrm{Der}}_k(R,k)$. A derivation kills constants and ${\mathfrak{m}}^2$, so it factors through $a\mapsto a-a(P)$; conversely every linear functional on $\mathfrak{m}/{\mathfrak{m}}^2$ defines such a derivation by the product rule. The universal property of Kähler differentials therefore identifies this space withIn particular, ${\Omega }_{V,P}^1/{\mathfrak{m}}_P{\Omega }_{V,P}^1\cong {\mathfrak{m}}_P/{\mathfrak{m}}_P^2$, the algebraic cotangent space.

A point is a smooth point of a variety when its local ring is a regular local ring; over this algebraically closed field this says ${\dim }_k{T}_{V,P}=n=\dim V$. Tensor the finite presentation of a module above with $k(P)$. The tangent space is the kernel of the Jacobian matrix $J(P)$, hence has dimension $N-\mathrm{rank}J(P)$. This proves the Jacobian criterionFor a smooth irreducible variety $V$, choose at each point an invertible $(N-n)$-minor and shrink the affine chart so that it remains invertible. Its relations eliminate $N-n$ of the generators of ${\Omega }_{A/k}$, yielding a surjection $A^n\to {\Omega }_{A/k}$. At the generic point, its target has dimension $n$: over a perfect ground field, a separating transcendence basis of the function field has differentials forming a basis. Equivalently this follows from the assumed density of the smooth locus. The kernel therefore becomes zero over the fraction field of the integral domain $A$. As a submodule of $A^n$, it is a torsion-free module, so it is already zero. Thus these maps give local isomorphisms with ${\mathcal{O}}_V^n$, proving local freeness of differentials on a smooth variety with rank $n$.

For an affine chart $\mathrm{Spec}A\subseteq V$, write $W\cap \mathrm{Spec}A=\mathrm{Spec}(A/I)$ and $\mathcal{M}=\tilde{M}$. The restriction of a module sheaf to a closed subvariety is $\widetilde{M{\otimes }_{A}A/I}=\widetilde{M/IM}$. The Conormal exact sequence for Kähler differentials isIt follows from the generators and relations: passing to $A/I$ imposes precisely the additional relations $df=0$ for $f\in I$. The first map is well defined because $d(I^2)$ lies in $I{\Omega }_{A/k}$. Glue these exact module sequences, using exactness of localization, to obtainIn the final assertion, interpret locally principal subvariety as a proper local hypersurface. Its ideal on each chart is $(f)$ with $f\ne 0$. Since $V$ is an irreducible variety and reduced, $f$ is a non-zero-divisor, and $A/(f)\to (f)/(f^2)$, $\bar{a}\mapsto af$, is an isomorphism. These local rank-one descriptions make the conormal sheaf ${\mathcal{I}}_W/{\mathcal{I}}_W^2$ invertible.

Because $W$ is not contained in the singular locus of $V$, there is a dense open subset of $W$ where both $V$ and $W$ are smooth varieties. At a closed point there, $\dim W=\dim V-1$ by the Krull principal ideal theorem. The tangent description $T_{W,P}=\ker (df:T_{V,P}\to k)$ then forces $df\ne 0$. Hence the conormal map is injective at the generic point of $W$. Its kernel is a subsheaf of a line bundle on the integral variety $W$, so it is a torsion-free sheaf; a torsion-free sheaf with zero generic fibre is zero. This proves conormal injectivity for a generically smooth Cartier divisor. If zero equations were allowed in the phrase locally principal, $W=V$ would be a counterexample to invertibility; the proper-hypersurface convention is essential.

A point $P$ belongs to the smooth locus of a variety if its local ring ${\mathcal{O}}_{X,P}$ is a regular local ring, meaningEquivalently, the Zariski tangent space has the expected local dimension. Over the algebraically closed field, this agrees with smoothness for a reduced variety. In affine coordinates, the Jacobian criterion expresses it by the rank of the matrix of first derivatives of defining equations.

The Jacobian criterion on an affine chart $x_j\ne 0$ says that a point of the projective hypersurface is singular exactly when all derivatives of its dehomogenized equation vanish there. The Euler homogeneous function theorem identity ${\sum }_i{x}_i{f}_{x_i}=df$ then recovers the omitted $j$th derivative on $X$, without dividing by $d$. Thus the correct statement in every characteristic isIf the printed $Z(I)$ denotes a zero locus inside $X$, this proves it as written. If it means the ambient projective zero locus, the assertion is false when the characteristic divides $d$. When the characteristic does not divide $d$, the Euler homogeneous function theorem forces $f=0$ at every common zero of the derivatives, and the intersection with $X$ is redundant.

For a genuine counterexample to the unrestricted ambient interpretation, in characteristic $p$ takeThis is an irreducible homogeneous polynomial: it is primitive and linear in $X_2$ over $k[X_0,X_1]$, with coprime coefficients. At $[1:0:0]$ all partial derivatives vanish, including when $p=2$, but $f=1$. This point is outside $X$. Hence the ambient-locus version needs the displayed intersection or a restriction on the degree.

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

Homogenizing the affine equation gives the projective closureIts partial derivatives areIf all three vanished at a point of $X$, then $X=0$ and either $Y=0$ or $Z=0$. If $Y=0$, the equation $F=0$ forces $Z=0$; if $Z=0$, the equation $F_Z=0$ forces $Y=0$. Either conclusion gives the forbidden zero triple of homogeneous coordinates. The Jacobian criterion therefore shows that $X$ is a smooth projective curve; more specifically, it is the projective closure of y cubed equals x to the fourth plus one, a smooth plane quartic.

Suppose first that the affine variety $V$ is irreducible. If $fg\in I(V)$, thenBoth pieces are Zariski closed in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a prime ideal.

Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the Hilbert Nullstellensatz would giveChoose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of irreducibility and prime vanishing ideal:

For the semicubical parabolaconsiderCertainly $(y^2-x^3)\subseteq \ker \theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker \theta$, thenThe first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. ThereforeThe latter is a subring of the integral domain $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and

An affine variety of dimension $d$ is smooth at $p$ when its Zariski tangent space has dimension $d$; equivalently, the Jacobian criterion requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient isAt $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the coordinate ring and singularity of the semicubical parabola gives

PutBy the Jacobian criterion, a projective plane curve is singular only at a common projective zero of $F$ and all its partial derivatives. HereTheir simultaneous vanishing forces $X_0=X_1=X_2=0$, which is not a point of projective space. Hence the Fermat quartic $C$ is nonsingular. In particular it is a smooth plane quartic and has geometric genus $(4-1)(4-2)/2=3$.

Let $X$ be an irreducible variety of dimension $d$. A point $p\in X$ is nonsingular, or smooth, whenwhere $T_{p}X$ is the Zariski tangent space; it is a singular point when $\dim T_{p}X>d$. 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. 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. Forthe partial derivatives are $2X_0,\dots ,2X_{n-1},0$. They vanish simultaneously at the unique projective pointThus this point is the singular locus of the projective quadric cone.

For varieties $X,Y$,If $Y$ is smooth of dimension $m$, the product point $(x,y)$ is singular exactly when $x$ is singular. HenceandThis is the singular locus of a product with a smooth variety.

To construct the requested examples, put $r=n-k>0$. If $r\ge 2$, 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 cubicwhose only singular point is $[0:0:1]$. Let $W$ denote the chosen $r$-dimensional variety and embedin 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 isand no integer $d$ makes this number equal to two. Therefore $C$ cannot be smooth and must contain a singular point.

The primitive ray generators of $\sigma$ areand $|\det (v_0,v_3)|=3$. They do not form a lattice basis, so the smoothness criterion for a toric variety shows that $U_{\sigma }$ is singular. The presentation from part (i) gives the same conclusion by the Jacobian criterion: all derivatives of $B^3-AC$ vanish at $A=B=C=0$.

Insert the primitive raysLet $\Sigma$ consist of the conesand all their faces. Sinceevery cone of $\Sigma$ is smooth. Its support is $\sigma$, so this fan subdivision induces a proper birational toric morphismThe source is smooth, and $p$ is an isomorphism over the dense algebraic torus; it is therefore a toric resolution of singularities.

Over a perfect field, a point of a variety is smooth when its local ring is a regular local ring. For an irreducible variety of dimension $n$ at a closed point, this is equivalent to its Zariski tangent space having dimension $n$. The Jacobian criterion expresses the equality as a matrix-rank condition.