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Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

The composite of two finite type morphisms is of finite type: on suitable affine open subsets, take the algebra generators of the first algebra together with the images of the algebra generators of the second. To check separatedness, factor the diagonal morphism of the composite asThe first arrow is a closed immersion, and the second is the base change of the closed immersion ${\Delta }_g:Y\to Y{\times }_{Z}Y$. Hence their composite is a closed immersion.

Finally, after any base change $Z^{\prime }\to Z$, the composite is $X^{\prime }\stackrel{f^{\prime }}{\to }{Y}^{\prime }\stackrel{g^{\prime }}{\to }{Z}^{\prime }$. Both arrows take closed subsets to closed subsets, so their composite does as well. This is universal closedness. All three conditions hold, and therefore

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

Put $P={\mathbb{P}}_k^2$ and $S=k[t_0,t_1,t_2]$. A homogeneous polynomial $F$ of positive degree $d$ is nonzero, and $S$ is an integral domain. Multiplication by $F$ therefore gives an exact sequence of coherent sheaveswhere $i:X\hookrightarrow P$ is the closed immersion.

We compute the necessary cohomology of twisting sheaves on projective space. The standard three affine open subsets $D_{+}(t_i)$ and all their intersections are affine schemes. For a quasi-coherent sheaf, vanishing of quasi-coherent cohomology on an affine scheme and the acyclic cover theorem identify its sheaf cohomology with the Čech cohomology of this cover. In degree $m$, the Čech cochain complex isDecompose it into subcomplexes indexed by Laurent monomials $t_0^{a_0}{t}_1^{a_1}{t}_2^{a_2}$ with $a_0+a_1+a_2=m$. Let $N=\{i:a_i<0\}$. Such a monomial occurs in the summand for a nonempty index set $J$ precisely when $N\subseteq J$, since only the variables in $J$ are inverted.

If $N$ is empty, the monomial contributes the cochain complex of a full two-simplex: its degree-zero cohomology is $k$ and its higher cohomology vanishes. If $|N|=1$, the dimensions in degrees $0,1,2$ are $1,2,1$; the first map has rank one and the second map has rank one with the same kernel as the first map's image. If $|N|=2$, there is one copy of $k$ in each of degrees $1$ and $2$ and the map between them is an isomorphism. These two cases are acyclic over every field, including characteristic two. If $N=\{0,1,2\}$, only the degree-two term occurs, contributing one copy of $k$. It follows thatand that $H^0(P,{\mathcal{O}}_P)=k$, $H^0(P,{\mathcal{O}}_P(-d))=0$, and $H^2(P,{\mathcal{O}}_P)=0$. Moreover the Laurent-monomial description of top cohomology on projective space gives a basis of $H^2(P,{\mathcal{O}}_P(-d))$ consisting ofFor $d\ge 3$, subtracting one from each $b_i$ counts these as $\left(\genfrac{}{}{0ex}{}{d-1}{2}\right)$. For $d=1,2$ there are no such triples, and the formula $(d-1)(d-2)/2$ is also zero.

Use the long exact sequence in sheaf cohomology of the initial sequence, and cohomology under a closed immersion to identify the terms for $i_{\ast }{\mathcal{O}}_X$. The vanishings just proved giveThereforeThis proves the cohomology of a projective plane hypersurface without imposing smoothness, irreducibility or reducedness. In particular, the second number is an arithmetic genus, not a claim about the genus of a normalization.

For a divisor $D$ on a smooth projective curve, let $L(D)=\{f\in k(X):\mathrm{div}(f)+D\ge 0\}\cup \{0\}$, and write $\ell (D)={\dim }_{k}L(D)$. The Riemann-Roch theorem stateswhere $g$ is the genus, $K_X$ is a canonical divisor given by a nonzero rational differential, and the degree of a divisor is the sum of its coefficients since $k$ is algebraically closed.

For genus one, a nonzero regular differential has a nonnegative divisor of degree $2g-2=0$, hence no zeros; therefore $K_X\sim 0$. Negative-degree divisors have no nonzero sections. Thus $\ell (n{P}_{\infty })=n$ for $n>0$, $\ell (0)=1$, and $L(n{P}_{\infty })=0$ for $n<0$.

Choose $x\in L(2P_{\infty })\setminus L(P_{\infty })$ and $y\in L(3P_{\infty })\setminus L(2P_{\infty })$. Their unique poles have orders two and three respectively. Distinct pole orders prove linear independence in the following complete list:Since $y^2$ also lies in the last six-dimensional space, it satisfiesThe coefficient $a$ cannot vanish because only $x^3$ among the displayed basis has a sixth-order pole available to cancel the leading pole of $y^2$. Homogenizing gives a plane cubic relation, valid also in characteristics two and three without completing squares or dividing by three.

It remains to prove that the map actually embeds the curve. For $D=3P_{\infty }$, Riemann-Roch gives $\ell (D)=3$, $\ell (D-Q)=2$ for every point $Q$, and $\ell (D-Q-R)=1$, including $R=Q$. The first equality makes the linear system base-point-free; the second separates distinct points, and the double-point case separates tangent directions. Therefore $D$ is very ample and ${\varphi }_D$ is a closed immersion into ${\mathbb{P}}^2$. Its pullback of a line has degree three, so its image has degree three. The displayed cubic equation consequently defines the image, which is smooth because it is isomorphic to $X$. Thus $\boxed{{\varphi }_{3P_{\infty }}:X\cong \text{a smooth plane cubic}}$. The affine expression $(1:x:y)$ extends across $P_{\infty }$ to $(0:0:1)$ by the pole orders.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

A closed immersion $i:X\to Y$ is a morphism of schemes whose underlying map is a homeomorphism onto a closed subset and whose map of structure sheavesis surjective. Equivalently, $X$ is the closed subscheme defined by a quasi-coherent ideal sheaf of a closed subscheme on $Y$. The affine-local characterization is particularly useful: for every affine open subscheme $V=\mathrm{Spec}A\subseteq Y$,for an ideal $I\subseteq A$, with the map induced by the quotient homomorphism.

Work over an affine open subscheme $V=\mathrm{Spec}A$ of $Z$. Since $g$ is a closed immersion, its inverse image is $\mathrm{Spec}(A/I)$ for some ideal $I$. Since $f$ is a closed immersion, the inverse image of this affine scheme in $X$ is $\mathrm{Spec}((A/I)/J)$ for an ideal $J\subseteq A/I$. Let $\tilde{J}$ be the inverse image of $J$ in $A$. The composite on this open is induced by the surjectionThe affine-local characterization from part (a) therefore shows that $g\circ f$ is a closed immersion. In particular, closed-subscheme ideals compose by taking the inverse image of the second quotient ideal.

Over an affine open subscheme $\mathrm{Spec}A$ of the target, a closed immersion has the form $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$. Pulling it back along an affine scheme map $\mathrm{Spec}B\to \mathrm{Spec}A$ givesThe identification follows directly by tensoring the quotient presentation; it does not require $B$ to be flat over $A$. Since $B\to B/IB$ is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.

A morphism of schemes $f:X\to Y$ is a separated morphism if its diagonal morphismis a closed immersion. The diagonal morphism is induced by the pair of identity maps on $X$. Thus separatedness is a condition relative to the given target $Y$, not a requirement that $X$ be absolutely separated.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.