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Multiplication by the defining polynomial of a degree-$d$ projective hypersurface gives the short exact sequenceThus the ideal sheaf of $X_d$ is ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$.

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.

Set $n=0$ in part (ii). The constants give an injection $k\hookrightarrow H^0(Y,{\mathcal{O}}_Y)$ because the projective hypersurface $Y$ is nonempty, while part (ii) gives surjectivity. ThusA disconnected scheme has a nontrivial idempotent global regular function, equal to zero and one on its two clopen pieces. A field has no such idempotent, so the connectedness from global regular functions criterion gives

For the closed immersion $i:Y\hookrightarrow X$, the ideal sheaf of a closed subscheme givesBecause the projective hypersurface is cut out by one homogeneous polynomial $f$ of degree $d$, multiplication by $f$ identifiesThis is the ideal-sheaf sequence of a projective hypersurface.

Let the degree-$d$ projective hypersurface $X_d$ be defined by the homogeneous polynomial $F$. Multiplication by $F$ identifies its ideal sheaf with ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$, giving the ideal-sheaf sequence of a projective hypersurfaceIn particular, the kernel in the question is ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$.

For $n=2$, additivity of the Euler characteristic in this short exact sequence and the line-bundle formulagiveHere $H^2(X_d,{\mathcal{O}}_{X_d})=0$ by the preceding cohomological-dimension argument. Moreover $H^0({\mathbb{P}}^2,\mathcal{O}(-d))=H^1({\mathbb{P}}^2,\mathcal{O}(-d))=0$ for $d>0$, so the long exact sequence gives $H^0(X_d,{\mathcal{O}}_{X_d})\cong k$. Consequentlywhich is the genus-degree formula for a projective plane curve.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

Let $X\subseteq {\mathbb{A}}^n$ have ideal $I(X)$, and let $p\in X$. Its Zariski tangent space isIf $X$ is irreducible, $p$ is smooth whenotherwise it is singular. Equivalently, for generators of $I(X)$, the Jacobian has the maximum rank $n-\dim X$ at $p$.

Now let the irreducible affine plane cubic be $X=V(f)\subseteq {\mathbb{A}}^2$. Suppose distinct points $p,q\in X$ were both singular. Parametrize their joining line byand put $h(t)=f(\ell (t))$. Then $\mathrm{deg}h\le 3$. At $t=0$,so zero is a root of multiplicity at least two. The same argument at $t=1$ gives another root of multiplicity at least two. Therefore $h$ has at least four roots counted with multiplicity and must vanish identically. The line through $p,q$ is then contained in $X$, so its linear equation divides $f$, contradicting irreducibility. This proves singular points of an irreducible affine plane cubic are unique when they exist.

For the density statement, embed an irreducible affine variety $X$ of dimension $d$ in ${\mathbb{A}}^n$ and choose generators $f_1,\dots ,f_r$ of its ideal. By dimension from minimum tangent dimension, some point has tangent dimension $d$, so the Jacobian has rank $n-d$ there. Some $(n-d)\times (n-d)$ minor $\Delta$ is consequently nonzero on $X$. The distinguished open setis nonempty, and at each of its points the Jacobian rank is at least $n-d$. Since every tangent space has dimension at least $d$, the rank is also at most $n-d$. Thus every point of $X_{\Delta }$ is smooth. A nonempty open subset of an irreducible topological space is dense, proving density of the smooth locus.

Finally write the smooth irreducible projective hypersurface asfor an irreducible homogeneous polynomial $F$. The closure of ${\pi }^{-1}(X)$ is its affine cone over a projective hypersurfaceAt a nonzero point $y\in Y$, simultaneous vanishing of all partial derivatives of $F$ would make the projective point $[y]\in X$ singular. Since $X$ is smooth, this cannot happen. Hence every nonzero point of $Y$ is smooth, and the origin is its only possible singular point.

Both possibilities occur. If $X$ is a projective hyperplane, then $F$ is linear and $Y$ is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conicIts affine cone isAll first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

For an algebraically closed field $k$, the strong Hilbert Nullstellensatz says that every ideal $J\subseteq k[x_1,\dots ,x_n]$ satisfiesIts weak form says that $V(J)=\varnothing$ exactly when $J=(1)$. Now let $S=\mathbb{C}[x_0,\dots ,x_n]$ and let $I\subseteq S$ be a homogeneous ideal. The affine zero set of $I$ is the affine cone over $V_{+}(I)$ together with the origin. Hence, when the projective zero set is empty, eitheror the affine zero set is exactly the origin. In the latter case the Hilbert Nullstellensatz givesthe irrelevant ideal of projective space. Equivalently, some power $(x_0,\dots ,x_n{)}^N$ is contained in $I$. These are precisely the alternatives in the Projective Nullstellensatz.

Let $V\subseteq {\mathbb{P}}^3$ be a smooth quadric surface. After a projective linear transformation, the Segre embedding identifiesChoose distinct points $p,q\in {\mathbb{P}}^1$. The two fibresare the promised disjoint curves on a smooth quadric surface: each is a smooth projective line, and $C_p\cap C_q=\varnothing$. By the Bézout theorem, any two nonempty curves in the projective plane intersect. An isomorphism of algebraic varieties preserves intersections and takes projective curves to projective curves, so these disjoint curves prove

Next let $W$ be a smooth projective curve and let $f:W⇢Y\subseteq {\mathbb{P}}^N$ be a rational map to a projective variety. At a point $P$ outside its initial domain, writeusing elements of the function field $\mathbb{C}(W)$. The local ring of a smooth algebraic curve at $P$ is a discrete valuation ring. If $m={\min }_i{v}_P(f_i)$, multiplying every coordinate by a function of valuation $-m$ makes all coordinates regular at $P$ and at least one a unit there. They therefore define a morphism near $P$. Repeating this at the finitely many missing points proves the extension of a rational map from a smooth projective curve.

Smoothness is essential. Let $\nu :{\mathbb{P}}^1\to C$ be the normalization of a nodal curve for a rational nodal cubic $C$. The inverse is a rational map $C⇢{\mathbb{P}}^1$ on the smooth locus. If it extended to a morphism $g:C\to {\mathbb{P}}^1$, then $g\circ \nu$ would agree with the identity on a dense Zariski-open subset of ${\mathbb{P}}^1$ and hence everywhere. But the two distinct points $a,b\in {\mathbb{P}}^1$ above the node would satisfya contradiction. This is the failure of rational-map extension on a singular curve.

Finally take the Fermat cubic curveand define the projective hypersurfaceBecause $F$ and $xyz$ have no common irreducible factor, the bihomogeneous polynomial $sF+txyz$ is irreducible, so $Z$ is an algebraic variety. Letbe the second coordinate projection. Every homogeneous cubic in three variables has a projective zero over $\mathbb{C}$, so every fibre is nonempty and $\pi$ is a surjective morphism. Over $p=[1:0]$ the fibre is the Fermat cubic curve, which is smooth and has genusOver $q=[0:1]$ the fibre isthe union of exactly three distinct projective lines, hence exactly three irreducible components. This is the cubic pencil with a triangular member required by the question.

Given homogeneous polynomials $F,G$ of the same degree, their pencil is the one-parameter familyIts total space is the projective hypersurface $V(sF+tG)\subseteq {\mathbb{P}}^2\times {\mathbb{P}}^1$, and projection to $[s:t]$ has these plane curves as its fibres.

For a reduced projective hypersurface $X=V(f)$ over an algebraically closed field,The Euler homogeneous function theorem makes the intersection redundant only when the characteristic does not divide the degree. In characteristic $p$, the irreducible polynomial $x_0^p+x_1^{p-1}{x}_2$ has zero gradient at $[1:0:0]$, which is not on $X$.