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with 
are positive integers and 
is prime (
), we get: 
be a real numbers such that 
Prove that
Where 
be non-negative real numbers. Prove that
Looking for an algebraic solution!
be a positive integer, and let 
denote the integers modulo . Determine the number of functions 
satisfying 
for all 
.
to 
. The answer is 
. Write 
.
(commutative with unity) two -modules 
and 
, given any projective resolution of ,![\[ \cdots \stackrel{d_3}{\longrightarrow} P_2 \stackrel{d_2}\longrightarrow P_1 \stackrel{d_1}\longrightarrow P_0 \stackrel{\varepsilon}\longrightarrow A \longrightarrow 0, \]](http://latex.artofproblemsolving.com/b/b/4/bb4d99128d50f202a83535f0004b3c35486d1fd4.png)
we can consider the corresponding cochain complex given by applying the contravariant functor 
and ignoring :![\[ 0 \stackrel{d^0}\longrightarrow \operatorname{Hom}_R (P_0, B) \stackrel{d^1}\longrightarrow \operatorname{Hom}_R (P_1, B) \stackrel{d^2}\longrightarrow \operatorname{Hom}_R (P_2, B) \stackrel{d^3}\longrightarrow \cdots. \]](http://latex.artofproblemsolving.com/d/b/e/dbe3c208659d1773db190e1587e75a51a27db629.png)
It is a basic fact of homological algebra that the cohomology 
of this cochain complex does not depend on the choice of projective resolution; we denote this -module as 
.
as a ![$\mathbb Z[G]$](http://latex.artofproblemsolving.com/9/e/e/9ee0b3c260ab493dc59863e0bc61aee23ffc4361.png)
-module where 
acts trivially, there is a famous free resolution called the bar resolution![\[ \cdots \stackrel{d_3}{\longrightarrow} F_2 \stackrel{d_2}\longrightarrow F_1 \stackrel{d_1}\longrightarrow F_0 \stackrel{\varepsilon}\longrightarrow \mathbb Z \longrightarrow 0, \]](http://latex.artofproblemsolving.com/f/e/1/fe1f4f951f8e9fc9c21137ed198753f08bc6db43.png)
where 
is the module with group structure ![$\mathbb Z[G]^{\otimes (n+1)}$](http://latex.artofproblemsolving.com/7/c/3/7c30b34be12917ee4c0927e18588b2d13dbab87b.png)
and module structure on simple tensors given by 
, 
is the augmentation map (taking each 
to 
), and each 
is given by extending the map
Now consider the -module 
where acts trivially (the fact that this is the same group as is a coincidence), and apply the contravariant functor ![$\operatorname{Hom}_{\mathbb Z[G]} (-, B)$](http://latex.artofproblemsolving.com/8/4/0/840f51e1755bcc1b5eb8fd37ab29476f42e29555.png)
to get a cochain complex,![\[ 0 \stackrel{d^0}\longrightarrow \operatorname{Hom}_{\mathbb Z[G]} (F_0, B) \stackrel{d^1}\longrightarrow\operatorname{Hom}_{\mathbb Z[G]} (F_1, B) \stackrel{d^2}\longrightarrow \operatorname{Hom}_{\mathbb Z[G]} (F_2, B) \stackrel{d^3}\longrightarrow \cdots. \]](http://latex.artofproblemsolving.com/a/f/b/afb639e83d1da3155ef09b8e43d038dc31cd3853.png)
Considering each as a free -module with basis 
for 
, the group structure of ![$\operatorname{Hom}_{\mathbb Z[G]} (F_n, B)$](http://latex.artofproblemsolving.com/3/3/5/335c34f21045a56f65575e13c5c06534026879e3.png)
is isomorphic to abelian group of functions 
under pointwise addition, and the map 
is given by the map
And voilà, since 
acts trivially, we wish to find the size of 
. Now using the definition of 
and the first isomorphism theorem, we get![\begin{align*}
\# \ker d^5 &= \# \operatorname{Ext}_{\mathbb Z[G]}^4 (\mathbb Z, B) \cdot \#\operatorname{im} d^4 \\
&= \# \operatorname{Ext}_{\mathbb Z[G]}^4 (\mathbb Z, B) \cdot \frac{\# \operatorname{dom} d^4}{\#\ker d^4} \\
&\hdots \\
&= \prod_{n=0}^4 \left(\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B) \cdot \#\operatorname{dom} d^n\right)^{(-1)^{4-n}}.
\end{align*}](http://latex.artofproblemsolving.com/9/8/a/98aa21dfea1d9cd6e83bf8c8d694373f20ab9587.png)
Of course, for 
, we have ![$\operatorname{dom} d^n = \operatorname{Hom}_{\mathbb Z[G]} (F_{n-1}, B)$](http://latex.artofproblemsolving.com/3/e/3/3e3399fc9c297be1a263b092b34645e89f1310d2.png)
which has 
elements, and 
, so it suffices to compute ![$\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B)$](http://latex.artofproblemsolving.com/3/c/0/3c0130d5eab67a9ebcfaed74944ecebc430571c9.png)
. To do this, we'll consider a different projective resolution of the -module . Specifically, if 
is a generator of , then we can check that 
satisfies 
, so we have the free resolution![\[ \cdots \stackrel{\times (1 - \sigma)}{\longrightarrow} \mathbb Z[G] \stackrel{\times N}\longrightarrow \mathbb Z[G] \stackrel{\times (1 - \sigma)}\longrightarrow \mathbb Z[G] \stackrel{\varepsilon}\longrightarrow \mathbb Z \longrightarrow 0, \]](http://latex.artofproblemsolving.com/4/6/1/461641d7903c800c934af3f6c548a124fc1d929d.png)
where again takes any to (See the example on Dummit and Foote p. 801). This has corresponding chain complex under ,![\[ 0 \longrightarrow B \stackrel{0}{\longrightarrow} B \stackrel{0}\longrightarrow B \stackrel{0}\longrightarrow \cdots, \]](http://latex.artofproblemsolving.com/9/3/c/93c6ed6ded711437fd58277182e7c1b07aa6216e.png)
(every morphism is 
) under the identification ![$\operatorname{Hom}_{\mathbb Z[G]} (\mathbb Z[G], B) \cong B$](http://latex.artofproblemsolving.com/1/f/1/1f15dd442371087585a50eea0003aecb8a48cee5.png)
since acts as . Thus, ![$\# \operatorname{Ext}_{\mathbb Z[G]}^n (\mathbb Z, B) = |B|$](http://latex.artofproblemsolving.com/3/7/1/371c176fa5099b5f6bb6c10161498fca177b23f3.png)
for all , and we have![\[ \# \ker d^5 = m^{m^3 - m^2 + m^1 - m^0 + 0} \cdot m^{1 - 1 + 1 - 1 + 1} = m^{m^3 - m^2 + m}, \]](http://latex.artofproblemsolving.com/d/6/8/d68fcbe3626a385a7d4185d01d28196f3b10ae8b.png)
where the first term is the contribution from 
and the second term is the contribution from , as desired.
.
