Line bundles on projective three-space and their restrictions

On each standard chart $U_i\cong \mathrm{Spec}k[z_1,z_2,z_3]$, the coordinate ring is a UFD: start with the field $k$ and apply [F6] to $k[z_1]$. At each later variable, write a polynomial as its content times a primitive polynomial, factor the content in the old UFD, factor the primitive part in the fraction-field polynomial ring using [F6], and clear denominators to primitive factors. [F5] preserves primitivity under products and reflects irreducibility between the old UFD and its fraction field, so these factorizations exist uniquely up to units. Iterating [F4] gives the claim for three variables. Let $R=k[z_1,z_2,z_3]$, $K=\mathrm{Frac}(R)$, and let $\mathcal{L}$ be an invertible sheaf on $\mathrm{Spec}R$. Its local frames give a nonzero rational section $s$ in the one-dimensional generic fibre. Refine a trivializing cover to principal opens $D(f_a)$; [F7] gives a finite subcover. Write $s=g_a{e}_a$ on each member, where $e_a$ is a frame and $g_a\in K^{\times }$. For an irreducible $p\in R$, define $v_p(s)=v_p(g_a)$ using any member containing the generic point of $V(p)$. This is independent of the member: on an overlap containing that generic point, two frames differ by a unit, whose $p$-valuation is zero. Only finitely many $v_p(s)$ are nonzero, since each of the finitely many $g_a$ has finite factor support. Choose representatives for this finite support and set $g={\prod }_p{p}^{v_p(s)}\in K^{\times }$. On $D(f_a)$, the quotient $g_a\mathrm{/}g$ has valuation zero at every irreducible not dividing $f_a$. Unique factorization then writes it as a unit of $R_{f_a}$, because every remaining prime factor is inverted there. Thus $s\mathrm{/}g$ is a nowhere-zero regular frame on every member of the cover. The local sections agree on overlaps as the same rational section, so [F3] glues them to a global frame. Hence $\mathcal{L}$ is trivial on each $U_i$.

Choose a frame $e_i$ on each of the four charts. The units on $U_i\cap U_j=D(x_j\mathrm{/}{x}_i)$ are exactly $c_{ij}(x_j\mathrm{/}{x}_i{)}^{n_{ij}}$, with $c_{ij}\in k^{\times }$ and $n_{ij}\in \mathbb{Z}$: the chart ring is a polynomial UFD and its only units are constants, while the overlap inverts just the displayed coordinate. On a triple overlap, the two independent invertible ratios force $n_{ij}=n_{jk}=n_{ik}$ from the cocycle equation. Since every pair of the four indices occurs in such triples, there is one common integer $n$. The constants satisfy $c_{ik}=c_{ij}{c}_{jk}$. Set $a_0=1$ and $a_j=c_{0j}^{-1}$, and replace $e_i$ by $a_i{e}_i$; then every new transition constant is $c_{ij}{a}_j\mathrm{/}{a}_i=1$. The resulting transition functions are $(x_j\mathrm{/}{x}_i{)}^n$, exactly those defining $\mathcal{O}(n)$.

If $\mathcal{O}(n)\cong \mathcal{O}(m)$, restrict the transition cocycle to a coordinate line. The two standard affine charts have transition $t^{n-m}$, and units on either chart are constants. This transition is a coboundary only when $n-m=0$, so $n=m$.

For a $k$-line $L\subset {\mathbb{P}}_k^3$, extend a basis of its two-dimensional vector subspace to a basis of $k^4$. The resulting projective coordinate change carries $L$ to a coordinate line and preserves the hyperplane sheaf, since it changes the homogeneous coordinate sections by an invertible linear transformation. Restricting its two standard chart frames to that line gives transition $t$ for $\mathcal{O}(1)$ and $t^n$ for $\mathcal{O}(n)$, so $\mathcal{O}(n){\mathrm{\mid }}_L\cong {\mathcal{O}}_{{\mathbb{P}}^1}(n)$.

For the conic clause, choose a line $H$ in its plane ${\mathbb{P}}_k^2$. The conic is geometrically integral (a reducible plane conic is singular at the intersection of its line components), so its quadratic equation restricts to a nonzero binary quadratic on $H\cong {\mathbb{P}}_k^1$. Its zero divisor $D=C\cap H$ has degree two: factoring that binary quadratic into homogeneous irreducible factors counts each closed point with its residue degree and multiplicity, and the total factor degree is two. The equation of $H$ is a section of ${\mathcal{O}}_C(1)$ with zero divisor $D$. Pull it back along the given $k$-isomorphism $\varphi$; its zero divisor $D^{\mathrm{\prime }}$ on ${\mathbb{P}}_k^1$ still has degree two, and the pulled-back line bundle is $\mathcal{O}(D^{\mathrm{\prime }})$. For each closed point $Q\mathrm{\ne }\mathrm{\infty }$ of ${\mathbb{A}}_k^1\subset {\mathbb{P}}_k^1$, let $p_Q(t)$ be its monic irreducible polynomial. Then $\mathrm{div}(p_Q)=Q-\mathrm{deg}(Q)\mathrm{\infty }$. Thus every divisor of degree two on ${\mathbb{P}}_k^1$ is linearly equivalent to $2\mathrm{\infty }$, including when the two intersection points coincide or are not $k$-rational. Therefore ${\varphi }^{\ast }({\mathcal{O}}_C(1))\cong {\mathcal{O}}_{{\mathbb{P}}^1}(2)$. The transition definition gives $\mathcal{O}(n)=\mathcal{O}(1{)}^{\otimes n}$, also for negative $n$ using duals, so its pullback to $C$ is ${\mathcal{O}}_{{\mathbb{P}}^1}(2n)$.

Let $h:{\mathbb{P}}_k^3\hookrightarrow {\mathbb{P}}_k^N$ be a closed immersion. Restrict to a line. Its pullback hyperplane sheaf is ${\mathcal{O}}_{{\mathbb{P}}^1}(n)$ by step 3.2. The ambient homogeneous coordinates give sections of the pullback module [F9] with no common zero. Since [F8] makes $h$ injective on points, these coordinate sections separate two distinct $k$-points of the line. On the two standard affine charts, a global section of ${\mathcal{O}}_{{\mathbb{P}}^1}(n)$ is represented by $f_0(t)\in k[t]$ and $f_1(t^{-1})\in k[t^{-1}]$ with $f_0(t)=t^n{f}_1(t^{-1})$. Hence there are no nonzero sections if $n<0$, only constants if $n=0$, and the two-dimensional span of $1,t$ if $n=1$. For $n<0$ the coordinate sections cannot define a morphism; for $n=0$ they define a constant map, contradicting injectivity on the line. Thus $n>0$. The bound is sharp: the identity immersion of ${\mathbb{P}}_k^3$ has $n=1$. All choices made above are finite choices of frames or bases, not a choice function, so no AC is used.