## Suggestions about the exercises in Sec 3.3

Updated December 3, 2007

0. In all of the exercises, the basic approach is as follows:

• If working with a surface given parametrically as follows:

x = F(t,u),   y = G(t,u),   z = H(t,u),

then let  I = ⟨F(t,u) - x, G(t,u) - y, H(t,u) - z⟩  in  k[t,u,x,y,z],  and find a Groebner basis for  I  with respect to lex order with  t > u > x > y > z.  Then look at the second elimination ideal.

• If working with a space curve given parametrically as follows:

x = f(t),   y = g(t),   z = h(t),

then let  I = ⟨f(t) - x, g(t) - y, h(t) - z⟩  in  k[t,x,y,z],  and find a Groebner basis for  I  with respect to lex order with  t > x > y > z.  Then look at the first elimination ideal.

1.   In exercise 4, we start out exactly as in the discussion in the text of the tangent surface of the twisted cubic.  If we run Maple to check the calculation, we'll get exactly the same Groebner basis as in the text.  The implicit equation is the polynomial equation  g7(x,y,z) = 0.  The main point of the exercise is to show that a real solution  (x,y,z)  extends to a point  (t,u,x,y,z)  of  V(I)  with real values of  x  and  u.
Because  g1 = x -t - u,  the extension step from u to t presents no problem.

In the extension step from  (x,y,z)  to  u(x,y,z)  all of the equations  gi(u,x,y,z) = 0,  i = 2,...,6,  must be satisfied.  For instance,

g3(u,x,y,z) = u(x² - y) - x³ + (3/2)xy - (1/2)z.

And so, if  x² - y ≠ 0,  then we can express u as a quotient of two real expressions.   ¿But what do we do? if the coefficients of u in the equations   g3 = 0, ..., g6 = 0  are all equal to 0 at a given real point  (x,y,z)  of the tangent surface.

2.   In exercise 6 we're working with the following parametrization:

x = uv,   y = u²,   z = v².

Therefore, if the parameter values u and v are real, then two of the coordinates have to be positive.  But ¿do the equations allow those coordinates to be negative?  And ¿can those negative real coordinate values be realized by non-real parameter values?  Actually, the extension theorem gives a formal answer to this last question, but when you see the answer you may say that you could have guessed it.

Comments and questions to:  roberts@math.umn.edu

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