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For univariate polynomials, I can compute the squarefree variant with the same roots as $p / \text{gcd}(p, dp/dx)$, which is simple enough.

Then I learned about radical of ideals generated by multivariate polynomials, and it seems that it is non-trivial to get the generator set of the radical ideal, at least for the general case.

But it seems to me that if I could compute the squarefree version of each multivariate polynomial in the generating set, I would get the same variety, and since there is no $f$ such as $f^{m>1} = g$ if $g$ is squarefree, then this ideal must be radical.

Is my reasoning correct? If I can compute the squarefree variants of my polynomials with an algorithm such as this, then they generate the radical ideal?

user26857
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lvella
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1 Answers1

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No, this can't work in general. Try $(x,y^2+x)$: both are irreducible and therefore square-free, but $(x,y^2+x)=(x,y^2)$ which is not radical. Your impression that finding generator sets for radical ideals is somewhat difficult is correct.

Hank Scorpio
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