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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.)

user26857
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Eric Haengel
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3 Answers3

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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.

user26857
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Johannes Kloos
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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$(I)$.

Connor
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  • I see what you're saying, that sounds like a good idea. Do you know of any good way to factor the $f_i$ into irreducibles? – Eric Haengel May 17 '12 at 02:18
  • That's always tough... I don't know a straightforward method for an arbitrary polynomial, especially in multiple variables. Eisenstein's criterion can be very helpful in some cases for determining if a given factor is irreducible. – Connor May 17 '12 at 02:24
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    I think acoustician's method is wrong. Consider $I = (x + y^2, x + 2 y^2)$, its generators are irreducible, but $I$ is not radical: $y \notin I$ and $y^2 \in I$. However acoustician's method works for monomial ideals. – Andrea May 18 '12 at 07:56
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    You're definitely right, Andrea. Can you think of a description for non-monomial ideals? I'm stumped. – Connor May 19 '12 at 01:03
  • @Andrea: Right, that is an important example, but I think the problem is that while $x+y^2$ and $x+2y^2$ generate $I$, they are not a grobner basis. Instead we can write $I = <x, y^2>$, where $x$ and $y^2$ do form a grobner basis, and it is true that $\text{rad}(I) = <x, y>$ is generated by the square-free part of each generator. So, I think what acoustician was saying works if we start with a grobner basis for $I$. – Eric Haengel May 24 '12 at 17:14
  • No, even for a Gröbner basis one can't find the radical of ideal by taking the square-free parts of the polynomials. (-1) – user26857 Dec 14 '15 at 21:51
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    For instance, over $\mathbb Q$ the polynomials $xy,x^2+y^2,y^3$ form a Gröbner basis. By taking the square-free parts we get the ideal $(xy,x^2+y^2,y)=(x^2,y)$ which is not radical. In fact, the radical of ideal $(xy,x^2+y^2,y^3)$ is $(x,y)$. – user26857 Dec 14 '15 at 22:17
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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

panteha
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