Algebraic Geometry. Only do Exercise 0.12 please. Thank you!

C={(x1,x2,x3)=(t3,t4,t5):tC}C3 We have given this curve parametrically, but it is in fact easy to see that we can describe it equally well in terms of polynomial equations: C={(x1,x2,x3):x13=x2x3,x22=x1x3,x32=x12x2}. What is striking here is that we have three equations, although we would expect that a 1-dimensional object in 3-dimensional space should be given by two equations. But in fact, if you leave out any of the above three equations, you are changing the set that it describes: if you leave out e. g. the last equation x32=x12x2, you would get the whole x3-axis {(x1,x2,x3):x1=x2=0} as additional points that do satisfy the first two equations, but not the last one. So we see another important difference to linear algebra: it is not clear whether a given object of codimension d can be given by d equations - in any case we have just seen that it is in general not possible to choose d defining equations from a given set of such equations. Even worse, for a given set of equations it is in general a difficult task to figure out what dimension their solution has. There do exist algorithms to find this out for any given set of polynomials, but they are so complicated that you will in general want to use a computer program to do that for you. This is a simple example of an application of computer algebra to algebraic geometry. Exercise 0.12. Show that if you replace the three equations defining the curve C in Example 0.11 by x13=x2x3,x22=x1x3,x32=x12x2+ for a (small) non-zero number C, the resulting set of solutions is in fact 0-dimensional, as you would expect from three equations in 3-dimensional space. So we see that very small changes in the equations can make a very big difference in the resulting solution set. Hence we usually cannot apply numerical methods to our problems: very small rounding errors can change the result completely