Common Solution Sets of Real Polynomialsby Maciej Kurowski and Tomasz OsmanThis is an abridged version of the paper submitted to the VIII European Union Contest for Young Scientists in Helsinki, where it was awarded a second prize. It happens that various relations (of geometric nature) between vectors or other mathematical objects can be represented by an equation of the form W(x1,...,xn)=0, where W is a polynomial in n real variables. For instance, two straight lines are perpendicular only if their direction coefficients a and b satisfy the equation ab+1=0. Quite often we have to do with two different conditions, presented algebraically as
and it is known that one of the conditions implies the other (i.e. if W1(x1,...xn)=0, then also W2(x1,...xn)=0). It turns out then that under some additional assumptions, there is a simple relationship between the polynomials W1 and W2. Let R[x1,...,xn] (respectively, R[x1,...,x(n1)]) be the ring of real polynomials in n variables x1,...,xn (respectively, in (n1) variables x1,...,x(n1)). In the sequel we will make use of the fact that polynomials in each of these rings can be uniquely decomposed into "primes", i.e. represented as the product of prime polynomials. A polynomial is prime, if it is not the product of two polynomials of lower degree. For instance, for n=1 the polynomial is prime in R[x], whereas is the product of two prime polynomials:
Theorem. Suppose the polynomial W1 is prime in R[x1,...,xn] and assumes both positive and negative values. If W2 is a polynomial in R[x1,...,xn] such that for all (x1,...,xn),
then there is a polynomial G in R[x1,...,xn] such that . For onevariable polynomials the theorem follows easily from Bezout's theorem. In the proof of the theorem we will need the following Lemma 1. If the polynomials W1, W2 in R[x1,...,xn] are relatively prime (i.e. have no common nonconstant divisor), then there exist polynomials T1, T2 in R[x1,...,xn] and S in R[x1,...,x(n1)] such that for all (x1,...,xn) the following holds:
and . Note: If you follow carefully the sketch of the proof below, you may notice that the assumptions of the lemma may be made slightly weaker. In fact, the following is true: Corollary 1. The claim of Lemma 1 remains valid, when the greatest common divisor of the polynomials W1 and W2 does not depend on xn. Sketch of the proof. Consider the ring R(x1,...,x(n1))[xn] of polynomials in the variable xn with rational functions of x1,...,x(n1) as coefficients. The reader may wish to proof by himself that a polynomial Q in R[x1,...,xn] is prime in this ring if and only if it is prime in R(x1,...,x(n1))[xn]. (This can be proved in almost the same way as the well known fact that any onevariable polynomial with integer coefficients prime in Z[x] is prime in Q[x] too.) This implies that the polynomials W1 and W2 are relatively prime in R(x1,...,x(n1))[xn]. We may assume without loss of generality that the degree of W1 (with respect to the variable xn) is not less than the degree of W2 and apply Euclid's algorithm to find the greatest common divisor. We get a sequence of equations: W1=Q1W2 + W3, W2=Q2W3 + W4, ... , W(k2)=Q(k2)W(k1) + Wk, where all the polynomials Wj and Qj are in the ring R(x1,...,x(n1))[xn] and
Moreover, necessarily , for otherwise W(k1) of positive degree would be a common divisor of the relatively prime polynomials W1 and W2. Thus . We are now quite near the claim of the lemma. The equations Wj = QjW(j+1) + W(j+2) (satisfied for j=1,..., k2) imply
where Pi is in R(x1,...,x(n1))[xn]. Now the claim is obtained by multiplying both sides of the last equation by the least common multiple of the polynomials which are in the denominators of the coefficients of P1 and P2. Let's turn now to the proof of the theorem. Assume the claim is false. The polynomial W1 being prime, W1 and W2 are relatively prime. Corollary 1 implies that for any k in{1, 2, ..., n}, there exists a nonzero polynomial Sk which depends on n1 variables and T1(x1,...,xn)W1(x1,...,xn) + T2(x1,...,xn)W2(x1,...,xn)=Sk(x1,...,x(k1),x(k+1),...,xn). Seeking contradiction, we will prove that the solution set of the polynomial W1 is sufficiently large. More precisely, the following lemma holds: Lemma 2. For some k in {1, 2, ...., n} there exists an (n1)dimensional sphere K with centre (a1,...,a(k1), a(k+1), ..., an) and radius d with the following property: for any point (x1,...,x(k1), x(k+1), ..., xn) in K there is a real number xk such that W1(x1,...,xn)=0. The theorem follows from this lemma immediately. Indeed, since by assumption each solution of W1 is a solution of W2, the last equality implies that Sk disappears on the entire sphere K and therefore , a contradiction.
Sketch of the proof of Lemma 2.
Let A (and respectively, B) be the set of all those
points at which W1
is positive (respectively, negative) and let Z be the solution
set of W1. Clearly, both A and B are open
and .
Indeed, it is straightforward from the definition that . The image of K under the translation by the vector w=(0,...,0, akbk, 0,...,0) is contained in the sphere U(k1), which is in turn contained in the union of A and Z. Now it suffices to recall Darboux's property to see that by moving any point in K by vectors of the form tw for t in [0, 1] we must meet the set Z (see the picture below). And translation by a vector tw being nothing but an appropriate choice of the kth coordinate of a point, the proof of Lemma 2, and therefore of the entire theorem, is accomplished.
The small sphere V is contained
in the set B of points at which W1
assumes negative values, whereas the big sphere U(k1)
is contained in the set .
Therefore, the polynomial W1
being continuous on each line in ,
when moving the points of the "big circle" K
by vectors tw with t in [0, 1], for some
value of t the image is in the set Z of all points
at which W1
is zero.
