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Working over $R=\mathbb{C}[x_1,...,x_n]$, I'm given a ring homomorphism with $i\in{1,...,n}$ and $t\in \mathbb{C}$. $\phi_{i,t}(x_j)=x_j$ for $j\neq i$ to themselves. From this I've proven that an ideal $I$ is monomial iff $\forall i,t$ $\phi_{i,t}^{-1}(I)=I$. And subsequently that if $I$ is a monomial ideal then so is its radical.

I'm allowed to know Dickson's lemma, namely that we have a finite set of generators for a monomial ideal. But I'm now asked to find an algorithm that given a monomial ideal with generators will spit out generators for its radical. (And, of course to prove that it works.) I included the above because I thought maybe this homomorphism was set up to help me, but maybe once I proved all the above stuff it could serve me no longer. Oh, and I'm not allowed to know anything about multivariable division algorithms or Grobner bases. All that stuff to come. Anywho, I'm pretty stuck...

Edit: Wait, is it as simple as dividing the powers of the variables of each generator by their gcd? After I've done this is my new set a set of generators for the radical? Intuitively, it seems like a straight forward, nice approach, but perhaps something goes wrong?

AsinglePANCAKE
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    Take each generator of the monomial ideal and reduce the power of each variable to $1$, this should give you a generating set. – JSchlather Feb 14 '13 at 05:44
  • @JacobSchlather Oh, now I feel dumb for both the original question and the edit...Any tips for proving this is the right thing to do? – AsinglePANCAKE Feb 14 '13 at 05:48
  • I'm a little rusty on this stuff. It should be straightforward to show that those elements are in the radical. Then probably you can take any monomial in the radical and look at whatever power of it is in the ideal and then write it as one of the generators times something and conclude its in the generating set. – JSchlather Feb 14 '13 at 06:01
  • An answer is here. – Leo Oct 02 '13 at 02:38

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