In Mathlib.

Let \(g \in \operatorname{Gal}(K/F)\) be a generator. Since \(K/F\) is separable, the trace map \(K \to F\) is nonzero; we can thus choose \(y \in K\) with

Set

then \(z - g(z) = 1\), and so \(z^p-z\) is fixed by \(g\). By Lemma ??, \(z^p-z \in F\). Set \(a := z^p - z\); the isomorphism in question now carries \(z\) to the class of \(x\). Since the polynomials \(x^p-x\) and the minimal polynomial of \(z\) are monic of the same degree \(p\) (the latter because it must divide \([K:F] = p\) and cannot equal \(1\)) and the latter divides the former, they coincide.

Since \(K\) is algebraically closed, \(g\) admits a root \(c\) in \(K\). Let \(f(x)\) be the minimal polynomial of \(c\), which necessarily divides \(g\). Then \(\deg (f)\) equals the degree \([F(c):F]\), which divides \([K:F]\).

By Lemma 3, the polynomial \(g(x) = x^4 - 2ax^2 + (a^2+b)\) has a monic divisor \(f(x)\) of degree dividing \([K:F] = 2\). If \(\deg (f) = 1\), then \(f\) has a root \(c \in F\) and \(-b = (c^2-a)^2\) is a square in \(F\).

Suppose now that \(\deg (f) = 2\). If \(f\) has linear coefficient 0, then \(f(x) = x^2 + t\) for some \(t \in F\) and \(-b = (t+a)^2\). Otherwise, write \(g(x) = f(x) e(x)\); by comparing coefficients of \(x^3\), then coefficients of \(x\), we have \(e(x) = f(-x)\) and so \(a^2 + b = f(0)^2\).

If \(p=0\) in \(F\), then by Proposition 2, \(K = F[x]/(x^p-x-a)\) for some \(a \in F\). Since \(K\) is algebraically closed, there exists \(y \in K\) such that \(y^p-y=ax^{p-1}\). Write \(y = \sum _{i=0}^{p-1} y_i x^i\) with \(y_i \in F\); then

In particular, comparing coefficients of \(x^{p-1}\) in the equation

yields \(y_{p-1}^{p} - y_{p-1} = a\) and so \(x-y_{p-1} \in \mathbb {F}_p\), a contradiction.

Suppose by way of contradiction that \(p \neq 2\). By Lemma 5, \(F\) cannot have characteristic \(p\); consequently, \(K\) contains a primitive \(p\)-th root of unity \(\zeta _p\). By Lemma 3, the minimal polynomial of \(\zeta _p\) over \(F\) has degree dividing \([K:F] = p\), but also bounded above by \(p-1\) since \(\zeta _p\) is a root of \((x^p-1)/(x-1)\). Hence the degree is 1, that is, \(\zeta _p \in F\).

By Kummer theory, \(K \cong F[x]/(x^p-a)\) for some \(a \in F\). Lemma 1 implies that \(x^{p^2}-a\) is irreducible over \(F\); however, since this polynomial splits in \(K\), its degree must divide \([K:F] = p\), a contradiction.

The extension \(K/F\) is normal, and by hypothesis it is separable, so it is Galois; let \(G\) be the Galois group. If \(G\) is trivial, then \(F=K\) is algebraically closed. Otherwise, for any prime factor \(p\) of \([L:K]\), by Cauchy’s theorem \(G\) contains a nontrivial cyclic subgroup of prime order \(p\), whose fixed field is a subfield \(E\) of \(K\) containing \(F\). Applying Lemma 6 to \(K/E\) yields \(p=2\).

By Lemma 5, \(F\) is not of characteristic \(2\). We can thus find \(a \in F\) which is not a square; since \(F\) contains a square root of \(-1\), \(-a\) is not a square either. However, this contradicts Lemma 4.

It will suffice to check that \(F\) is perfect. Let \(p\) be the characteristic of \(F\). If \(p = 0\), then \(F\) is automatically perfect. If \(p {\gt} 0\), then \([K:F] = p^n\) for some nonnegative integer \(n\). For any \(a \in F\), we can find \(x \in K\) with \(x^{p^{n+1}} = a\) because \(K\) is algebraically closed. Since \(K/E\) is purely inseparable, the minimal polynomial of \(x\) over \(F\) has the form \(x^{p^e}-c\) for some \(c \in F\) and some nonnegative integer \(e\), which must be at most \(n\) because the degree of the minimal polynomial is at most \([K:F] = p\). Now \(c^{n-e}\) is a \(p\)-th root of \(a \in E\); since \(a\) was arbitrary we conclude that \(F\) is perfect (in the ring-theoretic sense, and hence also in the field-theoretic sense).

Let \(E\) be the separable closure of \(F\) within \(K\). Then \(K/E\) is purely inseparable, so by Lemma 8, \(E\) is algebraically closed. Meanwhile, \(E/F\) is separable, so by Lemma 7 \(F\) is algebraically closed.

By contrapositive: if every irreducible factor has even degree, then these degrees sum to an even number which is the degree of the original polynomial.

For \(x \in F\), Lemma 4 implies that one of \(x\) or \(-x\) is a square in \(F\).

If \(f\) is a polynomial of odd degree, by Lemma 3 its irreducible factors all have degrees 1 or 2, but by Lemma 11 there must be an irreducible factor of odd degree; this must have degree 1, and its unique root is a root of \(f\) in \(F\).

We show that the sum of any two squares in \(F\) is a square. For \(a,b \in F\), by Lemma 4, one of \(a^2+b^2\) or \(-b^2\) is a square; the latter cannot happen with \(b \neq 0\). By induction, it follows that any sum of squares in \(F\) is a square. Since \(-1\) is not a square in \(F\), it follows that \(F\) is real.

Let \(K\) be an algebraic closure of \(F\). Let \(i \in K\) be a square root of \(-1\). By Lemma 9, \(F(i)\) is algebraically closed; so either \(i \in F\) and \(F\) is algebraically closed, or \(i \notin F\) and \(F\) is real closed by Lemma 12.

Pick a nonempty open affine subspace \(U\) of \(X\), let \(Z\) be the reduced divisor \(X \setminus U\), and choose a finite set of generators of \(\Gamma (U, G)\) as a module over \(\Gamma (U, \mathcal{O}_X)\); these generators can be viewed as elements of \(G(mZ)\) for some nonnegative integer \(m\). We thus have a surjection \(\mathcal{O}_X^{\oplus d} \to G(mZ)\) for some positive integer \(d\). In particular, we can choose a nonzero (hence injective) map \(\mathcal{O}_X(mZ) \to G\) such that the composition \(\mathcal{O}_X(mZ)_x \to G_x \to F_x\) is nonzero. We may then take \(H\) to be the saturation of the image of \(\mathcal{O}_X(mZ) \to G\), which will also be saturated in \(F\).

We prove the theorem for a general vector bundle \(F\) of rank \(n\) by induction on \(n\), the case of rank 0 serving as a trivial base case. For the induction step, apply Lemma 19 with \(G = F\) to construct a saturated rank-1 subbundle \(F_1\) of \(F\), then apply the induction hypothesis to \(F/F_1\) to conclude.

Let \(H \neq 0\) be the image of \(f\). Let \(H'\) be the saturation of \(H\) in \(G\). Then

so all of these inequalities must be equalities. Since \(G\) is stable this means that \(H = H' = G\), and since \(F\) is stable the kernel of \(f\) must be zero.

By Lemma 25, every nonzero endomorphism of \(\mathcal{O}_X\) is invertible.

We prove the claim by strong induction on \(n+i\). We may subsume all cases with \(i=0\) into the base case.

Let \(G\) be an arbitrary subbundle of \(F\) of rank \(i\); we then have an exact sequence

and hence an equality

Now \(G \cap F_1\) is isomorphic to a subbundle of \(F_1\) of some rank \(m \in \{ 0,\dots ,i\} \); because \(F_1\) is semistable,

Meanwhile, \(G/(G \cap F_1)\) is a subbundle of \(F/F_1\) of rank \(i-m\), and so by the induction hypothesis

This proves the claim.

This follows from Lemma 27 because every line bundle is stable.

We proceed by induction on \(n = \operatorname{rank}(F)\). Let \(S \subseteq \mathbb {Q}\) be the set of slopes of nonzero subbundles of \(F\). The set \(S\) is nonempty (it contains \(\mu (F)\)), bounded above (by Lemma 20 and Lemma 28), and consists only of rational numbers with denominator bounded by \(n\). It thus admits an upper bound \(\mu _1\); let \(F_1\) be a subbundle of \(F\) of \(\mu _1\) of maximal rank among such subbundles. We then continue by induction.

Let \(F_1, G_1\) be the first steps in the respective HN filtrations of \(F\) and \(G\); then \(\mu (F_1) \geq \mu (F)\), \(\mu (G_1) \geq \mu (G)\). Then \(F_1 \otimes G_1\) is a nonzero subbundle of \(F \otimes G\) of slope \(\mu (F_1)+\mu (G_1)\); since \(F \otimes G\) is semistable, we must have

We conclude that \(\mu (F_1) = \mu (F), \mu (G_1) = \mu (G)\) and so \(F\) and \(G\) are semistable.

By recursing along the HN filtration, we reduce to the case where \(F\) is semistable of slope \(\mu {\lt} 0\). The claim holds in this case because any nonzero element of \(H^0(X,F)\) would correspond to a nonzero morphism \(\mathcal{O}_X \to F\), the saturation of whose image would be a submodule of \(F\) of positive degree in violation of the semistability condition.

By recursing along the HN filtration and applying Lemma 32, we reduce to the case where \(F\) is semistable of degree 0. In this case, we prove that \(\dim _E H^0(X, F) \leq \operatorname{rank}(F)\) by induction on \(\operatorname{rank}(F)\): any nonzero element of \(H^0(X,F)\) corresponds to a nonzero morphism \(\mathcal{O}_X \to F\), the saturation of whose image must be saturated (as otherwise it would violate the semistability condition) and thus sits in a filtration

in which \(F/\mathcal{O}_X\) is again semistable of degree 0.

We prove the claim by induction on \(j-i\). The base case \(j-i=1\) is covered by the second part of 1. For the induction step, suppose by way of contradiction that \(m_j{\gt} m_i+1\) for some \(i,j\). Set \(m := m_i + 1\). For \(k = i+1, \dots , j-1\) we have \(m_{k} \leq m_i+1\) and \(m_j \leq m_{k}+1\) by the induction hypothesis; combining with \(m_j \geq m_i + 2\) shows that both of these must be equalities and so \(m_k = m, m_j = m+1\). Now \((m_i, \dots , m_j) = (m-1, m, \dots , m, m+1)\), which contradicts 1.

For \(n = 1\), note that (a) directly forces \(m_1 = 0\). For \(n {\gt} 1\), we have \(m_n = (m_1 + \cdots + m_n) - (m_1 + \cdots + m_{n-1}) \geq 0\) by (a) and (b). Moreover, if \(m_n \geq 1\), then by (c) we have \(m_1,\dots ,m_{n-1} \geq 0\) and hence \(m_1 + \cdots + m_n \geq 1\), a contradiction. Hence \(m_n \leq 0\) as claimed. By (b), we have

We thus have \(m_n \leq \lfloor \frac{c+n-1}{n}\rfloor \).

Define the function \(f\colon S \to \mathbb {Z}\) by

from conditions (b) and (c), we see that \(\sum _{j=1}^i m_i \leq c_i\) for each \(i\), so \(f\) is bounded above by \(\sum _{i=1}^{n-1} c_i\). (We also have the simpler expression \(f(m_1,\dots ,m_n) = \sum _{i=1}^n (n-i) m_i\) but we will not need it.)

Since \(S\) is nonempty, we can choose a tuple \((m_1,\dots ,m_n) \in S\) on which \(f\) achieves its maximum. There then cannot exist indices \(i,j\) satisfying 2, as then the new sequence \((m'_1,\dots ,m'_n)\) would have \(\sum _{i=1}^{n-1} (n-i)m'_i {\gt} \sum _{i=1}^{n-1} (n-i)m_i\). Consequently, we can apply Lemma 34 to deduce that \(m_j \leq m_i + 1\) for \(1 \leq i {\lt} j \leq n\), then Lemma 35 to conclude.

By condition (c) of Definition 45, we can choose a closed point \(x \in X\) of degree \(1\); let \(\kappa _x\) denote the residue field of \(x\). By condition (a) of Definition 45, the ideal sheaf \(J_x\) of \(x\) is isomorphic to \(\mathcal{O}_X(-1)\); we thus get an inclusion \(\mathcal{O}_X \cong J_x(1) \to \mathcal{O}_X(1)\). Let \(j_x \colon x \to X\) be the canonical inclusion. By taking cohomology in the short exact sequences

and using condition (b) of Definition 45, we get exact sequences

we deduce the claim from this.

This should be treated as a black box.

Set \(n := \operatorname{rank}(F)\) and \(c = \deg (F)\). Let \(S\) be the set of nondecreasing sequences which occur as slope sequences of complete flags of \(F^\vee \). We check that \(S\) satisfies the conditions of Lemma 36.

1. By Lemma 20, there exists a complete flag.

2. We have \(m_1 + \cdots + m_n = \deg (F^\vee ) = -\deg (F) = -c\).

3. This holds by Lemma 20.

4. Let \(0 = F_0 \subseteq \cdots \subseteq F_n = F\) be a complete flag giving rise to the sequence \((m_1,\dots ,m_n)\) and suppose there exist indices \(1 \leq i {\lt} j \leq n\) for which 2 holds. Set \(m := m_i + 1\). Since \((F_j/F_{j-1})(-m) \cong \mathcal{O}_X(m_j-m)\) and \(m_j \geq m + 1\), we can find a copy of \(\mathcal{O}_X(1)\) inside \((F_j/F_{j-1})(-m)\); pick one such copy and let \(G\) be its preimage in \((F_j/F_{i-1})(-m)\). Then \(G\) satisfies the conditions of Lemma 47 and so \(H^0(X, G) \neq 0\). We conclude that \(F_j\) admits a saturated subbundle \(F_{i}'\) of rank 1 containing \(F_{i-1}\) such that \(\deg (F_{i}'/F_{i-1}) {\gt} m_i\). We fill in a complete flag between \(F_{i}'\) and \(F_j\) by projection (that is, take the sequence \(F_i', F_i' + F_i, \dots , F_i'+F_{j-1}, F_j\) and remove any redundancies); let \(m'_i\) be the associated slope sequence. We then have:

   • \(m'_k = m_k\) for \(k \notin \{ i,\dots ,j\} \);

   • \(m'_i {\gt} m_i\);

   • \(m'_i + \cdots + m'_j = m_i + \cdots + m_j\);

   • for \(k=i+1,\dots ,j-1\), \(m'_i + \cdots + m'_k \geq m_i + \cdot + m_k\). To check this, note that \(F_k'/F_{i-1}\) is either isomorphic to \(F_k/F_{i-1}\), in which case \(m'_i + \cdots + m'_k = m_i + \cdots + m_k\), or to \(F_{i}'/F_{i-1} \oplus F_{k-1}/F_{i-1}\), in which case

     \begin{align*} m’_i + \cdots + m’_k & = m’_i + (m_i + \cdots + m_{k-1}) \\ & \geq m_i + 1 + (m_i + \cdots + m_{k-1}) \\ & = m_k + m_i + \cdots + m_{k-1}. \end{align*}

   Together these points imply that \(m'_1 + \cdots + m'_k \geq m_1 + \cdots + m_k\) for \(k=1,\dots ,n-1\) with strict inequality when \(k=i\).

By Lemma 36, there is a tuple \((m_1,\dots ,m_n) \in S\) with \(m_n \leq \lfloor \frac{-c-n+1}{n} \rfloor \). This means that \(F\) admits a complete flag with \(\deg (F_1) \geq - \lfloor \frac{-c-n+1}{n} \rfloor \geq 0\) and so \(H^0(X, F_1) \neq 0\). Since \(H^0(X, F_1) \to H^0(X, F)\) is injective, it follows that \(H^0(X, F)\neq 0\).

Let \(F\) be a semistable bundle of degree \(0\) and rank \(n\); we proceed by induction on \(n\). By Lemma 48, there exists a nonzero morphism \(\mathcal{O}_X \to F\). Because \(F\) is semistable, of slope \(0\), this morphism must have saturated image and the quotient \(F/\mathcal{O}_X\) must also be semistable of degree \(0\), and hence trivial by the induction hypothesis (or vacuously in the base case \(n=1\)). Since \(H^1(X, \mathcal{O}_X) = 0\), we conclude that \(F\) is also trivial. Consider an exact sequence as in Lemma 50.

We proceed by induction on \(n = \operatorname{rank}(F)\), with the case \(n=0\) being vacuous. Set \(c = \deg (F)\); by the condition on the HN slopes, we have \(0 \leq c \leq n\). If \(c = 0\) (resp. \(c = n\)), then \(F\) (resp. \(F(-1)\)) is semistable of degree 0 and we may deduce the claim from Proposition 49; we may thus assume that \(0 {\lt} c {\lt} n\).

Let \(F_1\) be the maximal subbundle of \(F\) with \(\deg (F_1) = \operatorname{rank}(F_1)\). Since \(c {\lt} n\) we have \(F_1 \neq F\), so by Lemma 48 we have \(H^0(X, F/F_1) \neq 0\). We can thus choose a nonzero map \(f \colon \mathcal{O}_X \to F/F_1\); the saturation of the image of this map is a rank-1 subbundle of \(F/F_1\) of degree \(\geq 0\). However, this degree cannot be \(\geq 1\) because \(F/F_1\) has all slopes less than 1. Consequently, if we let \(F_2\) be the preimage of the image of \(f\) along \(F \to F/F_1\), we have an exact sequence

which splits because \(H^1(X, \mathcal{O}_X) = 0\). We conclude that there exists an exact sequence

in which \(F'\) itself fits into an exact sequence

It will suffice to check that \(F/F_2\) has HN slopes in \([0,1)\), as then \(F'\) will have HN slopes in \([0,1]\) and we may conclude using the induction hypothesis (plus the fact that any extension of trivial bundles is trivial). Let \(G'\) be a nontrivial proper subbundle of \(F/F_2\) and let \(H\) be the preimage of \(G'\) in \(F\). Then \(\operatorname{rank}(H) = 1 + \operatorname{rank}(F_1) + \operatorname{rank}(G')\) and \(\deg (H) = \operatorname{rank}(F_1) + \deg (G')\). From the definition of \(F_1\) we have \(\deg (H){\lt} \operatorname{rank}(H)\), or in other words \(\deg (H) \leq \operatorname{rank}(H) - 1\); this implies \(\deg (G') \leq \operatorname{rank}(G')\); this proves the claim.

By replacing \(F\) with the first step of its HN filtration, we may reduce to the case where \(F\) is semistable. By twisting, we may further ensure that \(\mu (F) \in [0,1)\). We then deduce both claims from Lemma 50 and the assumption \(H^1(X, \mathcal{O}_X) = 0\) from Definition 45.

Using the assumption that \(H^1(X, \mathcal{O}_X) = 0\), we reduce to the case where \(F\) is stable with slope in \([0,1)\). In this case, suppose by way of contradiction that \(F^\vee \otimes F\) is not semistable, and let \(G\) be the first step of its HN filtration. Since \(G \otimes G\) has slope \(2\mu (G) {\gt} \mu (G)\), the composition map \((F^\vee \otimes F) \otimes (F^\vee \otimes F) \to F^\vee \otimes F\) restricts to zero on \(G \otimes G\). By Corollary 51, \(G\) is generated by global sections; consequently, \(G \otimes G\) is generated by nonzero global sections of the form \(s_1 \otimes s_2\). Any such section corresponds to a pair of nonzero endomorphisms \(F \to F\) which compose to zero; this contradicts Lemma 25. We deduce that \(F^\vee \otimes F\) is semistable of slope \(0\), and hence trivial by Proposition 49.

Let \(F,G\) be two semistable vector bundles on \(X\). By Lemma 52,

is semistable. By Lemma 31, \(F \otimes G\) must be semistable.

Let \(F,G\) be two such bundles. By Proposition 53, \(F^\vee \otimes G\) is semistable; since it has slope 0, it is trivial by Proposition 49. We can thus find a nonzero morphism \(f\colon F \to G\), which must be an isomorphism by Lemma 25.

By Proposition 53, \(F_2^\vee \otimes F_1\) has nonnegative HN slopes. Now \(\operatorname{Ext}^1_X(F_2, F_1) = H^1(X, F_2^\vee \otimes F_1)\) which vanishes by Corollary 51.

Both assertions follow directly from Lemma 55.

By Lemma 46, the condition \(d {\lt} \dim _{E} H^0(X, \mathcal{O}_X(1))\) is equivalent to \(d-1 \leq \dim _E H^1(X, \mathcal{O}_X(-1))\). We may thus choose an element of \(H^1(X, \mathcal{O}_X(-1)^{\oplus (d-1)}) \cong H^1(X, \mathcal{O}_X(-1))^{\oplus (d-1)}\) consisting of \(d-1\) elements of \(H^1(X, \mathcal{O}_X(-1))\) which are linearly independent over \(E\). This element corresponds to an exact sequence

and it will suffice to show that \(F\) is stable.

Let \(G\) be a subbundle of \(F\) of rank \(e \in \{ 1,\dots ,d-1\} \). Then

Since \(G \cap \mathcal{O}_X^{\oplus d-1}\) is a subbundle of \(\mathcal{O}_X^{\oplus d-1}\) which is semistable of degree 0, we have \(\deg (G \cap \mathcal{O}_X^{\oplus d-1}) \leq 0\). Since \(G / (G \cap \mathcal{O}_X^{\oplus d-1})\) is isomorphic to a subbundle of \(\mathcal{O}_X(1)\), we have \(\deg (G / (G \cap \mathcal{O}_X^{\oplus d-1}) \leq 1\). We deduce that \(\deg (G) \leq 1\), and it will suffice to show that equality cannot occur.

If \(\deg (G) = 1\), then we have a filtration of \(F\) with successive quotients

Choose a complement \(\mathcal{O}_X^{d-e}\) of \( G \cap \mathcal{O}_X^{\oplus d-1}\) in \(\mathcal{O}_X^{\oplus d-1}\); then this complement maps injectively to \(F/G\), but since both source and target have degree 0 this map must be an isomorphism. Consequently, the exact sequence

splits; but this unpacks to the statement that a certain \(E\)-linear combination of the chosen elements of \(H^1(X, \mathcal{O}_X(-1))\) vanishes, a contradiction.

We proceed by induction on \(d\), the case \(d=1\) being evident because \(\mathcal{O}_X(c)\) is stable for all \(c \in \mathbb {Z}\).

We will produce several examples of semistable bundles \(F\) of slope \(\frac{c}{d}\). For each such bundle, we may apply Proposition 56 to decompose \(F\) as a sum of stable bundles. Let \(G\) be one such bundle; by Lemma 54, all of the other summands are isomorphic to \(G\), and so \(\operatorname{rank}(G)\) divides \(\operatorname{rank}(F)\). By Lemma 54 again, the isomorphism class of \(G\) does not depend on the choice of \(F\) either. It will thus suffice to produce enough examples so that the gcd of the resulting values of \(\operatorname{rank}(F)\) is equal to \(d\).

• For \(d{\gt}1\), we can choose \(b \in \{ 1,\dots ,d-1\} \) with \(bc \equiv 1 \pmod{d}\). By Lemma 57, we can find a stable bundle \(G_1\) of rank \(bd\) and degree \(1\). By the induction hypothesis, we can find a stable bundle \(G_2\) of rank \(b\) and degree \((bc-1)/d\). By Proposition 53, \(G_1 \otimes G_2\) is semistable of rank \(b^2 d\) and slope

  \[ \frac{1}{bd} + \frac{(bc-1)/d}{b} = \frac{c}{d}. \]

• By Lemma 57 again, we can find a stable bundle \(G'_1\) of rank \(d\) and degree \(1\). By Proposition 53, \((G'_1)^{\otimes c}\) is semistable of rank \(d^c\) and slope \(\frac{c}{d}\).

We deduce that there exists a stable bundle \(G\) of slope \(\frac{c}{d}\) and degree dividing

On the other hand, since \(\deg (G) \in \mathbb {Z}\), \(\operatorname{rank}(G)\) must be divisible by \(d\). Hence \(G\) is the desired stable bundle of rank \(d\) and degree \(c\).

Apply Theorem 13.

If \(\dim _{P_0} P_1 = \infty \), then we may deduce (a) and (b) from Lemma 58 and Lemma 54. Otherwise, by Lemma 59 we may argue as follows.

• In the classical case, \(H^1(X, \mathcal{O}_X(-1)) = 0\). Hence any exact sequence of the form

  \[ 0 \to \mathcal{O}_X \to F \to \mathcal{O}_X(1) \to 0 \]

  splits, as then does any filtration as in Lemma 50, proving (a). From this, (b) is obvious.

• In the twistor case, by identifying \(P \otimes _{P_0} P_0(\sqrt{-1})\) with \(P_0(\sqrt{-1})[x^2,xy,y^2]\) and applying the classical case, we see that \(S \subseteq \tfrac {1}{2}\mathbb {Z}\). We deduce (a) and (b) by applying Lemma 57 to construct a stable bundle of rank \(2\) and degree \(1\), then applying Lemma 54 for uniqueness.

Given (a)–(b), (c) follows from Lemma 25 and Lemma 52; (d) is immediate from Proposition 53 and Proposition 56; and (e) is immediate from Proposition 56.

Exactness at the left and middle is a consequence of the factorial property. To prove exactness on the right, we interpret \(P/sP\) using 6 as in Lemma 62. Then \(t\) corresponds to \(\lambda T^n\) for some nonzero \(\lambda \in L_s\); similarly, any given \(w \in P_{n+1}\) corresponds to \(\mu T^{n+1}\) for some \(\mu \in L_s\). Choose \(u \in P_1\) corresponding to \(-(\mu /\lambda )T\); then \(tu + w\) maps to zero in \(P/sP\) and so can be written as \(sv\) for some \(v \in P_n\).

If \(a\) and \(b\) have a common factor, then we can choose \(u,v\) not both zero with \(au+bv =0\), and then take \(w=0\). Similarly if \(a\) and \(c\) have a common factor or if \(a\) and \(b\) have a common factor. We may thus assume hereafter that none of these cases occur.

By hypothesis, \(c\) has two coprime factors \(c_1,c_2\). We can find \(d_1, d_2,e_1,e_2 \in P_1\) such that

here \(e_1,e_2\) must be coprime to \(c_2\) (as otherwise \(c_2\) would divide \(a\) or \(b\)) and \(d_1,d_2\) must be coprime to \(c_1\). The condition on \(u,v,w\) becomes

Now it is sufficient (and even necessary) to choose \(u,v \in P_1\) such that

Using the coprimality conditions from above, we can rephrase as

for some \(d'_2, e'_2, f \in P_1\) with \(f\) coprime to \(c\), then introduce variables \(w_1, w_2 \in P_1\) with

If \(e'_2 = d'_2\) then we can enforce \(w_1 = c_2\), \(w_2 = c_1\). Otherwise, we can eliminate \(u\) to get

and separately eliminate \(v\) to get

with the proviso...

By condition (i) of Definition 61, we can pick a nonzero \(s \in P_1\). By condition (iii) of Definition 61, we can find \(t \in P_1\) with nonzero image in \(P_1/sP_0\). This proves the claim.

For any \(s,t \in P_1\) which are linearly independent over \(P_0\) (as in Lemma 68, by 6 we have \(P_2 = P_1 s + P_1 t\). By induction on \(m+n+k\) we deduce that \(P_{m+n+k} = P_{m+k} s^n + P_{n+k} t^m\) for any \(m,n {\gt}0\) and \(k \geq 0\), and hence

Pick \(s,t\) as in Lemma 68; by Lemma 69, \(X\) is covered by the distinguished open subspaces corresponding to \(s\) and \(t\). By Proposition 66, each of these subspaces is regular of dimension 1. By condition (iii) of Definition 61, each closed point of either subspace has algebraically closed residue field.

The map \(\operatorname{Pic}(X) \to \mathbb {Z}\) carries \(\mathcal{O}_X(n)\) to \(n\), so we need only check injectivity. This follows from Proposition 66.

By Proposition 70, \(X\) is an abstract complete curve. Of the conditions from Definition 45, (i) comes from Proposition 71; (ii) comes from Lemma 69; and (iii) comes from the description of the degree function in Proposition 70.