When we use universal coefficient theorem, we only need to compute Tor$(A,B)$. But suddenly, later on, like in more advanced homological algebra. We start to construct Tor$_n$, Ext$_n$… Can't we just use a short free resolution to substitute compute only one Tor or Ext as before?
When can I have “short” projective resolution
commutative-algebrahomological-algebra
Related Solutions
For a ring $R$ and $R$-modules $A$ and $B$, projective resolutions can be used to compute both $\operatorname{Tor}_n^R(A,B)$ and $\operatorname{Ext}_R^n(A,B)$. Let $$\cdots\xrightarrow{\ \ \ }P_2\xrightarrow{\ \ \ }P_1\xrightarrow{\ \ \ }P_0\xrightarrow{\ \varepsilon \ }A\xrightarrow{\ \ \ }0$$ be a projective resolution of $A$. Then, applying $-\otimes_RB$ to the above gives you a chain complex: $$\cdots\xrightarrow{\ \ \ }P_2\otimes_RB\xrightarrow{\ \ \ }P_1\otimes_RB\xrightarrow{\ \ \ }P_0\otimes_RB\xrightarrow{\ \ \ }0$$ and the homology of this complex gives you $\operatorname{Tor}_n^R(A,B)$. You can also choose a projective resolution of $B$, apply $A\otimes_R-$, and compute the homology of the resulting complex.
Similarly, applying $\operatorname{Hom}_R(-,B)$ to the resolution above gives rise to a cochain complex: $$0\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_0,B)\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_1,B)\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_2,B)\xrightarrow{\ \ \ }\cdots$$ and the cohomologyof this complex is $\operatorname{Ext}_R^n(A,B)$.
Edit: As Tobias mentions below the fact that you can compute these groups using both methods is non trivial. A proof of this can be found in Weibel's book, in the "Balancing Tor and Ext" section.
Let $X^{(Y)}$ mean a direct sum of copies of $X$, indexed by the set $Y$. I'll assume left $R$-modules are being dealt with.
The trick, given a module $M$, is to set $F(M)=R^{(M)}$, with the map $\pi_M\colon F(M)\to M$ defined by $$ \pi_M\Bigl(\sum_{x\in M}r_xe_x\Bigr)=\sum_{x\in M}r_xx $$ where $e_x$ is the vector having $1$ at the $x$-component and $0$ elsewhere. In other words, we define $\pi_M$ on the basis by sending $e_x$ to $x$.
If $f\colon M\to N$ is a module homomorphism, then $f$ induces a module homomorphism $\tilde{f}\colon F(M)\to F(N)$ by $$ \tilde{f}\Bigl(\sum_{x\in M}r_xe_x\Bigr)=\sum_{x\in M}r_xe_{f(x)} $$ In other words, we define $\tilde{f}$ on the basis by sending $e_x$ to $e_{f(x)}$.
Now $\pi_N\tilde{f}(e_x)=\pi_N(e_{f(x)}=f(x)$ and $f\pi_N(e_x)=f(x)$. Therefore $$ \pi_N\tilde{f}=f\pi_M $$ because both homomorphisms coincide on a basis of $F(M)$.
In a more abstract way, this is the adjunction “free module”–“forgetful functor”.
Best Answer
The classical universal coefficient theorem is a theorem about $\mathbb{Z}$-modules (or abelian groups, if you prefer). The ring $\mathbb{Z}$ has the very special property that every submodule of a free $\mathbb{Z}$-module is free. As a result, any $\mathbb{Z}$-module $M$ has a "short" free resolution: we can pick an epimorphism $f:F\to M$ from a free module $F$, and then $\ker f=G$ is another free module giving a free resolution $$0\to G\to F\to M\to 0.$$ As a result, $\operatorname{Tor}_n^\mathbb{Z}$ and $\operatorname{Ext}^n_\mathbb{Z}$ are trivial for $n>1$ (and for $n=1$, they are sometimes abbreviated as just $\operatorname{Tor}$ and $\operatorname{Ext}$).
However, none of this works in general over an arbitrary ring $R$. A submodule of a free $R$-module need not be free, and so an $R$-module need not have a short free resolution. So, $\operatorname{Tor}_n^R$ and $\operatorname{Ext}^n_R$ may be nontrivial for $n>1$.