1
$\begingroup$

Let $k$ be an algebraically closed field, let $I, J \subseteq k[x_1,\dots, x_n, y_1,\dots, y_m]$ be ideals, and let $i,j$ be the minimum number of generators of $I$ and $J$, respectively. I have two questions:

  1. Is the minimum number of generators of the product ideal $IJ$ at least $\max\{i,j\}$?

  2. If $I$ is generated by polynomials that depend only on $x_1,\dots, x_n$ and $J$ is generated by polynomials that depend only on $y_1,\dots, y_m$, then is the minimum number of generators of the product ideal $IJ \subseteq k[x_1,\dots, x_n, y_1,\dots,y_m]$ equal to $i j$?

Question two is a repost from MSE.

$\endgroup$

1 Answer 1

4
$\begingroup$
  1. No. $k[x,y]$ we can find an ideal $I$ with $m$ generators for any $m\geq 5$ but $I^2$ has $9$ generators. See this paper.

  2. Yes, at least in the graded case. There is probably a better way to show this, but from the top of my head here is one approach: show that $R/I, R/J$ are Tor-independent. It follows that the natural map from $I\otimes_R J\to IJ$ is an isomorphism. Finally, it is easy to see that $\mu(I\otimes J) = \mu(I)\mu(J)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.