Let R = Q[m,a,F,v,E,c]. --------------------- This means: ----------- This R is the ring of all polynomials with coefficients in Q, and the variables are m,a,F,v,E,c. In this ring R there exist no relations between these variables. Here are examples of elements of R: ----------------------------------- 0, 1, 123/568, a, a^100, E, F+17*m, (a+v)/99 The following are not elements of R: ------------------------------------ sqrt(-1), sqrt(2) (because: Q is the set of all rational numbers and does not contain these two numbers. If we want to have these elements in our ring, we must replace Q by C, the complex numbers). F/m, F^3/a (because: in R, variables are not allowed in the denominator) x, y, z (because: these are not in our list of variables) An example of an ideal in R. ---------------------------- Lets take the following two elements of R: f1 := F - m*a f2 := E - 1/2 * m * v^2 and let I be the ideal: I = (f1,f2). This is an ideal in R. If we think of f1 and f2 as equations: F - m*a = 0, and E - 1/2 * m*v^2 = 0, then I is the set of all equations that we can obtain from them using: *) multiply an equation by any element of R. *) add equations. So I = { a1*f1 + a2*f2 | a1, a2 in R }. The following are examples of elements of I: f1, f2, 0, 5*f1 + a^3*E*f2, f1^3, (17 + m/5 + E^2)*f2, .. Here is a trickier example of an element of I: 2*a*E - F*v^2 (this is in (f1,f2) because it is equal to 2*a*f2 - v^2*f1) The following are examples of things that are *NOT* in I: sqrt(-1)*f1 (this is not even in R because Q does not have an element sqrt(-1) x*f1 (again this is not even in R because we have no x) E - m*c^2 (this is not in I because if all you do is adding equations and multiplying them by elements of R, then starting with f1,f2 there is no way you can obtain this equation. In other words: E-m*c^2 is not equal to a1*f1+a2*f2 for any a1,a2 in R). Any ideal I in a ring R has the following properties: ----------------------------------------------------- I1) If g1 and g2 are in I, then so is g1+g2. I2) If g is in I, and k is any element of R, then k*g is in I. In mathematics, this is in fact the definition of an ideal. Any nonempty subset of R that satisfies these two properties I1) and I2) is called an ideal. There are two very special ideals: I = {0}. This is the smallest ideal in R. and I = R. This is the largest ideal in R. Notice that I = R if and only if 1 is an element of I. Why? Well, if I=R then 1 is an element of I. Conversely, if 1 is an element of I, then every multiple of 1 is an element of I because of property I2). But any k in R equals k=k*1 so every k in R is a multiple of 1. The ring R/I. R modulo I. -------------------------- Let again I = (f1,f2) where f1,f2 as before. What is R/I = R/(f1,f2) ?? How to compute in this ring? (remark: ring means: we can add and multiply). Computing in the ring R/(f1,f2) is like computing in R, while using two relations: f1=0 and f2=0. In the original ring R, the polynomials f1+c*E-3*m and c*E-3*m are different polynomials. But if we are working in the ring R/I, then they are considered the same because then we can replace f1 and f2 by 0. So in R/I we can have things that may look different, but that really are the same. For example 2*a*E is completely different than F*v^2 in the ring R. But in R/I they are the same because their difference is 2*a*E - F*v^2 = 2*f2 - v^2*f1 = 2*0 - v^2*0 = 0 Here we used that when working in R/(f1,f2) we may always replace f1,f2 by 0. So in the ring R/I we have F*v^2 = 2*a*E. In general, if A,B are two polynomials in R, then viewed as elements of R/I they are the same if and only if their difference A-B is in I. So: ------------------------------------------------------------------ | In order to decide if two polynomials A,B are equal to each | | other when viewed as elements of R/I, we have to be able to | | decide if A-B is in I or not. | ------------------------------------------------------------------ For example, in the ring R/I we have that E is not equal to m*c^2 because E-m*c^2 is not in I. Recall that this was so because I=(f1,f2) and using the operations I1) and I2) above, we can not obtain the equation E-m*c^2 from the equations f1, f2. Solutions of the equations: --------------------------- Let R be a ring of polynomials, say R = Q[x1,..,xn] and I be an ideal. Then the solutions of the equations in I are denoted by: V(I) = { (c1,..,cn) | c1..cn are complex numbers, such that subs(x1=c1, x2=c2, .., xn=cn, f) = 0 for all f in I}. For example, if R = Q[x1,x2,x3] and I = (x1+x2+2, x1-x2) then V(I) = { (-1, -1, c3) | c3 is any complex number }. If R = Q[x] and I = (x^3*(x^2+1)*(x-4), x*(x^2+1)*(x-7) ) then (use Euclidean Algorithm) I = (x*(x^2+1)) and so V(I) = { 0, sqrt(-1), -sqrt(-1) }. If R = Q[x,y] and I = (x+5*y, x^3*y+y^2, x-3*y, y^3*x+3*x) then V(I) = { (0,0) }. Notice that x,y are both in I because we can write x and y as combinations of x+5*y and x-3*y. In this example the equations in I can be simplified and we find I = (x,y). If R = Q[x,y,z] and I = (x-1, y-3, x+y+6) then there are no solutions. You can not find any complex numbers for which all three equations would simultaneously hold. So V(I) = {}, the empty set. What happens if the equations contradict themselves? ---------------------------------------------------- Suppose R = Q[x,y,z] and that I = (x-1, y-3, x+y+6). We say that the equations contradict themselves when V(I) is empty. The simplest possible set of equations that contradict themselves is this set: {1}. You see, no matter what complex number values you choose for x,y,z, the equation 1=0 is not going to hold. Thus, if I = (1) then the equations in I contradict themselves. But we have seen before that the ideal (1) is the entire ring R because (1) = {f*1 | f in R} = {f | f in R} = R. So, if I = (1) then I = R, and then V(I) = {}. Note: If I = (1), then in R/I we have 1 = 0. Now if we take any element a in R/I, then a*1 = a*0 = 0. So if I = (1) then every element of R/I is 0. So if the ideal I is as large as possible (I = the entire ring R) then R/I is as small as possible (has one element, namely 0). Hilbert's Nullstellensatz: -------------------------- This mathematical theorem says that if V(I) = {} then I=R. So whenever we have contradictory equations, then the ideal contains the element 1 (and is therefore equal to R). Lets check this in an example: I = (x-1, y-3, x+y+6) has no solutions. Is 1 then really an element of I? Remember that I = {a1*(x-1) + a2*(y-3) + a3*(x+y+6) | a1,a2,a3 in R}. So our question is: Do there really exist a1,a2,a3 in R such that: 1 = a1*(x-1) + a2*(y-3) + a3*(x+y+6) ? The answer is yes, and since all given equations are linear we can find such a1,a2,a3 with Gaussian elimination, and we get: a1 = -1/10, a2 = -1/10, a3 = 1/10, so yes, these a1,a2,a3 indeed exist, and Hilbert is right. How to test if a given set of equations contradict themselves? -------------------------------------------------------------- If I = (f1,..,fm) then the equations contradict if 1 is in I. So the equations f1=0,..,fm=0 have solution(s) if and only if 1 is not an element of I. If we have an algorithm that can test that, then we are able to decide if the system f1=0,..,fm=0 has solutions or not.