Generators for the radical of an ideal I am interested in finding a generating set of the radical of an ideal given a set of generators for the ideal itself, but after a lot of thought I cannot figure out a good way to do it. Specifically:

Let $k$ be an algebraically closed field, and $I \subset k[x_1, ..., x_n]$ an ideal. If $I = (f_1, ..., f_m)$, is there any good way to find a set of generators for $\text{rad}(I)$?

Edit: One may assume that $f_1, ..., f_m$ form a Gröbner basis for $I$. (Given any set of generators for an ideal in $k[x_1,..., x_n]$ it is always possible to find a Gröbner basis, so this assumption is without loss of generality.)
 A: There are algorithms that find the generators for the radical of an ideal, e.g.:
Gianni, P.; Trager, B.; Zacharias, G.: Gröbner Bases and Primary Decomposition of Polynomial Ideals. J. Symb. Comp. 6, 149–167 (1988).
Note though that these methods are non-trivial; as far as I know, there is no simple characterization of the generators of a radical ideal. The construction by Connor only gives a subset of a generating set.
A: one way to solve the problem ( of finding generating set for radical ideal ,having the generator set of ideal ) is either to reduce it into principal ideal (by means of finding its reduced groebner basis) or factorizing it into power monomials).
Then for example:
if having    I=< xy, x^2+xy> , applying the routine leads to G={xy} as the reduced groebner basis for I.
(x^2+xy is eliminated since it's monomial xy € LT (xy) ) 
Now we use the theorem that says : Radical(f)= f/ GCD(f,df/d x1,d x2,...)
So: Radical(I)=xy/y=x
considering Lex ordering x>y
Thus    Radical(xy,x^2+xy)=x
A: If each $f_i$ is a product of powers of irreducibles, I think the generator for rad$(f_i)$ will be the product of the irreducibles for $f_i$ raised only to the first power. Do this for each $f_i$ and you have a generating set for rad$(I)$.
