|  |  D.4.2.7 is_surjective Procedure from libraryalgebra.lib(see  algebra_lib).
 
Example:Usage:
is_surjective(phi); phi map to basering, or ideal defining it
Return:
an integer, 1 if phi is surjective, 0 if not
Note:
The algorithm returns 1 if and only if all the variables of the basering are
contained in the polynomial subalgebra generated by the polynomials
defining phi. Hence, it tests surjectivity in the case of a global odering.
If the basering has local or mixed ordering or if the preimage ring is a
quotient ring (in which case the map may not be well defined) then the return
value 1 needs to be interpreted with care.
 |  | LIB "algebra.lib";
ring R = 0,(x,y,z),dp;
ideal i = x, y, x2-y3;
map phi = R,i;                    // a map from R to itself, z->x2-y3
is_surjective(phi);
==> 0
qring Q = std(ideal(z-x37));
map psi = R, x,y,x2-y3;           // the same map to the quotient ring
is_surjective(psi);
==> 1
ring S = 0,(a,b,c),dp;
map psi = R,ideal(a,a+b,c-a2+b3); // a map from R to S,
is_surjective(psi);               // x->a, y->a+b, z->c-a2+b3
==> 1
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