Let $k$ be a perfect field (so reduced = geometrically reduced) and $f:X\rightarrow \mathrm{Spec}(k)$ a Cohen-Macaulay morphism. Denote by $i:X_{red}\rightarrow X$ the underlying reduced subscheme and by $\omega_{X}$ and $\omega_{X_{red}}$ the relative dualizing sheaves of $X$ and $X_{red}$ over $k$. What can one say about the relationship between these two sheaves?

One might hope that there is an "adjunction formula" relating them, but I only know the adjunction formula in the context of a pair of maps $g:Y\rightarrow X$ and $f:X\rightarrow Z$ that are flat, of finite type, and CM, so this doesn't apply to the closed immersion $i:X_{red}\rightarrow X$ unless $X$ is already reduced (failure of flatness).

Certainly one has a trace morphism $i_*: i_*\omega_{X_{red}}\rightarrow \omega_X$. Can one describe the image and kernel of $i_*$, say in terms of the ideal sheaf defining $i$?

  • $\begingroup$ A basic comment: it is somewhat difficult to be CM and not reduced. This is because CM implies S_1, basically by definition. A result of Serre is that (R_0 and S_1) implies reduced. So any example must fail to be R_0, that is to say, it must not be generically reduced. $\endgroup$ – David E Speyer Feb 3 '10 at 23:19
  • $\begingroup$ On the other hand, every zero dimensional noetherian ring is CM, and almost none of them are reduced. So there are examples. $\endgroup$ – David E Speyer Feb 3 '10 at 23:20
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    $\begingroup$ Sure, and in fact there are lots of examples that arise "in nature". For example, the regular proper model of the modular curve $X_1(N)$ over $\mathbf{Z}_p$ for $p$ dividing $N$ to high order (and $N$ divisible by some other sufficiently large prime, say, to avoid stacky issues) has closed fiber that is not reduced. However, CM is stable under base change, and regular implies CM, so... $\endgroup$ – B. Cais Feb 3 '10 at 23:51

Dear Bryden,

Hopefully I have things straight, and there is a general formula $i^!\omega\_X = \omega\_{X\_{red}}$. One then has the functorial isomorphism (of sheaves on $X$) $RHom_{\mathcal O_{X\_{red}}}({\mathcal F},\omega_{X\_{red}}) = Rhom_{\mathcal O_X}(i_\*{\mathcal F}, \omega\_X),$ for a coherent sheaf $\mathcal F$ on $X_{red}$. (Normally we would have to apply an $Ri_*$ to the source of this isomorphism, to put the RHom sheaves on the same space, and would have to have an $Ri_*$ in the formula on the RHS. But $i_*$ is exact, and in fact just identifies sheaves on $X_{red}$ with sheaves on $X$ via the identification of their underlying topological spaces.

Now $RHom_{\mathcal O_X}(i_*{\mathcal F},\omega_X) = Hom_{\mathcal O_X}(i_*{\mathcal F}, {\mathcal J}^{\bullet})$, where ${\mathcal J}^{\bullet}$ is an injective resolution of $\omega_X$, which in turn equals $Hom_{\mathcal O_{X_{red}}}(\mathcal F,{\mathcal J}^{\bullet}[\mathcal I]),$ where $\mathcal I$ is the ideal sheaf of $X_{red}$ in $X$.

Finally, this last complex can be identified with $RHom_{\mathcal O_{X_{red}}}(\mathcal F, RHom_{\mathcal O_X}(\mathcal O_{X_{red}}, \omega_X)).$

So we get the formula $\omega_{X_{red}} = RHom_{\mathcal O_X}(O_{X_{red}}, \omega_X).$ (And the derivation shows that this should be valid for any closed immersion, provided one is in a context where the dualizing complex formalism is satisfied, except that probably there should be some shifts in dimension in general, because the dualizing complex probably coincides with the dualizing sheaf place not in degree 0, but in degree $-dim X$. However, in our case the dimensions of $X$ and $X_{red}$ coincide, so this shift can be ignored.)

Note that, as this formula shows, $\omega_{X_{red}}$ could be a complex, not just a sheaf. This is reasonable, I guess; in general, even if $X$ is CM, this needn't imply that $X_{red}$ is (I imagine).

If in fact $X_{red}$ is CM, then I guess we find just one non-zero term in the formula for $\omega_{X_{red}},$ and so have $\omega_{X_{red}} = \omega_X[\mathcal I].$

With a bit of luck, the above is not bogus, and answers your question.

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    $\begingroup$ An example where X is CM and X_{red} is not: Let k be a field of characteristic 2 and consider the subvariety of A^4 cut out by w^3*z=x^4 and w*z^3=y^4. This is a complete intersection, hence CM. On X_{red}, (wz)^4=(xy)^4 so, since char(k)=2, wz=xy. Similarly, x^{12}=w^9*z^3=w^8*y^4 so x^3=w^2 y and y^2=z^2 x. It is somewhat well known that k[w,x,y,z]/(wz-xy, x^3-w^2 y, y^3-z^2 x) is isomorphic to k[t^4, t^3 u, t u^3, u^4]. The latter ring is not normal (t^2 u^2 is missing), so X_{red} is not normal. Working a little harder, X_{red} is not S_2 so X_{red} is not CM. $\endgroup$ – David E Speyer Feb 3 '10 at 23:56
  • $\begingroup$ Dear Matt, Thank you for the response; everything looks good to me. Your formula for $\omega_{X_{red}}$ as an $RHom$ holds (with suitable modifications) for any proper morphism $f:Y\rightarrow Y$ of noetherian schemes with $Y$ admitting a dualizing complex (RD, VII 3.4c). Thanks again! $\endgroup$ – B. Cais Feb 4 '10 at 0:07

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