Singular

Thanks you very much for the answer.
Now it works !

For the computation of the Hilbert Series only the leading monomials
(power product without coefficients) of the standard basis are considered: i.e. every
expression in a is considered to be non-zero if not identical to 0.
If you need to distinguish the different cases for different values of a
you have to compute the different standard bases first:
see for example grobcov.lib [url]http://www.singular.uni-kl.de/Manual/4-0-1/sing_953.htm[/url]

Hello,
I have a problem when I try to compute Hilbert Series for S=Q[x,y,z]/I when I is an homogeneous ideal with parameter

Let f = x^3+y^3+z^3-3*a*x*y*z be a polynomial.
Is known that :
-for a^3=1, {f=0} is singular and Hilbert Series is infinite
-for a^3<>1 {f=0} is non-singular (smooth) and Hilbert Series is finite
[b]Singular package can work with parameters but does not discriminate in this case...[/b]

[b]The question is Why does not work the following program, without fix the parameter with minpoly[/b]
If we do not fix the parameter,
ring R=(0,a),(x,y,z),dp;
poly f;
f=x^3+y^3+z^3-3*a*x*y*z;
ideal G=std(jacob(f));
hilb(G,2);
//--> Hilbert series for S is H(t)= 1+3t+3t^2+t^3 (polynomial)
the result is wrong for some values (a^-1=0)....

If we fix the parameter with minpoly, we get the correct result:
ring R=(0,a),(x,y,z),dp;
minpoly =a^3-1; // or minpoly=a-1 or minpoly=a^2+a+1
poly f;
f=x^3+y^3+z^3-3*a*x*y*z;
ideal G=std(jacob(f));
hilb(G,2);
//--> Hilbert series for S is H(t)=
(1+2t)/(1-t)=1+3t+3t^2+3t^3+3t^4+... (infinite)

Without parameter, everything is OK:
for a = 1
ring R=0,(x,y,z),dp;
poly f;
int a; a=1;
f=x^3+y^3+z^3-3*a*x*y*z;
ideal G=std(jacob(f));
hilb(G,2);
//--> Hilbert series for S is H(t)=
(1+2t)/(1-t)=1+3t+3t^2+3t^3+3t^4+... (infinite)
Also for a^3<>1 eg a= 2 we get the correct result.
//--> Hilbert series for S is H(t)= 1+3t+3t^2+t^3 (polynomial)

[b]Is possible to work with parameters without fix them to compute Hilbert Series?
[/b]

Thank's in advance