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Fix a field $k$ and let $R = k[x_1,x_2,\ldots]$. Say that an ideal $I \subset R$ is generated in finite degree if there exists a generating set $S$ for $I$ (possibly infinite) and an integer $n$ such that $S$ consists of elements of degree at most $n$.

Question: If $I$ and $J$ are ideals of $R$ that are generated in finite degree, is $I \cap J$ generated in finite degree?

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    $\begingroup$ rather "in bounded degree"? $\endgroup$
    – YCor
    Commented Nov 19, 2016 at 22:42
  • $\begingroup$ If this is true, then there exists $u(d)$ such that for every $n$ and every pair $I,J$ of ideals in $k[x_1,\dots,x_n]$ generated in degree $\le d$, the intersection $I\cap J$ is generated in degree $\le u(d)$. I don't know if the latter statement is true but I would expect that it's known whether it holds (it's a natural issue, e.g., in the algorithmic problem of computing generators for an intersection of ideals). $\endgroup$
    – YCor
    Commented Nov 19, 2016 at 22:52
  • $\begingroup$ Analogy for groups: a group $G$ is boundedly presented by a generating subset $S$ if the kernel of $F_S\to G$ is generated, as normal subgroup, by elements of bounded length. When $S$ is finite this means that $G$ is finitely presented (and does not depend on $S$). In this context the condition is not stable under intersection (even with a finitely generated free group), because there exist fibre products of 2 finitely presented groups that is not finitely presented. Anyway it's a very different context. $\endgroup$
    – YCor
    Commented Nov 19, 2016 at 23:03
  • $\begingroup$ One more remark: if $I$ is generated by monomials of degree $\le d$ and $J$ by monomials of degree $\le e$, then $I\cap J$ is generated by monomials of degree $\le d+e$. $\endgroup$
    – YCor
    Commented Nov 20, 2016 at 1:43

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