COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA. COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
386 Hans Zassenhaus For this purpose we must assign to each polynomial U of E[Xj a sign function SIGN (U, Z, P) assuming one of the 3 values 1, 0, - 1 such that the operational rules mu m + ww, PI) = ww, P>> (13) wu P))VK 9) = WqI, PI> (14) if tJ(xj+ V(X) = W(X), U(X)V(X) = T(X), define an ordered extension E(R(Z, P)) consisting of all symbols U(R(Z, P)), according to the positivity rule Utw, m > 0 (15) if and only if SIGN (U(R(Z, P)))= 1, and the equality definition Utw, PI> = ww, p>> (16) if and only if SIGN ((U- V) (R(Z, P))) = 0 when (U- V)(X) = U(X) - V(X); also the conditions (ll), (12) must be fulfilled. As usual the expression Z(R(Z, P)) is identified with R(Z, P) when Z(X) = X. By definition the complexity of the ordered extension E(R(Z, P)) is 1 more than the complexity of E. On the degree level 1 the construction with the desired properties is simple enough. The algebraically ordered extensions of Pto be considered are Ffor each complexity.t If P is a constant polynomial over F, then NR(P) = 0. If P is the linear polynomial AXfB of F[X], and U is any polynomial of E[fl, then we have the defining equation SIGN (U, 1, P) = sign (U(-B/A)), (17) and the symbol U(R( 1, P)) is canonically identified with U( - B/A)). Theorems 1-6 will be verified readily in case the degree of P is not greater than 1. We assume now that D=-1, that all constructions on the degree level D- 1 of any prescribed complexity can be performed as specified above, and that Theorems l-6 are demonstrated for polynomials P of degree smaller than D and for any field (in place of F) that can be constructed on the ievel D - 1. We begin with a proof of Rolle’s theorem for polynomials of degree D. The assumption of Theorem 6, viz. leads to a factorization P(A) = 0 = P(B), P(X) = (X- A)L(X- B)MQ(X) t As A. Hollkott stresses correctly, in reality we do get new ordered fields even here in as much as the collection of symbols to be considered expands with increasing complexity. But in our case a canonical order-preserving isomorphism with Fis set up at each stage. A real root calculus with positive exponents L, M such that Q(A) + 0, Q(B) $I 0, and certainly the degree of Q is less than D - 1. By the induction assumption there is an ordered extension of F generated by a root Z? of P satisfying A-=B-= B such that NR (Q, A, Z?) = 1. It su5ces then to prove Rolle’s theorem under the additional assumption that there is no root of Q between A and B. By the between value theorem Q(A)Q(B)=-0. Upon differentiation we have P’(S) = (X-A)L-l(X- B)“+(X), &(X) = (L(X-B)+M(X--4))Q(X)+(X-A)(X-B)&’(X), $(A) = W -B>QW, &@I = WB- A)&(B), &(A)&(B) = --LMtA - Bj”QtA)Q@), &4$(B) -c 0. By the between value theorem applied to Q(X) there is an ordered extension of F generated by a root of Q(X) between A and B. This root also is a root of P’(X) between A, B. The mean value theorem follows in the customary way by application of Rolle’s theorem to the polynomial P(X)-P(B)(P(X)--P(A))I(B--A)-P(A)tP(X)--P(B))I(A -B>. We proceed to the proof of the between value theorem for a polynomial P of degree D. For convenience sake let A-= B. If at any stage of the ensuing construction we should meet an element R in an ordered extension E of F that was obtained on the D- 1 level such that A-= R-= B, P(R)= 0, then the elements U(R) (U cF[Xj) with the operational rules as defined in E provide the required collection of symbols forming an ordered extension of F with a root of P between A and B. It will be assumed in the ensuing construction that this will not happen. For example, if it should happen that there is a non-trivial factorization P(X) = M(X)L(X) in E[X] such that both M and L are non-constant, then either M(A)M(B) -C 0 or L(A)L(B) < 0 so that either M or L will have a root in an ordered extension of E on the D- 1 level. Henceforth we assume that we will not meet non-trivial factorizations of P in E[X& This implies that P is separable, because P/GCD (P, P’) cannot be a proper divisor of P. Now let E be an ordered extension of F generated by NR(P’, A, B) distinct roots R(J), . . . ,R(K) of P’ belonging to [A, B], according to Theorems 2, 3. Let A = A(0) cf A(1) -== . . . -=c A(S) = B the set formed by the NR (P’, A, B) roots of P’ belonging to [A, B) and 387
388 Hans Zassenhaus the elements A, B in order of magnitude. There is a first index J such that P(A(J))P(A(J+ 1)) -Z 0. By definition (and by Theorem 4) there is no root of P’ between A(J) and A(J+ 1) either in E or in any ordered extension. It follows from the between value theorem that the sign of P’ between A(J) and A(J+ 1) is constant + 0. It follows from the mean value theorem that P is strictly monotone in E as well as in any ordered extension of E. Hence for any chain A(J) = B(0) -== B(1) < . . . -== B(K) = A( J+ 1) (18) there is precisely one index H such that 04 HIK, P(B(H))P(B(H+l))
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388 Hans Zassenhaus<br />
the elements A, B in order of magnitude. There is a first index J such that<br />
P(A(J))P(A(J+ 1)) -Z 0.<br />
By definition (and by Theorem 4) there is no root of P’ between A(J)<br />
and A(J+ 1) either in E or in any ordered extension.<br />
It follows from the between value theorem that the sign of P’ between<br />
A(J) and A(J+ 1) is constant + 0.<br />
It follows from the mean value theorem that P is strictly monotone in<br />
E as well as in any ordered extension of E. Hence for any chain<br />
A(J) = B(0) -== B(1) < . . . -== B(K) = A( J+ 1) (18)<br />
there is precisely one index H such that<br />
04 HIK, P(B(H))P(B(H+l))
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