# Wednesday, February 22

:::{.proposition title="Kunneth formula"}
For $C, D \in \Ch(\kmod)$,
\[
H_n(C\tensor D) = \bigoplus _{i+j=n} \qty{ H_i(C) \tensor_k H_j(D) }
.\]
:::

:::{.example title="?"}
For $C = k[y]t^0 + k[y]t$ and $D = k[x]t^0 + k[x]t$ with connecting maps $\cdot y, \cdot x$ respectively, the total complex is 

\begin{tikzcd}
	&& 0 && 0 \\
	\\
	0 && {k[y]\tensor k[x]} && {k[y]\tensor k[x]} && 0 \\
	\\
	0 && {k[y]\tensor k[x]} && {k[y]\tensor k[x]} && 0 \\
	\\
	&& 0 && 0
	\arrow[from=5-3, to=7-3]
	\arrow[from=5-1, to=5-3]
	\arrow[from=5-3, to=5-5]
	\arrow[from=5-5, to=5-7]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=1-3, to=3-3]
	\arrow[from=3-3, to=5-3]
	\arrow[from=1-5, to=3-5]
	\arrow[from=3-5, to=5-5]
	\arrow[from=5-5, to=7-5]
	\arrow[from=3-5, to=3-7]
\end{tikzcd}
> [Link to Diagram](https://q.uiver.app/?q=WzAsMTIsWzIsMiwia1t5XVxcdGVuc29yIGtbeF0iXSxbNCwyLCJrW3ldXFx0ZW5zb3Iga1t4XSJdLFsyLDQsImtbeV1cXHRlbnNvciBrW3hdIl0sWzQsNCwia1t5XVxcdGVuc29yIGtbeF0iXSxbNiwyLCIwIl0sWzQsMCwiMCJdLFsyLDAsIjAiXSxbMCwyLCIwIl0sWzAsNCwiMCJdLFsyLDYsIjAiXSxbNCw2LCIwIl0sWzYsNCwiMCJdLFsyLDldLFs4LDJdLFsyLDNdLFszLDExXSxbNywwXSxbMCwxXSxbNiwwXSxbMCwyXSxbNSwxXSxbMSwzXSxbMywxMF0sWzEsNF1d)

Taking diagonals yields
\[
(k[y] \otimes k[x])t^2 + (k[y]\tensorpowerk 2 \tensor k[x] \tensorpowerk 2) t + (k[y] \tensor k[x])t^0
.\]
Since $k[x]\tensor_k k[y] \cong k[x,y]$, this is isomorphic to

\[
k[x,y]t^2 + k[x,y]\sumpower 2 t + k[x,y] t^0
.\]

Note that $H_*(C) = kt^0 = H_*(D)$, so $H_*(C\tensor D) = (k\tensor_k k) t^0 = kt^0$.
:::

:::{.remark}
There is a free and thus projective resolution of $k$ in $\mods{k[x,y]}$: 
\[
0 \to k[x,y] \to k[x,y]\sumpower{2} \to k[x,y]\to k \to 0
.\]
:::

:::{.remark}
For $A, B\in \gr\kalg$, define $A\tensor_k B\in \kalg$ by 
\[
(a_1 \otimes b_1)(a_2 \otimes b_2) \da (-1)^{\abs{a_1} \abs{b_1} }a_1 a_2 \otimes b_1 b_2
.\]
If $M\in \bimods{A}{A}$ and $N\in \bimods{B}{B}$, note that $M\tensor N\in \bimods{A\tensor_k B}{A\tensor_k B}$ using 
\[
(a_1 \otimes b_1)(m\tensor n) \da a_1 m a_2 \otimes b_1 n b_2
.\]
:::

:::{.lemma title="?"}
If $M\in \mods{A^e}$ and $N\in \mods{B^e}$ are projective, then $M\tensor_k N \in \mods_{(A\tensor B)^e}$ is again projective.
:::

:::{.proof title="?"}
Write $M \oplus M' = (A^e)\sumpower I$ and $N \oplus N' = (B^e)\sumpower J$, then
\[
M \oplus N \oplus (M' \otimes N \oplus M \otimes N' \oplus M' \otimes N') 
&= (A^e \tensor_k B^e)\sumpower{I\times J} \\
&\cong ( (A\tensor B)^e )\sumpower{I\times J}
.\]
:::

:::{.lemma title="?"}
If $P\covers A$ is a projective resolution of $A$ in $\mods{A^e}$ and $Q\covers B$ a projective resolution in $\mods{B^e}$, then $P\tensor Q \cover A\tensor B$ is a projective resolution in $\mods{(A\tensor B)^e}$.
:::

:::{.proof title="?"}
By the previous lemma, $P_i \otimes Q_j\in \mods{(A \otimes B)^e}$ is projective for each $i$ and $j$, and 
by Kunneth, $H_0(P \otimes Q) = H_0(P) \otimes H_0(Q) \cong A \otimes B$.
:::

:::{.theorem title="?"}
Let $A, B\in \kalg$ and suppose $\exists P\covers A, Q\covers B$ are free resolutions in $\mods{A^e}$ and $\mods{B^e}$ respectively with each $P_i\in \mods{A^e}^\fg$ and $Q_i\in \mods{B^e}^\fg$.
Then 
\[
\HoH^*(A\tensor B) \cong \HoH^*(A) \tensor \HoH^*(B)
.\]
:::

:::{.proof title="?"}
Since $P\tensor Q \covers A\tensor B$ is a free resolution, we have $\HoH^*(A\tensor B) = H_*( \Hom_{(A \otimes B)^e}(P\tensor Q, A\tensor B))$.
Note that $\Hom_{A^e}(A\tensor S \tensor A, \wait) \cong \Hom_k(S, \wait)$ for any $S$ and we can write $P_i \tensor Q_j \cong ((A\tensor B)^e)\sumpower{I\times J}$.
There are some finitely-generated $P_i'$ and $Q_i'$ such that $P_i \cong A \tensor P_i' \tensor A$ and $Q_j \cong B \tensor Q_j' \tensor B$, so 
\[
\Hom_{(A\tensor B)^e}(P_i \otimes Q_j, A\tensor B) \cong \Hom_k(P_i' \otimes Q_j', A \otimes B) \cong \Hom_k(P_i^', A) \otimes \Hom_k(Q_j', B)
.\]
This gives $\Hoh(A) \tensor \HoH(B)$ as a $k\dash$module, and the claim is that this is an isomorphism of algebras.
Let $\Delta_A: P\to P\tensorpower{A}{2}$ and $\Delta_B: Q\to Q\tensorpower{B}{2}$, then one can define $\Delta: P \otimes Q \to (P \otimes Q)\tensorpower{A\tensor B}{2}$ using 
\[
P \otimes Q \mapsvia{\Delta_A \times \Delta_B} P \tensorpower A 2 \tensor Q\tensorpower B 2 \iso (P\tensor Q) \tensorpower{A\tensor B}{2} \\
(x \otimes _A x') \tensor (y \otimes _B y') \mapsto (-1)^{\abs y \abs x} (x \otimes y) \otimes _{A \otimes B} (x' \otimes y')
.\]

Now for $f,g$ homogeneous elements in degrees $m', n'$ resp. in $\Hom_{B^e}(Q, B)$, consider the element in $\HoH(A) \otimes \HoH(B)$ represented by $f \otimes f'$ in
\[
\Hom_{A^e}(P_m, A) \otimes \Hom_{B^e}(Q_{m'}, B) \cong \Hom_k(P_m', A) \otimes \Hom_k(Q'm, B) \cong \Hom_k(P_m' \tensor Q_{m'}', A \otimes B) \cong \Hom_{A \otimes B}(P_m \otimes Q_{m'}, A \otimes B)
.\]
Letting $x\in P_r$ and $y\in Q_s$, suppose \( \Delta_A(c) = \sum_r x_r' \otimes x_r'' \) in $P_m \tensor P_n$ and \( \Delta_B(y) = \sum_t y_t' \otimes y_t'' \).
:::