MATH 392 -- Seminar in Computational Commutative Algebra
Lab 2/Problem Set 5 Solutions
> | with(Groebner); |
> | Plist:=[x^2*y + x^2 + 1, x^3 + x*y + 2]; |
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> | B:=Basis(Plist,plex(x,y)); |
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1. <LT(I)> = <x,y^4>. The monomial ideal <LT(f_1),LT(f_2)> = <x^2y, x^3> is
contained in this.
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2.
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3. x has been eliminated from the first equation, so we can solve that,
substitute back for y in the second, and solve for x.
4.
> | yroots:=[fsolve(B[1]=0,y,complex)]; |
> | for i to nops(yroots) do
Bs:=subs(y=yroots[i],B); solve(Bs[2],x); end do; |
5. Same idea, but y is eliminated from the first equation:
> | Basis(Plist,plex(y,x)); |
6. The grevlex Groebner basis does not have the same elimination
property (also contains 3 polynomials)
> | Bgrev:=Basis(Plist,tdeg(x,y)); |
> | for i to 3 do LeadingTerm(Bgrev[i],tdeg(x,y)); end do; |
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Book problems:
2.
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remainder is not zero, so the given f is not in the ideal.
3.
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8.
> | plot3d([(2+cos(t))*cos(u),(2+cos(t))*sin(u),sin(t)],t=0..2*Pi,u=0..2*Pi); |
> | Id:=[x-(2+ct)*cu,y-(2+ct)*su,z-st,ct^2+st^2-1,cu^2+su^2-1]; |
> | B8:=Basis(Id,plex(ct,st,cu,su,x,y,z)); |
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9. The torus knot:
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This says the curve K lies on the torus T. Here is a plot showing that:
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10. We want to maximize the (squared) distance f from the
point (1,1,1), at the poinst on the surface V(g) using Lagrange Multipliers
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Clearly the point with all positive coordinates is closer to (1,1,1) than the point
with all negative coordinates.
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