# an exercise in the Rotman book

I read the following exercise in the book of Rotman: "an introduction to homological algebra"

9.21: In the situation $(_RA, _SB_R, _SC)$ with $B$ $R$-projective, use the adjoint isomorphism to obtain isomorphisms $$\mathrm{Ext}^n_S(B \otimes_RA, C) \cong \mathrm{Ext}^n_R(A, \mathrm{Hom}_S(B, C)).$$

I consider the following example: $S = \mathbb{Z}, R = \mathbb{Z}/2\mathbb{Z}, A = B = C = R$. Then $$\mathrm{Ext}^1_S(B \otimes_RA, C) \cong \mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}.$$ And $$\mathrm{Ext}^1_R(A, \mathrm{Hom}_S(B, C)) \cong \mathrm{Ext}^1_{\mathbb{Z}/2\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}) \cong 0.$$

So what is the incorrect: the exercise or my example?

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Your example looks correct, so I'd say the exercise. – Fernando Muro Jun 28 '12 at 8:39
There is only a spectral sequence with $E_2=\mathrm{Ext}_R(A,\mathrm{Ext}_S(B,C))$ converging to $\mathrm{Ext}_S(B\otimes_RA,C)$ in that situation. – Mariano Suárez-Alvarez Jun 28 '12 at 8:51
(You can construct it picking an $R$-projective resolution $X\to A$ and an $S$-injective resolution $C\to Y$, and looking at the two standard spectral sequences for the double complex $\hom_S(B\otimes_RX,Y)$, which is isomorphic to $\hom_R(X,\hom_S(B,Y))$) – Mariano Suárez-Alvarez Jun 28 '12 at 8:53
$B$ has to be $R$-flat and $S$-projective for the assertion of this exercise to hold. – Leonid Positselski Jun 28 '12 at 9:48