Let $I \subseteq \mathbb{C}[x_1,\ldots,x_n]$ be an ideal generated by polynomials $f_1,\ldots,f_r$ of degree at most $d$. Is it possible to generate the radical $\sqrt{I}$ of this ideal with polynomials of degree at most $d$? If not, is there any other upper bound known in terms of $d$ for the degrees of a set of polynomials generating $\sqrt{I}$?
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In general, one can not bound the degree of the radical $\sqrt{I}$ using only $d$, even in the polynomial ring $R=\mathbb C[x,y,z,t]$. Consider the (complete intersection) ideal $I= (x^mt-y^mz, z^{n+2}-xt^{n+1})$, generated in degrees $m+1$ and $n+2$. Then the radical of $I$ has a generator of degree $mn+2$, as shown in Lemma 2.4 of this paper.
answered Dec 31, 2017 at 23:49
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$\begingroup$ So the the Castelnuovo-Mumford regularity of an ideal is equal to the tight degree bound of the generators? Meaning any polynomial set of degree less than $reg(I)$ can not generate $I$? $\endgroup$– PewMar 30, 2019 at 2:53
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1$\begingroup$ The regularity is in general an upper bound on the max degree of generators. However in that example they actually compute the degrees of the generators, not just regularity. $\endgroup$ Mar 30, 2019 at 3:39