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Combining computations: testing the
maximality of a left ideal
This application is particularly important for representation theory.
Suppose we have fixed a degree lexicographical ordering Dp
on U(sl2) over the
field of complex numbers. Given two
monomials, namely e and h2, we want to know, which maximal left ideals are
of the form < e+c, h2+t(h)>,
where c is a constant and
lm(t(h))<h2
these monomials as leading monomials of their generators.
Step 1. We are looking for such values of
parameters, that I is proper.
We introduce commutative variables a0,a1,a2,c0
as parameters, thus building an algebra A
and consider the ideal I, generated
by the polynomials p1=e+c0 and p2=h^2+a2*f+a1*h+a0.
ring A=(0,q),(e,f,h,a0,a1,a2,c0),Dp;
matrix d[7][7];
d[1,2]=-h;
d[1,3]=2*e;
d[2,3]=-2*f;
system("PLURAL",1,d);
option(redSB);
option(redTail);
We will eliminate e, f, h from I
and place the result in the ideal CI.
poly p1=e+c0;
poly p2=h^2+a2*f+a1*h+a0;
ideal I=p1,p2;
ideal CI=eliminate(I,e*f*h);
CI;
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==>
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CI[1]=a2*c0+4*c0^2
CI[2]=a1*c0^2-2*c0^2
CI[3]=a0*a2^2-16*a0*c0^2
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Step 2. Now we will compute the associated primes of the ideal CI
from the commutative subalgebra K[a0,a1,a2,c0] of
A.
LIB "primdec.lib";
list L=minAssChar(CI);
L;
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==>
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[1]:
_[1]=a2+4*c0
_[2]=a1-2
[2]:
_[1]=c0
_[2]=a0
[3]:
_[1]=c0
_[2]=a2
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Step 3. The output shows us there are three components (=associated primes). We will analyze the corresponding noncommutative ideals, substituting
the parameters with their values.
1st Component:
ideal J1=I;
J1=subst(J1,a1,2);
J1=subst(J1,a2,-4*c0);
We test for two subcases, depending on value of c0
print(std(subst(J1,c0,0)));
This is clearly not a maximal ideal, since the second polynomial
in one variable is not irreducible.
print(std(subst(J1,c0,q)));
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==>
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e+(q), h^2+(-4*q)*f+2*h+a0
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2nd Component:
ideal J2=I;
J2=subst(J2,c0,0);
J2=subst(J2,a0,0);
J2=std(J2);
J2;
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==>
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J2[1]=e
J2[2]=h*a2
J2[3]=f*a2+h^2+h*a1
J2[4]=h^3+h^2*a1+2*h^2+2*h*a1
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3rd Component:
ideal J3=I;
J3=subst(J3,c0,0);
J3=subst(J3,a2,0);
J3=std(J3);
J3;
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==>
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J3[1]=e
J3[2]=h^2+h*a1+a0
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This is clearly not a maximal ideal, since the second polynomial
in one variable is not irreducible.
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