Page:EB1911 - Volume 01.djvu/675

 linear factor which occurs to the second power in $$f$$. If, moreover, $$\Delta$$ vanishes identically $$f$$ is a perfect cube.

The Binary Quartic.&mdash;The fundamental system consists of five forms $$a_x^4=f$$; $$(f,f')^2=(ab)^2a_x^2b_x^2=\Delta_x^4$$; $$(f,f')^4=(ab)^4=i$$; $$(f,\Delta)^1=(a\Delta)a_x^3\Delta_x^3=(ab)^2(cb)a_x^2b_xc_x^3=t$$; $$(f,\Delta)^4=(a\Delta)^4=(ab)^2(bc)^2(ca)^2=j$$, viz. two invariants, two quartics and a sextic. They are connected by the relation

The discriminant, whose vanishing is the condition that f may possess two equal roots, has the expression j 2 - 6 i 3; it is nine times the discriminant of the cubic resolvent k 3 - 2 ik- 3j, and has also the expression 4(1, t') 6. The quartic has four equal roots, that is to say, is a perfect fourth power, when the Hessian vanishes identically; and conversely. This can be verified by equating to zero the five coefficients of the Hessian (ab) 2 axb2. Gordan has also shown that the vanishing of the Hessian of the binary n ic is the necessary and sufficient condition to ensure the form being a perfect n th power. The vanishing of the invariants i and j is the necessary and sufficient condition to ensure the quartic having three equal roots. On the one hand, assuming the quartic to have the form 4xix 2, we find i=j=o, and on the other hand, assuming i=j=o, we find that the quartic must have the form a o xi+4a 1 xix 2 which proves the proposition. The quartic will have two pairs of equal roots, that is, will be a perfect square, if it and its Hessian merely differ by a numerical factor. For it is easy to establish] the formula (yx) 2 0 4 = 2f.4-2(f y 1 ) 2 connecting the Hessian with the quartic and its first and second polars; now a, a root of f, is also a root of Ox, and con se uentl the first polar 1 of of q y p f? =y la xl -i-y2a x2 must also vanish for the root a, and thence ax, and a must also vanish for the same root; which proves that a is a double root of f, and f therefore a perfect square. When f = 6xix2 it will be found that 0 = -f. The simplest form to which the quartic is in general reducible is +6mxix2+x2, involving one parameter m; then Ox = 2m (xi +x2) +2 (1-3m2) x2 ix2; i = 2 (t +3m2) ;j= '6m (1 - m) 2; t= (1 - 9m 2) (xi - x2) (x21 + x2) x i x 2. The .sextic covariant t is seen to be factorizable into three quadratic factors 4 = x 1 x 2, =x 2 1 - 1 - 2 2, 4) - x, which are such that the three mutual second transvectants vanish identically; they are for this reason termed conjugate quadratic factors. It is on a consideration of these factors of t that Cayley bases his solution of the quartic equation. For, since -2t 2 =0 3 -21f 2 ,6,-3j(-f) 3, he compares the right-hand side with cubic resolvent k 3 -21X 2 k - j 2. of f=0, :and notices that they become identical on substituting 0 for k, and -f for X; hence, if k1, k2, k 3 be the roots of the resolvent -21 2 = (o + k if) (A + k 2f)(o + k 3f); and now, if all the roots of f be different, so also are those of the resolvent, since the latter, and f, have practically the same discriminant; consequently each of the three factors, of -21 2, must be perfect squares and taking the square root 1 t = -' (1)ï¿½x4; and it can be shown that 0, x, 1P are the three conjugate quadratic factors of t above mentioned. We have A +k 1 f =0 2, O+k 2 f = x2, O+k3f =4) 2, and Cayley shows that a root of the quartic can be xpressed in the determinant form 1, k, 0.1y the remaining roots being obtained by varying 1, k, x the signs which occur in the radicals 2 u The transformation to the normal form reduces 1, k 3 ,? the quartic to a quadratic. The new variables y1= 0 are the linear factors of 0. If 4) = rx.sx, the Y2 =1 normal form of a:, can be shown to be given by (rs) 4 .a x 4 = (ar) 4s: 6 (ar) 2 (as) 2rxsy -I- (as) 4rx; 4) is any one of the conjugate quadratic factors of t, so that, in determining rx, sx from J z+k 1 f =o, k 1 is any root of the resolvent. The transformation to the normal form, by the solution of a cubic and a quadratic, therefore, supplies a solution of the quartic. If (Xï¿½) is the modulus of the transformation by which a2 is reduced to 3 the normal form, i becomes (X /2) 4 i, and j, (Ap) 3 j; hence ? 3 is absolutely unaltered by transformation, and is termed the absolute invariant. Since therefore ? 2 - 9 m 2 (1 3 m 2)) 2 we have a cubic equation for determining m 2 as a function of the absolute invariant.

Remark.&mdash;Hermite has shown (Crelle, Bd. lii.) that the substitution, $$z=\frac{i}{j}\frac{\Delta}{f}$$, reduces $$\frac{x_2\partial x_1-x_1\partial x_2}{\sqrt{j}}$$ to the form

The Binary Quintic.&mdash;The complete system consists of 23 forms, of which the simplest are f =a:; the Hessian H = (f, f') 2 = (ab) 2axbz; the quadratic covariant i= (f, f) 4 = (ab) 4axbx; and the nonic co variant T = (f, (f', f") 2) 1 = (f, H) 1 = (aH) azHi = (ab) 2 (ca) axbycy; the remaining 19 are expressible as transvectants of compounds of these four.

There are four invariants (i, i')2; (13, H)6; (f2, 151c.; (f t, 17)14 four linear forms (f, i 2) 4; (f, i 3) 5; (i 4, T) 8; ( 2 5, T)9 three quadratic forms i; (H, i 2)4; (H, 23)5 three cubic forms (f, i)2; (f, i 2) 3; (13, T)6 two quartic forms (H, i) 2; (H, 12)3. three quintic forms f; (f, i) 1; (i 2, T)4 two sextic forms H; (H, 1)1 one septic form (i, T)2 one nonic form T.

We will write the cubic covariant (f, i) 2 =j, and then remark that the result, (f,j) 3 = o, can be readily established. The form j is completely defined by the relation (f,j) 3 =o as no other covariant possesses this property.

Certain convariants of the quintic involve the same determinant factors as appeared in the system of the quartic; these are f, H, i, T and j, and are of special importance. Further, it is convenient to have before us two other quadratic covariants, viz. T = (j, j) 2 jxjx; 0 = (iT)i x r x; four other linear covariants, viz. a = - (ji) 2 jx; s = (ia)ix; Y = (ra)r x: (3= (T0)T x . Further, in the case of invariants, we write A= (1, i') 2 and take three new forms B = (i, T) 2; C = (r, r`) 2; R = (/y). Hermite expresses the quintic in a forme-type in which the constants are invariants and the variables linear covariants. If a, a be the linear forms, above defined, he raises the identity ax(0) =ax(aJ3) - (3x(aa) to the fifth power (and in general to the power n) obtaining (aa) 5 f = (a13) 5 az - 5 (a0) 4 (aa) ax?3 -F... - (aa) and then expresses the coefficients, on the right, in terms of the fundamental invariants. On this principle the covariant j is expressible in the form R 2 j =5 3 + BS 2 a+4ACSa 2 + C(3AB -4C)a3 when S, a are the above defined linear forms.

Hence, solving the cubic, R 2 j = (S -m i a) (S - m 2 a) (S - m3a) wherein m 1 m2, m 3 are invariants.

Sylvester showed that the quintic might, in general, be expressed as the sum of three fifth powers, viz. in the canonical form

f=k1(px)5 +k2(gx) 5 +k3(rx) 5.

Now, evidently, the third transvectant of f, expressed in this form, with the cubic pxgxrx is zero, and hence from a property of the covariant j we must have j = pxgxrx; showing that the linear forms involved are the linear factors of j. We may therefore write I. / f = k1.(S-mia)5+k2(S-m2a)5+k3(6-m3a)5; and we have merely to determine the constants k1, k2, k3. To determine them notice that R = (a6) and then (f, a 5) 5 = - R 5 (k1 +k2+k3) (f, a 4 5) 5 = - 5R5 ( m 1 k 1+ m 2 k 2+ m 3 k 3), (f, a352) 5 = -10R5 (m21ke +m2k2+m3k3) three equations for determining k 1, k2, k3. This canonical form depends upon j having three unequal linear factors. When C vanishes j has the form j = pxg x, and (f,j) 3 = (ap) 2 (aq)ax = o. Hence, from the identity ax (pq) = px (aq) -qx (ap), we obtain (pet' = (aq) 5px - 5 (ap) (aq) 4 pxg x - (ap) 5 gi, the required canonical form. Now, when C = o, clearly (see ante) R 2 j = 6 2 p where p = S +2 B a; and Gordan then proves the relation 6R 4 .f = B65ï¿½5B64p - 4A2p5, which is Bring's form of quintic at which we can always arrive, by linear transformation, whenever the invariant C vanishes. Remark.-The invariant C is a numerical multiple of the resultant of the covariants i and j, and if C = o, p is the common factor of i and j. The discriminant is the resultant of ax and ax and of degree 8 in the coefficients; since it is a rational and integral function of the fundamental invariants it is expressible as a linear function of A 2 and B; it is independent of C, and is therefore unaltered when C vanishes; we may therefore take f in the canonical form

6R 4 f = BS5+5BS4p-4A2p5.