Does anyone knows how to prove or disprove that if $(R,m)$ is a Noetherian local domain of dimension $d$ one has that $\mu(I)\leq d$ for every ideal $I\subset m$?
Where the arithmetic rank of $I$ is $\mu(I)$ and $\subset$ means strictly contained.
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yo yo
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1hi yo yo; just to clarify, by arithmetic rank, do you mean that $\mu(I)$ is the minimal possible size of a generating set for $\sqrt{I}$? – Atticus Stonestrom Sep 18 '21 at 15:10
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1Yes, exactly that – yo yo Sep 18 '21 at 15:12
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Really? I thought arithmetical rank is the minimal number of elements in $I$, generating an ideal $J$ with $\sqrt{J}=\sqrt{I}$. If so, the standard proof is the following (sketch). Inductively, choose $a_1,\ldots, a_r\in I$ such that any prime ideal containing $a_i, 1\leq i\leq r$ and not containing $I$ has height at least $r$. – Mohan Sep 18 '21 at 19:49
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I don't understand, can you please are more details? – yo yo Sep 20 '21 at 12:33
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I think the two definitions are equivalent and the reason is that $\sqrt{(a_1,...,a_t)}=\sqrt{(a_1^s,...,a_t^s)}$ the proof is what follows: one inclusion is clear. For the other take $g^n=\sum g_ia_i$ and than take $g^{nsr+1}$. The polynomial $g^{nsr}$ is in $(a_1^s,...,a_t^s)$ becouse an homogeneous polynomial of degree $sr+1$ in $a_1,...,a_t$ must contain at least one $a_i^s$. With this fact it is clear that $min{a_1,..,a_t\in R|\sqrt{(a_1,...,a_t)}=\sqrt{I}}=min{ a_1,..,a_t\in I|\sqrt{(a_1,...,a_t)}=\sqrt{I}}$ – yo yo Sep 20 '21 at 12:53