I need HELP!!! I don\'t know how to prove this !! 12.4 Prove Bezout\'s Theorem \

Question

I need HELP!!! I don't know how to prove this !! 12.4 Prove Bezout's Theorem ' For a field K,let a(x) and b(x) be polynomials in K[x] and let d(x) be the gcd ofa(x) and b(x) produced by the Euclidean algorithm. There existspolynomials r(x) and s(x) in K[x] such that d(x) = a(x)r(x)+ b(x)s(x).
    This is most easily done by induction onthe number of steps in the Euclidean Algorithm for a(x) and b(x).Ifthe algorithm terminates right away, which means a(x) divides b(x),the result is easily proved. Next assume that you know the resultfor pairs of polynomials for which the algorithm terminates after ksteps and deduce the result for pairs of polynomials for which thealgorithm terminates after k+ 1 steps. I need HELP!!! I don't know how to prove this !! 12.4 Prove Bezout's Theorem ' For a field K,let a(x) and b(x) be polynomials in K[x] and let d(x) be the gcd ofa(x) and b(x) produced by the Euclidean algorithm. There existspolynomials r(x) and s(x) in K[x] such that d(x) = a(x)r(x)+ b(x)s(x).
    This is most easily done by induction onthe number of steps in the Euclidean Algorithm for a(x) and b(x).Ifthe algorithm terminates right away, which means a(x) divides b(x),the result is easily proved. Next assume that you know the resultfor pairs of polynomials for which the algorithm terminates after ksteps and deduce the result for pairs of polynomials for which thealgorithm terminates after k+ 1 steps.

Explanation / Answer

Algebraic geometry is concerned with sets of solutionsto polynomical equations. Let F be a field and Fn theset of ordered n-tuples from F. The set of polynomials in nvariables; x1, ..., xn; is denoted asF[x1, ..., xn]. A polynomial f(x1,..., xn) in F[x1, ..., xn] mapsFn into F. The zeroes of a polynomial f are the pointsof Fn that f maps into the additive identity of F, 0. Ifwe have a mapping g, AB then the elements of the field Awhich are mapped into the additive identity of the field B arecalled the kernel of the mapping.

A variety V is a set of points in Fnsuch that there exists a set of polynomials {f1, ...,fm} in F[x1, ..., xn] where apoint p belongs to V if and only if p is a common zero of{f1, ..., fm}; i.e., p is in the intersectionof the kernels of {f1, ..., fm}. Thus avariety is a function of a set of polynomials in F[x1,..., xn] and may be donoted as V(f1, ...,fm). The degree of a variety V is the maximum degree ofthe polynomials generating V.

The homogeneous coordinates for a point p =(x,y) in F2 is the triple (X,Y,Z) such thatx=XZ-1 and y=YZ-1. For the fields of real orcomplex numbers this would be denoted as x=X/Z andy=Y/Z.

Homogeneous coordinates of the form (X,Y,0), where X andY are not both zero, represent ponts on the "horizon" or the "lineat infinity." Each point in F2 corresponds to a linethrough the origin in F3. The set of points inF3 with the point (0,0,0) deleted is called aprojective space. A "point" in projective space is a linethrough the origin, but not including the origin.

As an illustration of an intersection "at the horizon"consider two parallel lines, L1 and L2, givenby the equations:

where a1 is not equal to a2. Inhomogeneous coordinates these become:

If Z=0, then any point such that Y=bX satisfies bothequations. Therefore any point in projective space of the form(X,bX,0) is on both L1 and L2 and is thus anintersection point of L1 and L2.

Now consider the general case for lines. LetL1 and L2, given by theequations:

In homogeneous coordinates these become:

This latter set of equations is a system of twoequations in two unknowns, X and Y, with Z as a parameter. If Z=0then the two equations have a nontrivial solution if and only ifthe determinant of the coefficient matrix is zero. Thatdeterminnant is given by

If a1=a2 then the lines arecoincident. If a1a2 then the lines areparallel and have an intersection at infinity, asindicated previously.

If Z0 then the system of equations will have asolution if and only if the determinant of the coefficient matrixis not equal to zero. Thus if b1b2then there will be a solution. Let X* and Y* be the solution forZ=1. Then the general solution will be (x*Z,Y*Z,Z). Therefor ifb1=b2 there is either a coincidence of linesor one intersection (at infinity). Ifb1b2 there is one intersection (not atinfinity).

The equations are

When (X/Z) and (Y/Z) are substituted for x and y and theresults multiplied by the appropriate power of Z the equationsbecome

For the case of Z=0 these reduce to

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