## 11. Homological algebra

1. Let A be an abelian group.
   Show that the kernel of the natural map $A \to A \otimes \mathbb { Q }$ (sending a to $a \otimes 1 )$ is the subgroup of torsion elements (elements of finite order) Tors(A). (Don’t confuse Tors with Tor!)

2. Suppose $A  B$ is an injective map of abelian groups.
   Show that the natural map $A \otimes \mathbb { Q } \to B \otimes \mathbb { Q }$ is also injective.
   This shows that tensoring with $\mathbb { Q }$ is left exact as well as right exact; that is, Q is $f l a t$ as a Z-module.

3. Find a free resolution for $\mathbb { Q }$ as a Z-module.
   (Hint.
   $\textstyle { \frac { 1 } { n ! } } . { \big ) }$

4. Let A be an abelian group.
   Compute $\operatorname { T o r } ( \mathbb { Q } , A )$ . Now consider the group $\mathbb { Q } / \mathbb { Z } ,$ given by the exact sequence

$$
0 \to \mathbb { Z } \to \mathbb { Q } \to \mathbb { Q } / \mathbb { Z } \to 0 .
$$

(It can be thought of as the group of roots of unity, by identifying $[ p / q ]$ with $e ^ { 2 \pi i p / q } )$ . Compute $\operatorname { T o r } ( \mathbb { Q } / \mathbb { Z } , A )$

5. Extend the table of values of $\mathrm { T o r } ( A , B )$ and Ext(A, B) to the cases where A and B may be $\mathbb { Q }$ and $\mathbb { Q } / \mathbb { Z }$

6. Let $R = \mathbb { Z } [ x , y ]$ be the polynomial ring in two variables.
   Consider the quotients $M = R / ( x - y )$ and $N = R / ( x , y )$ as R-modules.
   Construct free resolutions of M and N and thereby compute $\mathrm { E x t } _ { R } ^ { * } ( M , M )$ , Ext∗R(M, N), Ext∗R(N, M), and $\mathrm { E x t } _ { R } ^ { * } ( M , M )$

7. Consider the exterior algebra $\Lambda ^ { * } .$ , generated by anti-commuting symbols $d x _ { 1 } , d x _ { 2 } , \ldots , d x _ { n }$ , over some field k. Show that taking “wedge product with $d x _ { 1 } { } ^ { , ; }$ as differential gives a chain complex

$$
0 \to \Lambda ^ { 0 } \to \Lambda ^ { 1 } \to \cdots \to \Lambda ^ { n } \to 0
$$

with trivial homology.
What happens if $d x _ { 1 }$ is replaced by an arbitary non-zero element $v \in \Lambda ^  1 \}$

8. Consider the group ring $R = \mathbb { Z } [ \mathbb { Z } _ { 2 } ]$ , and let x denote the non-trivial element of $\mathbb { Z } _ { 2 }$ . There is an action of R on $\mathbb { Z } ,$ given by letting x act by multiplication −1; call this module M. Construct a free resolution of M and use it to compute the groups $\mathrm { T o r } _ { * } ^ { R } ( M , M )$

9. Show that the functor To $\mathbf { \partial } _ { \cdot \ast } R _ { \left( - , - \right) }$ is symmetric, for any commutative ring $R ,$ as follows.
   Given modules M and $N ,$ take free resolutions $M _ { * }$ and $N _ { * }$ . There is a chain map $M _ { * } \to M$ , if we view M as a complex whose only non-zero chain group is M in dimension zero.
   Tensoring with the complex $N _ { * }$ gives us a chain map $\psi : M _ { * } \otimes _ { R } N _ { * } \to M \otimes N _ { * }$ . By considering the exact sequence of complexes

$$
0  ( Z _ { * } , 0 )  ( N _ { * } , d )  ( B _ { * - 1 } , 0 )  0
$$

and a technique similar to that of the K¨unneth theorem, conclude that $\psi$ gives an isomorphism on homology.
By now switching the roles of M and N, conclude that Tor is symmetric.

10. An exact sequence of abelian groups

$$
0  A  B  C  0
$$

is called an extension of C by A. (This is the topologist’s convention: we think of B as resembling the total space of a bundle over the base space $C$ with fibre A, in other words, as $^ { 6 6 } \mathrm { a }$ family of ${ } ^ { 6 } A { } ^ { \circ } \mathrm { s }$ parametrised by $C ^ { 9 9 }$ . Algebraists point out that it is A which is a subgroup of $B ,$ , so they call B an extension of A by C! Admittedly, logic is on their side, but the first convention is more useful and is prevalent these days.)
Extensions of C by A form a category in a fairly obvious way; in particular an equivalence is an isomorphism $\theta : B \to B ^ { \prime }$ making a commuting diagram

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Follow the steps below to prove that the group Ext(C, A) classifies such extensions up to equivalence, and hence deserves its name!

(a). Show how to associate to an (equivalence class of) extension $E = ( 0  A  B  C  0 )$ a (well-defined) characteristic element $c ( E ) \in \operatorname { E x t } ( C , A )$ . (Hint: let $0  F _ { 1 }  F _ { 0 }  C  0$ be a free resolution of $C ;$ then construct a map $F _ { 1 } \to A . )$

(b). Show how to associate to an element $c \in \operatorname { E x t } ( C , A )$ an equivalence class of (group B and) extension $E ( c ) = ( 0  A  B  C  0 )$ ：

(c). Check that these two procedures are mutually inverse.

(d). Show that a split extension corresponds to the zero element of $\operatorname { E x t } ( C , A )$ . We can therefore view the class $c ( E ) \in \operatorname { E x t } ( C , A )$ as measuring the obstruction to splitting of E.

(e). (If you’re still having fun!)
We know that $\operatorname { E x t } ( C , A )$ is itself an abelian group.
Work out how to define the sum $E + F$ of two extensions of C by A in such a way that $c ( E + F ) = c ( E ) + c ( F )$ so that $\operatorname { E x t } ( C , A )$ may be identified with the abelian group of equivalence classes of extensions.

It’s a remarkable fact that higher Ext-groups classify equivalence classes of longer exact sequences: for example, $\operatorname { E x t } ^ { 2 } ( C , A )$ classifies equivalence classes of sequence $0  A  X _ { 1 }  X _ { 2 }  C$ Moreover, there is a multiplication (or composition) $\operatorname { E x t } ^ { m } ( C , B ) \times \operatorname { E x t } ^ { n } ( B , A ) \to \operatorname { E x t } ^ { m + n } ( C , A )$ called the Yoneda product obtained by splicing them!