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Let $R$ be a (not necessarily commutative) ring and $M$ a simple right $R$-module. Then $\mathfrak{m}=Ann(M)$ is a maximal ideal of $R$. It is seems known that $$ pdim_{R}(M)=pdim_{R_{\mathfrak{m}}}(M_{\mathfrak{m}}) $$ where $R_{\mathfrak{m}}$ is the localization of $R$ with respect to $\mathfrak{m}$. "$\ge$"-part is obvious as localization is an exact functor mod-$R$ to mod-$R_{\mathfrak{m}}$. Could anyone give me a proof or a reference of this equality?

Thanks.

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Is localization always possible in your situation ? (usually one requires that Ore conditions or the like are satisfied) –  Ralph Aug 12 '12 at 8:10
    
I should have mentioned about localization. It depends on $M$. But please assume that it is possible. Examples in my mind are mostly not very noncommutative; they have plenty of normaliizng elements. –  M Simon Aug 12 '12 at 16:45
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up vote 7 down vote accepted

In general, it is not true that $Ann_R(M)$ is a maximal two-sided ideal of $R$ if $M$ is a simple $R$-module. For example, let $k$ be a field of characteristic zero, let $\mathfrak{g} = \mathfrak{sl}_2(k)$ and let $A = U(\mathfrak{g}) / \langle C \rangle$ where $C$ is the Casimir element. Then the Verma module $V$ of highest weight $-2$ for $\mathfrak{g}$ is a simple faithful $A$-module (so $Ann_A(V) = 0$), but the zero ideal in $A$ isn't maximal since it's contained in the proper two-sided ideal $\mathfrak{g}\cdot A$.

Annihilators of simple modules are called primitive ideals; whilst any maximal two-sided ideal $I$ in a ring $R$ is necessarily primitive (simply pick a maximal left ideal $J$ containing $I$ and consider the simple module $R/J$), the converse is not true as the above example shows.

As Ralph notes, in the general setting of non-commutative rings, the localisation $R_{\mathfrak{m}}$ does not exist: one needs certain Ore conditions to be satisfied by the set $S$ of elements in $R$ that are non-zero-divisors modulo $\mathfrak{m}$.

Even if the Ore localisation $R_\mathfrak{m}$ exists, it could happen that the Ore localisation $M_{\mathfrak{m}}$ is zero. This is actually the case for the $A$-module $V$ considered above: since $A$ is a Noetherian domain, the set $S$ of all non-zero elements of $A$ is an Ore set, but $V$ is a torsion $A$-module so $V_S = 0$. It's known that $pd_A(V) = 1$ in this case, so the inequality $pd_R(M) \geq pd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$ is strict in general.

In the positive direction, however, it is known that flat dimension behaves well under Ore localisation. More precisely: suppose that $S$ is a two-sided denominator set in a ring $R$ and let $N$ be an $R_S$-module. Then the flat dimension of $N$ as an $R_S$-module is equal to the flat dimension of $N$ as an $R$-module; this is Proposition 7.4.2(iii) of the book Noncommutative Noetherian Rings by McConnell and Robson.

For a finitely generated module $M$ over a Noetherian ring $R$, flat dimension coincides with projective dimension (see section 7.1.5 of the book cited above). This is enough to give a positive answer to your question under the following extra hypotheses:

  • the annihilator $\mathfrak{m}$ of your simple module $M$ is localisable on both sides
  • the ring $R$ is Noetherian
  • the set $S$ of regular elements mod $\mathfrak{m}$ acts invertibly on $M$.

If this is the case then the Ore localisation $M_{\mathfrak{m}} = M_S$ coincides with $M$, so we can deduce that

$pd_R(M) = fd_R(M) = fd_R(M_{\mathfrak{m}}) = fd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}}) = pd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$

using the results cited above.

The third condition is known to hold for example whenever $R / \mathfrak{m}$ is an Artinian ring, since in this case the image of $S$ in $R /\mathfrak{m}$ is precisely the set of units of $R / \mathfrak{m}$ and therefore acts invertibly on $M$.

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Thank you for the detailed answer. This helps me a lot. I now understand that whether or not the inequality holds depends not only $R$ but alos $M$. I will check the book you referred to. Thanks again! –  M Simon Aug 12 '12 at 19:28
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