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Projection $X\times {\mathbb{P}}^n\to X$ sends closed subsets to closed subsets, with geometric points or schemes understood. On an affine base, choose finitely many homogeneous equations for a closed subset. Its fiber is nonempty exactly when every graded multiplication map for a projective fiber fails to be surjective, by the Projective Nullstellensatz. The image is thus an intersection of closed rank-defect loci. An affine open cover proves the general case. This is the universally closed part of the proper morphism property of projective space.

Let $K$ be a number field of degree $d$, and normalize its absolute values so that the product formula is ${\prod }_v|a{|}_v^{n_v}=1$ for $a\ne 0$, where $n_v=[K_v:{\mathbb{Q}}_v]$; at a complex place use the ordinary modulus with weight two. For $P=[a_0:\cdots :a_N]$, defineOnly finitely many factors differ from one. Rescaling all coordinates multiplies the local maxima by $|a{|}_v$, whose weighted product is one, so the projective height is well-defined. It is at least one by comparison with any nonzero coordinate. On enlarging $K$ to $L$, the identity ${\sum }_{w\mid v}[L_w:{\mathbb{Q}}_v]=[L:K][K_v:{\mathbb{Q}}_v]$ makes the normalized product unchanged. Thus this is an absolute height on algebraic projective points. For coprime rational integer coordinates it is simply their maximum absolute value.

For a projective variety $V/K$ and an embedding $i:V\hookrightarrow {\mathbb{P}}^N$, take $h_i=h\circ i$. Changing the basis of the same projective linear system changes the height by a bounded amount: the coordinate transformation and its inverse have fixed local coefficient bounds, equal to one outside finitely many places. If $L$ is ample and $L^m$ is very ample, set $h_L=m^{-1}{h}_i$ for an embedding from $L^m$, up to a bounded function. A general line bundle can be written $(L\otimes A)\otimes A^{-1}$ with both $A$ and $L\otimes A$ very ample, defining its height as the difference of their embedding heights.

This is the Weil height machine, withThe first relation follows from the Segre embedding, whose local maximum is the product of the two maxima; tensor powers similarly use the Veronese embedding. The second follows by pulling back the defining sections. These constructions make the choices equivalent modulo bounded functions. For a fixed homogeneous morphism with no base point, upper local height bounds come from its coefficients, and lower bounds from the Projective Nullstellensatz expressing coordinate powers in the ideal of its defining forms. The constants are one outside a finite set of places. This gives the uniform bounded errors and shows why heights belong to line-bundle classes rather than being independent of every projective embedding.

The Northcott lemma states that for fixed $B$ and $D$, the algebraic points of ${\mathbb{P}}^N$ with $H(P)\le B$ and $[\mathbb{Q}(P):\mathbb{Q}]\le D$ form a finite set. In particular the $K$-rational points of bounded projective height are finite. Here is a proof, including the degree bound.

First let $\alpha$ have degree $e\le D$ and $H(\alpha )=H([1:\alpha ])\le B$. Write its primitive minimal polynomial as $a_e{\prod }_{j=1}^e(T-{\alpha }_j)$, with positive integer $a_e$. The height-Mahler measure formula isIts finite-place part follows from multiplicativity of the nonarchimedean polynomial coefficient norm: a primitive integer polynomial has norm one at each prime, so factoring it bounds the product of the roots' local maxima by the inverse leading-coefficient norm. The archimedean part is the displayed product over conjugates. Combining them with the product formula gives the identity.

Expanding the polynomial now bounds each integer coefficient byThere are finitely many degrees and finitely many such integer coefficient lists, each with finitely many roots. Hence there are only finitely many possible $\alpha$. This proves the minimal-polynomial coefficient proof of Northcott finiteness for algebraic numbers.

On the projective chart $a_j\ne 0$, each ratio ${\alpha }_i=a_i/a_j$ has degree at most the point's field degree andThe equality uses the product formula. Each coordinate ratio therefore belongs to a finite set, and there are finitely many charts. This proves the projective Northcott theorem. Restricting to $V$ gives the result for a chosen embedding. For an ample line-bundle height, the comparison $m{h}_L=h_i+O(1)$ reduces its bounded-height sets to this case.

Both qualifications matter: a degree bound cannot be dropped, since infinitely many roots of unity have height one; and the finiteness assertion need not hold for a nonample line bundle, such as the trivial bundle's bounded height on a variety with infinitely many rational points.

Put $I=(f_1,\dots ,f_r)$ and $\mathfrak{m}=(y_0,\dots ,y_n)$. If ${\mathfrak{m}}^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne 0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.

Conversely, over an algebraically closed field, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the Hilbert Nullstellensatz puts $y_j$ in $\sqrt{I}$. Choose integers $e_j\ge 1$ with $y_j^{e_j}\in I$, and setEvery degree-$N$ monomial has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those monomials generate ${\mathfrak{m}}^N$, provingThe argument includes the unit-ideal case. This proves the Projective Nullstellensatz criterion directly from its affine counterpart.

Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ monomial descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a vector space with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its ideal cannot contain any power of $\mathfrak{m}$.

For $P=[x_0:\cdots :x_N]\in {\mathbb{P}}^N(\mathbb{Q})$, choose coprime integer coordinates and define the projective heightWrite the morphism $F:{\mathbb{P}}^1\to {\mathbb{P}}^1$ as $F=[F_0:F_1]$, where $F_0,F_1\in \mathbb{Z}[X,Y]$ are homogeneous of degree $d$ with no common zero. Bounding each polynomial by the sum of the absolute values of its coefficients gives

Because $F_0$ and $F_1$ have no common projective zero, the Projective Nullstellensatz gives an integer $m\ge d$ and homogeneous polynomials $A_{ij}$ such that suitable nonzero integer multiples of $X^m$ and $Y^m$ lie in the ideal $(F_0,F_1)$. Evaluating at primitive coordinates and using the same coefficient bound givesafter absorbing the bounded common divisor of $F_0(x,y)$ and $F_1(x,y)$ into $C$. ThereforeThis is height growth under a morphism of the projective line.

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.