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Suppose $R$ noetherian and $M$ a finitely generated $R$-module. If you have a projective presentation of $M$: $P_1\rightarrow P_0\rightarrow M\rightarrow 0$, then by dualizing you obtain the following exact sequence: $0\rightarrow M^*\rightarrow P_0^*\rightarrow P_1^*\rightarrow D_R(M)\rightarrow 0$, $D_R(M)$ is called the auslander dual of $M$, of course it is not unique up to isomorphism but it can be proved that it is unique up to projective equivalence.

Now take an extension of rings $R\rightarrow R^\prime$. How can I prove that $D_{R^\prime}(M\otimes R^\prime)$ is projective equivalent to $D_R(M)\otimes R^\prime$?

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General case:

We may assume that $P_{i}$ are finitely generated because $M$ is and $R$ is Noetherian.

From $\mathrm{Hom}(R^{n},R)\cong R^{n}$ we deduce that for finitely generated free module taking dual commute with any base change. Furthermore the same remains true for finitely generated projective modules as: $\mathrm{Hom}(-,R)$ and $-\otimes R'$ preserve split exact sequences, and there is always a natural morphism $\mathrm{Hom}_{R}(N,R)\otimes R'\to \mathrm{Hom}_{R'}(N\otimes R',R')$.

This being said let's consider a commutative diagram: $$ \begin{matrix} P_{0}^{*}\otimes R' &\to & P_{1}^{*}\otimes R' & \to & D(M)\otimes R' & \to & 0\\ \downarrow \cong & & \downarrow \cong & & \downarrow & & \\ (P_{0}\otimes R')^{*} &\to & (P_{1}\otimes R')^{*} & \to & D(M\otimes R') & \to & 0\\ \end{matrix} $$ where:

  • the upper line is exact by right-exactness of tensor product,

  • the bottom line is exact by the definition of $D(M\otimes R')$,

  • maps $(P_{i}\otimes R')^{*}\to P_{i}^{*}\otimes R'$ are natural isomorphisms from above,

  • rightmost vertical arrow is an induced map of cokernels.

Since the two indicated vertical arrows are isomorphisms so is the third (this can be viewed as a special case of five lemma, but also follows immediately from the universal property of cokernel).

The first attempt (uncomplete):

Warning: the proof below works only for flat ring extensions (what might be considered as a trivial case).

Since $R$ is noetherian $M$ is finitely presented, and so you can take $P_1, P_0$ to be finitely generated (hence likewise finitely presented). But for finitely presented modules taking dual commutes with flat base change. Therefore multiplying $0\to M^{*}\to P_{0}^{*}\to P_{1}^{*}\to D_{R}(M)\to 0$ by $R'$ we obtain an exact sequence $$0\to (M\otimes R')^{*}\to (P_{0}\otimes R')^{*} \to (P_{1}\otimes R')^{*}\to D_{R}(M)\otimes R'\to 0$$ where exactness on the left follows by right-exactness of the tensor product and left exactness of the dual.

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