# Wednesday, February 01

:::{.exercise title="?"}
Show $\HoH^* k[x] = k[x]\sumpower{2}$ and find $\HoH_* k[x]$.
Use the complex
\[
k[x]\sumpower{2} \injects k[x]\sumpower{2} \surjects k[x]
.\]
:::

:::{.example title="?"}
Let $A = k[x]/\gens{x^n}$ and consider
\[
\cdots \mapsvia{v} A^e \mapsvia{u} A^e \mapsvia{v} A^e \mapsvia{u} A^e \mapsvia{\pi} A \to 0
\]
where $u = (x\tensor 1 - 1\tensor x)\cdot$ and $v = \qty{ (x^{n-1}\tensor x^0) + (x^{n-2} \tensor x^1) + (x^{n-3}\tensor x^2) + \cdots + (x^0 \tensor x^{n-1})} \cdot$.
Compute $uv(x^i \tensor x^j) = 0$ and $vu= 0, \pi u = 0$ to verify that this is a complex.
Show it is exact using the contracting homotopy $s_{-1}(1) = 1\tensor 1$ and
\[
s_{2m}(1\tensor x^j) = - \sum_{\ell =1}^j x^{j-\ell} \tensor x^{\ell-1},\qquad s_{2m-1}(1\tensor x^j) = \delta_{j, n-1}\tensor 1
.\]
Apply $\Hom_{A^e}(\wait, A)$ to get 
\[
0 \to A \mapsvia{u^*} A \mapsvia{v^*} A \mapsvia{u^*} \cdots
,\]
using $\Hom_{A^e}(A^e, A) \cong A$ via $f\mapsto f(1\tensor 1)$.
Show that $u^*(a) = 0$ for $a\in A$ corresponding to $f_a$ where $f_a(1\tensor 1) = a$: 
\[
u^*(a) = u^*(f_a(1\tensor 1)) = (u^* f_a)(1\tensor 1) = f_a(u(1\tensor 1)) = f_a(x\tensor 1 - 1\tensor x) = x f_a(1\tensor 1) - f(1\tensor 1)x = xa-ax = 0
\]
and similarly
\[
v^*(a) = v^*(f_a(1\tensor 1)) = (v^* f_a)(1\tensor 1) = f_a v(1\tensor 1) = f_a(x^{n-1}\tensor 1 + \cdots + 1\tensor x^{n-1}) = x^{n-1} f_a(1\tensor 1) + \cdots + f_a(1\tensor 1) x^{n-1} = x^{n-1} a + x^{n-2}ax + \cdots + ax^{n-1} = nx^{n-1} a
.\]
This yields
\[
0 \to A \mapsvia{0} A \mapsvia{nx^{n-1}\cdot} A \mapsvia{0} A \to \cdots
.\]

So the homology depends on if $\characteristic k \divides n$: 

- If so, $HH^* A = A + \sum_{n\geq 0} \gens{x}t^{2n+1} + \sum_{n\geq 0} A/\gens{x^{n-1}}$.
- If not, check!

:::


:::{.exercise title="?"}
How can you interpret $\HoH(A; M)$ in low degrees?

:::