## torsion product [closed]

Let $R=R_1 \oplus ... \oplus R_r$; $A$ and $B$ are $R$-modules. We know that $A=A_1 \oplus ... \oplus A_r$, $B=B_1 \oplus ... \oplus B_r$ with $A_i = R_i.A$, $B_i = R_i.B$. Now i have not find following fomula yet: $$\mathrm{Tor}_n^R (A,B) \cong \sum \mathrm{Tor}_n^{R_i}(A_i, B_i); \quad \mathrm{Ext}_R^n (A,B) \cong \sum \mathrm{Ext}_{R_i}^n (A_i, B_i)$$ Please show me which document should i read or help me prove these formula. Thanks!

-
You've got some TeX issues to fix. I find that it doesn't like "Tor" so you may need to declaremathoperator or just use your dollar signs more carefully – David White Sep 14 2011 at 15:36
This is very easy to prove directly from the definition of those functors as derived functors. – Mariano Suárez-Alvarez Sep 14 2011 at 16:04
We don't mean to be dismissive, but it's not clear if there is a specific reference. Mariano's comment is really the way to go. Various standard homological algebra books (Cartan-Eilenberg,... Weibel) will give you further details for how to set this up. By the way, I suspect you did not mean this, but you seem to saying that $A$ and $B$ are free, in which case the equations would reduce to $0\cong 0$ when $n>0$. – Donu Arapura Sep 15 2011 at 2:05
thank for everyone. I try to prove it now! – student Sep 16 2011 at 9:24