Codex @codex  0

A complete quadrangle consists of four projective points in the projective plane, no three collinear, and all six joining projective lines. Its three diagonal points are the intersections of opposite joining lines. Over a field of characteristic different from two, the line joining two diagonal points cuts either side through the third diagonal point in a pair of harmonic conjugate points. In a projective frame this follows by using the coordinate points and $[1:1:1]$ and computing the intersections.

For four distinct projective points on the complex projective line, use the conventionwhere $[p_i,p_j]$ is the determinant of chosen representative vectors. Rescaling any representative cancels, and applying an invertible matrix multiplies every bracket by its determinant, which also cancels. Thus the expression is a projective invariant. The projective general linear group action on the projective line has a unique element $T$ taking the first three points to $\infty ,0,1$. In those coordinates the displayed expression is exactly $T(P_3)$. Consequently the cross-ratio completely determines the orbit of the ordered quadruple under the projective linear group. Its value lies in $\mathbb{C}\setminus \{0,1\}$.

For the convention in cross-ratio normalization of four projective points, permuting four distinct projective points produces the six displayed rational expressions. Interchanging the first pair sends $\lambda$ to $1/\lambda$; interchanging the middle pair sends it to $1-\lambda$. The four permutations consisting of the identity and the three double transpositions form a Klein four-group preserving the cross-ratio. The resulting six-element action gives every displayed expression.

For real $\lambda \notin \{0,1\}$, repeated values occur exactly for $\lambda \in \{-1,1/2,2\}$, and then the orbit has size three. To prove this, a repetition gives a nonidentity stabilizer after composing one operation with the other's inverse. Its fixed-point equations give $-1$, $1/2$, $2$, or ${\lambda }^2-\lambda +1=0$, with the last case nonreal. With one point at infinity, $\lambda =(x_1-x_3)/(x_1-x_2)$; the three real exceptional values mean that one finite point is the midpoint of the other two. Thus these harmonic conjugate points configurations have three collinear equally spaced finite points in that affine coordinate.

An injective self-map of the complex projective line preserving all cross-ratios is a projective transformation; neither continuity nor prior surjectivity is needed. Choose the unique projectivity agreeing with the map on three distinct projective points and compose with its inverse. The resulting map fixes those three points. Normalize them to $\infty ,0,1$. Preservation of the cross-ratio $(\infty 0:1z)=z$ now fixes every other point as well, proving that the normalized map is the identity map. Undoing the composition proves the assertion.

For two projective triangles in a projective plane over a field, perspectivity from a point implies perspectivity from a line. The usual nondegenerate form assumes that the corresponding-side intersections are well defined and distinct. Its dual is the converse, by point-line duality. The result holds in the real projective plane; it is not an axiom valid in every abstract incidence plane.

One linear algebra proof explains the collinearity. Choose a representative $o$ of the center. Rescale the paired representatives to $p^{\prime }=p+\alpha o$, $q^{\prime }=q+\beta o$, and $r^{\prime }=r+\gamma o$, with nonzero coefficients in a nondegenerate configuration whose vertices differ from the center. The first corresponding-side intersection is represented by $x=\beta p-\alpha q=\beta p^{\prime }-\alpha q^{\prime }$. Similarly choose $y=\gamma q-\beta r$ and $z=\alpha r-\gamma p$. Then $\gamma x+\alpha y+\beta z=0$, so the three representatives are linearly dependent. Their projective points are consequently collinear.

Distinct projective points $C,D$ on the projective line are harmonic conjugates relative to distinct $A,B$ when $(AB:CD)=-1$. Use the cross-ratio convention $(AB:CD)=[a,c][b,d]/([a,d][b,c])$, where the brackets are two-by-two determinants. In an affine coordinate with $A=0$, $B=\infty$, the condition is $D=-C$. A complete quadrangle constructs the fourth point using joins and intersections only. Characteristic two is excluded because then $-1=1$, and the four points could not be distinct.

The homogeneous coordinates $[x_0:\cdots :x_n]$ of a projective point are defined only up to simultaneous multiplication by a nonzero scalar.

In the standard graded homogeneous coordinate ring $S=k[x_0,\dots ,x_n]$ of ${\mathbb{P}}^n$, the irrelevant ideal isIt vanishes only at the origin of the corresponding affine cone, which does not represent a projective point.

It is the polynomial-ring instance of the irrelevant ideal of a graded ring.

Write $C=V(P)$, where $P\in \mathbb{C}[x,y,z]$ is a squarefree polynomial and a homogeneous polynomial of degree $d$. A projective point $p\in C$ is a singular point precisely whenIf $p=[p_0:p_1:p_2]$ is a nonsingular point of an algebraic curve, its projective tangent line isThe coefficients are not all zero, and the Euler theorem for homogeneous functions gives $p_0{P}_x(p)+p_1{P}_y(p)+p_2{P}_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the projective point only rescales the equation.

The strong form of Bézout's theorem says that two projective plane curves $C,D$ of degrees $d,e$, with no common irreducible component, have finitely many intersections andEach intersection multiplicity is a positive integer. At $p\in C\cap D$,This is the criterion for a transverse intersection of plane curves. In particular, an intersection at a singular point of either curve contributes at least two.

For $F=x^3+y^3+3x{z}^2$, the first partial derivatives areTheir simultaneous vanishing would force $y=0$, $xz=0$ and $x^2+z^2=0$, hence $x=y=z=0$, which is not a projective point. This also verifies that the given plane cubic is nonsingular.

The Hessian matrix and its determinant areThus the Hessian curve is the union of $y=0$, $x=z$ and $x=-z$. On $y=0$, the equation of $C$ reduces to $x(x^2+3z^2)=0$, givingOn $x=z$ one must have $z\ne 0$, so put $z=x=1$ and obtain $y^3=-4$. On $x=-z$ similarly put $z=1$, $x=-1$, obtaining $y^3=4$. Choose $r\in \mathbb{C}$ with $r^3=4$ and let $\omega =\exp (2\pi \mathrm{i}/3)$. The remaining points areAll nine listed projective points are distinct. They exhaust the intersection with the Hessian curve, so the Hessian criterion for a flex proves that these, and only these, are the flexes in this example. There are none at infinity: $z=0$ and $H_F=0$ would force $x=y=0$ on $C$.

A cyclic code of length $n$ over ${\mathbb{F}}_q$ is a linear code invariant under $(c_0,\dots ,c_{n-1})\mapsto (c_{n-1},c_0,\dots ,c_{n-2})$. Associate the word with $c(T)={\sum }_{i=0}^{n-1}{c}_i{T}^i$. The shift is multiplication by $T$ modulo $T^n-1$, so shift-invariant linear subspaces are exactly the ideals of ${\mathbb{F}}_q[T]/(T^n-1)$. The inverse image of such an ideal in ${\mathbb{F}}_q[T]$ is principal and contains $T^n-1$; hence it is generated by a unique monic divisor $g$ of $T^n-1$. The code has basis $g,Tg,\dots ,T^{n-\mathrm{deg}g-1}g$, and dimension $n-\mathrm{deg}g$.

The defining zeros of a cyclic code are the roots of its generator $g$ in a splitting field. Since $g\mid T^n-1$, each is a root of unity over a finite field. Every codeword vanishes at each such root. If the characteristic divides $n$, root multiplicities must additionally be retained; in the Hamming construction below the modulus has no repeated roots.

The unrestricted ternary cyclic-equivalence assertion is false. At $s=2$ it would give a cyclic $[4,2,3{]}_3$ code. Butover ${\mathbb{F}}_3$, with the quadratic irreducible. A dimension-two cyclic code needs a degree-two generator, so its only possibilities are $T^2-1$ and $T^2+1$. Each is a weight-two codeword. Therefore neither has minimum distance three, and no code equivalent to the ternary $[4,2,3]$ Hamming code can be cyclic. A valid general version of the asserted construction assumes odd $s$. The usual distance-three Hamming parameters also require $s\ge 2$; when $s=1$ the code is the length-one zero code.

Here is the construction with its precise qualification. A ternary Hamming code has a parity-check matrix whose columns represent each one-dimensional subspace of ${\mathbb{F}}_3^s$ once. There are $n=(3^s-1)/2$ such subspaces. Its rank is $s$, because the coordinate lines are included. Its minimum distance is at least three, because no column is zero and no two are proportional. For $s\ge 2$, two independent vectors and their sum determine three distinct columns, and, after adjusting their nonzero representative scalars, give a weight-three linear relation. Thus its parameters are $[n,n-s,3{]}_3$. Permuting columns and changing their nonzero representatives produces a monomial equivalence of linear codes.

Take odd $s\ge 3$ and choose $\alpha \in {\mathbb{F}}_{3^s}^{\times }$ of order $n$. Since $n$ is odd, $-1$ is not in the cyclic subgroup generated by $\alpha$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ lie in distinct one-dimensional ${\mathbb{F}}_3$-subspaces: a proportional pair would have ratio $1$ or $-1$, neither possible for two distinct powers in this list. By their number they represent all projective points. Consequentlyis monomially equivalent to the ternary Hamming code. Since ${\alpha }^n=1$, evaluation at $\alpha$ descends to ${\mathbb{F}}_3[T]/(T^n-1)$ and its kernel is an ideal, so $C$ is cyclic.

The degree of $\alpha$ over ${\mathbb{F}}_3$ is $s$. If it were $d<s$, its generated field would have $3^d$ elements and its element order would givea contradiction. Its minimal polynomial of an algebraic element therefore has degree $s$. Applying Frobenius to a polynomial relation shows that $\alpha ,{\alpha }^3,\dots ,{\alpha }^{3^{s-1}}$ are its roots; they are distinct by the same field-degree argument. Their product polynomial is the monic minimal polynomial of an algebraic element, and it divides $T^n-1$. ThusThis identifies all the defining zeros. For $s=3$, $n=13$, the defining exponents are $1,3,9$ modulo $13$.

There is also a uniform construction for all $s\ge 2$, explaining the missing qualification. Let $\beta$ generate ${\mathbb{F}}_{3^s}^{\times }$, which has order $2n$. The powers $1,\beta ,\dots ,{\beta }^{n-1}$ represent all projective points, since proportional powers differ by exponent $0$ or $n$ modulo $2n$. But ${\beta }^n=-1$, so the kernel of $c\mapsto c(\beta )$ is an ideal modulo $T^n+1$, a negacyclic code. Multiplication by $T$ gives $(-c_{n-1},c_0,\dots ,c_{n-2})$, preserving its kernel. This code is always equivalent to the ternary Hamming code. If $s$ is odd, $n$ is odd too, and the substitution $T\mapsto -T$, or coordinate scaling $c_i\mapsto (-1{)}^i{c}_i$, changes this negacyclic construction into the cyclic one. In fact $\alpha =-\beta$ then has order $n$. These explicit constructions prove the qualified assertion rather than relying on a theorem name.

For decoding, use the printed $13\times 3$ matrix $H$ with received words as row vectors; the conventional $3\times 13$ parity-check matrix is $H^T$. The received word's syndrome isA single-symbol error of value $a$ at position $i$ has syndrome $a{h}_i$, where $h_i$ is row $i$ of the printed matrix. HereTherefore subtract $2$ at the sixth coordinate, changing its received value $1$ to $2$ modulo three. The decoded codeword isIts syndrome is zero. The thirteen projective rows and the two nonzero scalars give all $26$ nonzero syndromes exactly once, so this is the unique codeword at distance one. It is the transmitted word under the usual assumption of at most one symbol error. If positions $1,2,5$ are designated check positions, their independent unit rows make the other ten symbols systematic data, namely $(2,0,2,0,0,2,1,1,2,0)$.

Let $K$ be a number field and $E/K$ an elliptic curve. Heights measure the arithmetic size of points, and turn the finite quotient supplied by the Weak Mordell-Weil theorem into finite generation. For a projective point $[X:Z]\in {\mathbb{P}}^1(K)$ define the Absolute logarithmic Weil heightusing the usual normalized absolute values and local degrees $n_v=[K_v:{\mathbb{Q}}_v]$. The product formula makes this independent of scaling and of enlarging $K$. For a Short Weierstrass form put $h_x(P)=h([x(P):1])$ and $h_x(O)=h([1:0])=0$. Over $\mathbb{Q}$, if $x=A/B$ in lowest terms, $h(x)=\log \max (|A|,|B|)$.

The finiteness property is Northcott theorem: for bounded height and bounded degree over $\mathbb{Q}$ there are only finitely many algebraic numbers. One sees the mechanism from the height-Mahler measure formula. If $\alpha$ has degree $d\le D$ and height at most $H$, its primitive integral minimal polynomial has Mahler measure $e^{dh(\alpha )}\le e^{DH}$. Expanding its product of linear factors bounds every coefficient by $2^D{e}^{DH}$. Only finitely many integral polynomials satisfy this bound, and each has finitely many roots. Thus a height bound gives only finitely many $x$-coordinates in the fixed number field $K$, and each gives at most two points of $E$, besides $O$.

To understand multiplication, the doubling formula givesIts homogeneous numerator and denominator have degree four. They have no common projective zero: at infinity the numerator is nonzero, and at a root of $f(t)=t^3+at+b$ the numerator equals $f^{\prime }(t{)}^2$, since it is $f^{\prime }(t{)}^2-8tf(t)$. Nonsingularity excludes a common root of $f,f^{\prime }$. The coordinate doubling map is therefore a degree-four morphism of the projective line.

The general height growth under a morphism of the projective line is $h(\Phi (t))=dh(t)+O(1)$ for a fixed degree-$d$ morphism. Here is why the error is bounded independently of $t$. At a place $v$, normalize its homogeneous coordinates to have maximum norm $1$. The fixed homogeneous forms have uniformly bounded norm. They also have a positive lower bound: the normalized coordinate space is compact over $K_v$, and their common zero set is empty. Outside finitely many places their coefficients are integral and their reduction has no common zero, making both bounds $1$. Homogeneity and summation over places then give the asserted height estimate. Consequently there is a fixed $C$ with

Define the canonical height of an elliptic curve by the conventional normalizationThe successive approximants differ by at most $C/(2\cdot 4^{n+1})$, so the limit exists andThus canonical height retains Northcott finiteness, while replacing the bounded-error doubling rule by an exact one.

We also need the height parallelogram identity. It follows from the symmetric-square proof of the naive height parallelogram estimate, which we give explicitly using the supplied addition formulae. For $x_i=X_i/Z_i$, the unordered pair $x(P+Q),x(P-Q)$ is represented by the binary quadratic $C{T}^2-BTS+A{S}^2$, whereThese forms define the biquadratic morphism for elliptic sums and differences. They have no common zero. If $C=0$, the two projective coordinates agree. On the finite diagonal $x_1=x_2=t$, $B=4f(t)$ and $A=f^{\prime }(t{)}^2-8tf(t)$, so simultaneous vanishing would violate nonsingularity. At $(\infty ,\infty )$ the homogeneous $A$ is nonzero. This also handles $P=\pm Q$ and sums or differences equal to $O$, for which the affine fractions alone would be inappropriate.

The same local norm argument as above, now using bidegree $(2,2)$, givesOn the other hand the binary quadratic is the product of two linear factors for $x(P+Q)$ and $x(P-Q)$. At nonarchimedean places coefficient maximum norms multiply exactly: scale each factor to norm $1$ and reduce; the product of the two nonzero residue polynomials remains nonzero. This is multiplicativity of the nonarchimedean polynomial norm. At archimedean places the norms are comparable up to fixed constants, by the coefficient triangle inequality and compactness of normalized linear factors. ThereforeCombining the estimates proves the naive-height parallelogram relation with a uniform error. Apply it to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and pass to the limit to obtainAs in the polarization argument of (a), this makes $\hat{h}$ quadratic, with the canonical height pairingA torsion point has canonical height zero because its multiples have bounded naive height. Conversely, if $\hat{h}(P)=0$, then all $2^{n}P$ have bounded naive height. Northcott finiteness makes two equal, giving $(2^j-2^i)P=O$ for some $j>i$. Thus canonical height vanishes exactly on torsion. In particular the torsion subgroup is finite, since the comparison with naive height bounds the heights of all its points.

We can now prove the Mordell-Weil theorem. The Weak Mordell-Weil theorem, as explained in Question 2(ii), makes $E(K)/2E(K)$ finite. Choose representatives $R_1,\dots ,R_s$ and put $M={\max }_i\hat{h}(R_i)$. Every $P$ can be written $P=2Q+R_i$. The parallelogram identity and nonnegativity giveIterating this height descent makes the height at step $n$ at most $M+2^{-n}\max (\hat{h}(P)-M,0)$, so after finitely many steps it is at most $M+1$. There are only finitely many points of that height by Northcott and the naive-height comparison. Reversing $P=2Q+R_i$ shows that every point is an integer combination of the finitely many representatives and this finite bounded-height set. ThereforeOn the resulting free part the canonical height pairing is positive definite. Its nonnegativity extends from rational vectors to real vectors by density. If a nonzero real vector $v$ were in its kernel, in a basis of the free part round $nv$ to integer vectors $z_n$. Their rounding errors are uniformly bounded, and $\hat{h}(z_n)=\hat{h}(z_n-nv)$ would be uniformly bounded. The vectors $z_n$ are unbounded and therefore include infinitely many distinct vectors, contradicting Northcott finiteness. Thus there is no real null vector. Heights consequently support quantitative point searches, tests of independence, and the regulator of an elliptic curve, in addition to providing the descent that proves finite generation.

We prove the primitive three-cycle criterion through supports. First, suppose a subgroup contains $A_U$, the alternating group on a set $U$ of at least three letters, and a three-cycle with support $E$ meeting $U$.

If $E$ adds one letter, it has two letters in $U$. The group $A_U$ is transitive on unordered pairs of $U$: for three letters its cyclic action permutes the three pairs, and for at least four letters one can adjust the parity of a permutation sending one pair to another while preserving its target pair. Conjugating the new cycle, and using inverses, supplies every three-cycle on two old letters and the new letter. Together with $A_U$, these generate $A_{U\cup E}$.

If $E$ adds two letters, write its cycle as $\tau =(abc)$ with $a\in U$. Choose $a^{\prime }\in U\setminus \{a\}$ and conjugate within $A_U$ to obtain ${\tau }^{\prime }=(a^{\prime }bc)$. ThenThe preceding case first adds $b$; applying it again to $\tau$ adds $c$. This proves connected triple supports generate an alternating group, by successively merging the intersecting triples of a connected support hypergraph. We used that three-cycles generate alternating groups: pairs of transpositions with a common letter are three-cycles, while a pair of disjoint transpositions is a product of two three-cycles.

Now take all conjugates in $G$ of the given three-cycle. Their support hypergraph is $G$-invariant, and its connected components form a block system. Transitivity makes their union the whole set. There is an edge, so a component has more than one vertex. A primitive group action forces a single component. The preceding merging argument proves

For the first family of counterexamples, letIts elements are Möbius transformations $x\mapsto (ax+b)/(cx+d)$, modulo nonzero scalar matrices. Any ordered triple of distinct projective points is the image of $(\infty ,0,1)$ under exactly one such transformation: representatives of the first two target lines form a basis, and scaling the two columns sends their sum to the third target line. Thus the action is sharply three-transitive, hence two-transitive and primitive. Its degree is $p+1$ and order $p(p^2-1)$. It is divisible by $p$, butso it cannot contain $A_{p+1}$.

For the second example takeThe same projective-line construction over ${\mathbb{F}}_8$ is sharply three-transitive. Its order is divisible by 7 and is less than $|A_9|=181440$, so it does not contain $A_9$. For a concrete field one can use ${\mathbb{F}}_8={\mathbb{F}}_2[u]/(u^3+u+1)$.

Replace the determinant condition by preservation of a nondegenerate alternating bilinear form $B$ on $F^{2m}$. In a symplectic basis the symplectic group over a finite field isTo count symplectic bases, choose the first vector $e$ in $q^{2m}-1$ ways, then choose $f$ with $B(e,f)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate; its symplectic orthogonal complement has dimension $2m-2$ and supplies the rest of the basis. The resulting recurrence gives the finite symplectic group order

The elementary replacements are symplectic transvectionsExpanding $B(Tx,Ty)$ shows that the two cross terms cancel and the last term vanishes because $B(v,v)=0$. Thus these maps preserve $B$; their inverses are $T_{v,-c}$.

We need both generation and primitivity to use the argument from part (i). For nonzero vectors $v,w$ with $B(v,w)\ne 0$, the symplectic transvection with direction $w-v$ and parameter $1/B(v,w)$ sends $v$ to $w$. When $B(v,w)=0$, choose $z$ nonorthogonal to both and use two such steps. Such $z$ exists because two proper hyperplanes cannot cover a vector space: if the two linear forms are independent prescribe both values to be one, and if dependent prescribe either a nonzero value. Hence symplectic transvections act transitively on nonzero vectors.

They generate the whole symplectic group over a finite field. Indeed first use them to make a given symplectic transformation fix $e_1$. Its image $f^{\prime }$ of $f_1$ still satisfies $B(e_1,f^{\prime })=1$. A symplectic transvection with direction $f_1-f^{\prime }$ fixes $e_1$ and sends $f^{\prime }$ to $f_1$ if the pairing is nonzero. If that pairing is zero, pass through $f^{\prime }+c{e}_1$ for any $c\ne 0$: it has nonzero pairing with both partners, and both required directions lie in $e_1^{\perp }$. We have now fixed the first hyperbolic pair, and induction applies in its nondegenerate orthogonal complement. The final two-dimensional case is exactly the elementary generation of $S{L}_2(q)=S{p}_2(q)$ proved in part (i).

For a fixed projective point $⟨e⟩$, the symplectic transvections $T_{e,c}$ form an abelian subgroup normal in its stabilizer subgroup; conjugation scales $c$ by the square of the scalar multiplying $e$. Their conjugates supply all symplectic transvections. A central element commutes with every symplectic transvection, so it preserves every direction line and must be scalar. Conversely scalars commute with the whole group. The kernel of the projective action consists of scalar matrices, and preservation of $B$ restricts the scalar to ${\lambda }^2=1$. Thus the center of a group has size $\gcd (2,q-1)$ and the projective symplectic group over a finite field acts faithfully.

For $m\ge 2$ the stabilizer subgroup has precisely three orbits: its fixed point, the other points in $e^{\perp }$, and the points outside $e^{\perp }$. Transitivity on each follows by extending, respectively, an independent isotropic pair or a hyperbolic pair to a symplectic basis. WriteA nontrivial proper block through the fixed point would be a union of these stabilizer subgroup orbits, hence would have size either $h$ or $q^{2m-1}+1$. A block size divides $N$. But $N-qh=1$, so $h>1$ cannot divide $N$, and $N/2<q^{2m-1}+1<N$, excluding the other size. This proves primitivity. For $m=1$ the projective action is the two-transitive action of $PS{L}_2(q)$.

Here is a direct check of perfectness of finite symplectic groups, including the small fields where the diagonal group commutator needs replacement. If $q>3$, every symplectic transvection lies in an embedded $S{L}_2(q)$ on a hyperbolic plane containing its direction, and that subgroup is perfect by part (i). Since the symplectic transvections generate, the whole symplectic group over a finite field is perfect.

For $q=3$ and $m\ge 2$, all $T_{v,1}$ are conjugate, so their common image $z$ generates the abelianization and satisfies $3z=0$. Choose independent perpendicular vectors $e,f$. The four commuting symplectic transvections with directions $e,f,e+f,e-f$ and parameter one multiply to the identity: the sum of their rank-one matrices is zero in characteristic 3. Thus also $4z=0$, giving $z=0$.

For $q=2$ and $m\ge 3$, use the seven nonzero vectors of a three-dimensional isotropic subspace of a symplectic vector space. Their symplectic transvections commute, and their product is the identity. To see this explicitly, write a vector in three isotropic basis coordinates: in the sum of the outer products $v{v}^{\mathsf{T}}$ each diagonal coefficient occurs four times and each off-diagonal coefficient twice, all zero in characteristic 2. The common abelianization image $z$ therefore satisfies $2z=7z=0$. Again the group is perfect.

The simplicity argument from part (i) now applies to the projective symplectic group over a finite field. The remaining exceptional group besides $S{p}_2(2)$ and $PS{p}_2(3)$ is $S{p}_4(2)$. It is not simple: it is isomorphic to $S_6$. One can verify this identification without assuming it. On $F_2^4$ setThese are all 16 quadratic forms with polar form $B$; the Arf invariant of a quadratic form gives $\mathrm{Arf}(Q_a)=Q_0(a)$. There are six with invariant one. The symplectic group over a finite field permutes these six forms. Differences of their parameter vectors span $F_2^4$: writing $e_i,f_i$ for the symplectic basis, the six parameters areTheir pairwise differences include each of $e_1,e_2,f_1,f_2$. An element fixing all six forms therefore fixes all linear forms $B(a-b,-)$, hence is the identity. This gives a faithful action on six objects; the group order is 720, so its image is all of $S_6$. Consequently

Let $F={\mathbb{F}}_q$. The general linear group over a finite field consists of all invertible linear maps $F^n\to F^n$, and the special linear group over a finite field is its determinant-one subgroup. Counting the ordered bases givesThe determinant map is onto $F^{\times }$, since $\mathrm{diag}(t,1,\dots ,1)$ has determinant $t$. Its kernel is $S{L}_n(q)$, so

For $n>1$, commuting with all the elementary transvection matrices $I+t{E}_{ij}$ forces a matrix to be scalar. Therefore the center of a group of $S{L}_n(q)$ consists of $\lambda I$ with ${\lambda }^n=1$. The multiplicative group of a finite field is cyclic, so this center of a group has size $d=\gcd (n,q-1)$. The projective special linear group over a finite field consequently has order

Here is a proof of simplicity which isolates all the required ingredients. First we prove the Iwasawa simplicity lemma in the form needed. Suppose a group $H$ acts faithfully and primitively, is a perfect group, and has an abelian subgroup $U$ normal in a stabilizer subgroup $H_{\alpha }$, whose conjugates generate $H$. For a nontrivial normal subgroup $N$, its orbits form a block system. Faithfulness prevents all these orbits from being singletons, so primitivity makes $N$ transitive. Thus $H=N{H}_{\alpha }$. Modulo $N$, each conjugate of $U$ has exactly the same image as $U$, because we may write its conjugating element as $nh$ with $n\in N$ and $h\in H_{\alpha }$. The quotient $H/N$ is generated by this one abelian image and hence is abelian. Since $H$ is a perfect group, $H/N=1$. Thus every nontrivial normal subgroup is $H$.

Apply this to the action on projective points, that is, one-dimensional vector subspaces of $F^n$. Any ordered pair of distinct lines can be sent to any other by a linear map obtained by extending representatives to bases. Multiplying one target basis vector by a suitable nonzero scalar adjusts the determinant to one without changing either target line. Thus $S{L}_n(q)$ is two-transitive on these points. A linear map fixing every line is scalar: it is diagonal on a fixed basis, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. The projective quotient therefore acts faithfully, two-transitively, and primitively.

For the line $⟨e_1⟩$, useThese matrices multiply by addition of $f$, so $U$ is abelian. Conjugating by a matrix preserving $⟨e_1⟩$ preserves $U$, so it is normal in that stabilizer subgroup. Its conjugates include every elementary transvection matrix $I+t{E}_{ij}$.

These elementary transvection matrices generate $S{L}_n(q)$ by row elimination. For completeness, row addition is elementary, signed row interchange is elementary, and compensating diagonal scalings are elementary because, in a two-dimensional coordinate block,Elimination using these operations reduces a determinant-one matrix to the identity.

It remains to show that these groups are perfect groups. In the following matrix identities use $[x,y]=xy{x}^{-1}{y}^{-1}$; the alternative group commutator convention generates the same commutator subgroup. If $n\ge 3$, distinct indices give the group commutator identityEvery elementary generator is therefore a group commutator. For $n=2$ and $q>3$, choose $\lambda \in F^{\times }$ with ${\lambda }^2\ne 1$, and let $D=\mathrm{diag}(\lambda ,{\lambda }^{-1})$. ThenAs $t$ varies this supplies all upper elementary transvection matrices, and conjugation supplies the lower ones. Thus $S{L}_n(q)$, and its projective quotient, are perfect in the asserted cases. The preceding simplicity argument now applies.

The two exceptions are genuine: the projective actions identify $PS{L}_2(2)$ with $S_3$ and $PS{L}_2(3)$ with $A_4$, of orders 6 and 12 respectively. The first contains its normal subgroup of order 3; the second contains the normal Klein four group. We have proved

Let $R$ be a discrete valuation ring with uniformizer $\pi$, fraction field $K$, and normalized discrete valuation $\nu :K^{\times }\to \mathbb{Z}$. A map $\mathrm{Spec}K\to {\mathbb{P}}_{\mathbb{Z}}^n$ is a projective pointwith at least one nonzero coordinate. Put $m={\min }_{a_i\ne 0}\nu (a_i)$ and set $b_i={\pi }^{-m}{a}_i$. Then every $b_i$ lies in $R$, and at least one $b_j$ is a unit.

On the standard affine chart $U_j$ of projective space, the ratios $b_i/b_j$ all lie in $R$. They therefore define a map $\mathrm{Spec}R\to U_j\subseteq {\mathbb{P}}_{\mathbb{Z}}^n$ whose generic restriction is the original point. This proves existence in the DVR case. The assumed separatedness of ${\mathbb{P}}_{\mathbb{Z}}^n\to \mathrm{Spec}\mathbb{Z}$, through the valuative criterion for separatedness, gives uniqueness. Thus the morphism satisfies the requested DVR form of the valuative criterion for properness.

The required general linear position condition is that every $n+1$ of the $n+2$ representing vectors are linearly independent. Such an ordered configuration is a projective frame.

Choose representatives $p_0,\dots ,p_n$ of the first $n+1$ projective points; they form a basis. Write a representative of the last point as $p_{n+1}={\sum }_i{a}_i{p}_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to ${\tilde{p}}_i=a_i{p}_i$, so that $p_{n+1}={\sum }_i{\tilde{p}}_i$.

Do the same for the target configuration, obtaining representatives ${\tilde{q}}_i$ with $q_{n+1}={\sum }_i{\tilde{q}}_i$. The unique linear map sending the basis ${\tilde{p}}_i$ to ${\tilde{q}}_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced projective transformation sends every point to its prescribed partner. Thus any two ordered projective frames are projectively equivalent.

A projective quadric is the zero locus of a nonzero homogeneous quadratic polynomial $q(v)=v^{T}Bv$, with $B$ symmetric over a field of characteristic different from two. A projective point on it is singular when all first partial derivatives vanish. Since $\nabla q(v)=2Bv$, the quadric is nonsingular exactly when $B$ is invertible.

Over the complex numbers, a nonzero quadratic form in three variables can be diagonalized by an invertible change of basis. Its rank classifies a projective conic:

The zero polynomial is excluded from the definition of a conic. Thus these last two cases are precisely the singular conics.

Use part (a) to put the four projective points atA quadratic polynomial vanishing at the first three has no square terms, and vanishing at the fourth givesThese form a two-dimensional vector space. Choose the two membersTheir coefficient triples are independent and both satisfy the relation. Their symmetric bilinear form matrices are of the formFor $(A,B,C)=(1,1,-2)$ and $(1,-2,1)$, both determinants are $-1/2$, so both selected conics are nonsingular. Any coefficient triple with zero sum is uniquely a linear combination of these two. Hence every conic through the four points has matrixUndoing the projective transformation carries the entire family back to the original four points, preserving nonsingularity. This is a pencil of conics through four points.

Use the cross-ratio convention compatible with the permutation identities in this paper. Represent a projective point $P_i\in {\mathbb{CP}}^1$ by a nonzero vector $p_i\in {\mathbb{C}}^2$, and write $[p_i,p_j]=\det (p_i,p_j)$. DefineThe denominator is nonzero because the four projective points are distinct. Rescaling any one representative multiplies numerator and denominator by the same nonzero scalar, so the definition is independent of the chosen homogeneous coordinates. In finite affine coordinates it readswith the point at infinity of the complex projective line handled by the homogeneous formula.

A projective transformation is represented by an invertible matrix $A\in {\mathrm{GL}}_2(\mathbb{C})$. Since $[A{p}_i,A{p}_j]=\det (A)[p_i,p_j]$, the two factors of $\det (A)$ in the numerator cancel the two in the denominator. Therefore every projective transformation preserves this cross-ratio. This also shows that the formula descends from matrices to the projective linear group, whose representatives differ by a nonzero common scalar.