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Let $A$ be a $k$-algebra where $k$ is a field. Define $C^n(A):=\text{Hom}_k(A^{\otimes n+1}, k)$, where $A^{\otimes n+1}$ is the $n$-fold $k$ tensor product of $A$ with itself. Then the cyclic cohomology $HC^{\ast}(A)$ of $A$ is the cohomology of the total complex of $\require{AMScd}$ \begin{CD} \cdots @. \cdots @. \cdots \\ @AAA @AAA @AAA \\ C^2(A) @>B>> C^1(A) @>B>> C^0(A) \\ @AbAA @AbAA \\ C^1(A) @>B>> C^0(A) \\ @AbAA \\ C^0(A) \end{CD} where $b:C^n(A)\to C^{n+1}(A)$ is the Hochschild coboundary map and $B:C^{n+1}(A)\to C^{n}(A)$ is given by \begin{align} B(f)(a_0\otimes\dots\otimes a_n) &= \sum_{i=0}^n(-1)^{ni}f(1\otimes a_i\otimes\dots\otimes a_n\otimes a_0\otimes\dots\otimes a_{i-1}) \\ &\quad -(-1)^{ni}f(a_i\otimes 1\otimes a_{i+1}\otimes\dots\otimes a_n\otimes a_0\otimes\dots\otimes a_{i-1}) \end{align} I'm trying to figure out what $HC^{\ast}(k)$ is. According to Loday in Cyclic Homology (pg. 74) it is "immediate" that $HC^{2n}(k) = k$ and $HC^{2n+1}(k) = 0$ for $n\ge 0$. Unfortunately though, this isn't immediate to me and he doesn't give any other details. I know that when $A = k$ then the complex above becomes \begin{CD} \cdots @. \cdots @. \cdots \\ @A0AA @AidAA @A0AA \\ C^2(k) @>0>> C^1(k) @>0>> C^0(k) \\ @AidAA @A0AA \\ C^1(k) @>0>> C^0(k) \\ @A0AA \\ C^0(k) \end{CD} since $C^n(k)\simeq C^0(k)$ for all $n\ge 0$. So the Hochschild coboundary $b:C^n(k)\to C^{n+1}(k)$ is identity when $n$ is odd and the zero map when $n$ is even. The map $B$ is always the zero map.

I understand why the cyclic cohomology is $2$-periodic, and I understand why the odd cyclic cohomology is $0$. What I don't understand is why $HC^{2n}(k) = k$. Reading straight off the complex you get $HC^0(k) = \text{ker}(0) = C^0(k) = \text{Hom}_k(k, k)$, but I don't understand how to identify $\text{Hom}_k(k, k)$ with $k$. From Loday's use of the word "immediate" I'm guessing there is some quick trick that makes identifying these two a simple task? If there is one I don't know what it is. Any help would be appreciated!

Edit: I've had a thought about how to show that $\text{Hom}_k(k, k)\simeq k$. So since $\text{Hom}_k(k, k)$ has the structure of a $k$-bimodule and $k$ is a $k$-module, then for each $f\in \text{Hom}_k(k, k)$ and $a\in k$ we have

$$f(a) = f(a1) = af(1)$$

so that each $f$ is uniquely determined by how it acts on $1$. Then for any $a^\prime\in k$ define the map $f_{a^\prime}(a) = aa^\prime$. This is a well defined $k$-module homomorphism. Then we have

$$f_{a^\prime}(a) = af_{a^\prime}(1)$$

and since $k$ is a field then every $a\in k$ has a multiplicative inverse $a^{-1}$ and so

$$f_{a^\prime}(1) = a^{-1}af_{a^\prime}(1) = a^{-1}aa^\prime =a^\prime $$

So the mapping that sends $f\mapsto f(1)$ is a module isomorphism. Is this a valid argument?

Edit 2: Think I may have found a more elegant solution. So, Theorem 2.4 in these notes states that if $M$ is a free $k$-module of finite rank $n$ then the dual space $M^{\lor}:=\text{Hom}_k(M, k)$ is also a free $k$-module of rank $n$. Therefore $k^{\lor}=\text{Hom}_k(k, k)$ is a rank $1$ free $k$-module, which is isomorphic to $k$.

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Posting my second edit as an answer to this question.

Theorem 2.4 in these notes states that if $M$ is a free $k$-module of finite rank $n$ then the dual space $M^{\lor}:=\text{Hom}_k(M, k)$ is also a free $k$-module of rank $n$. Therefore $k^{\lor}=\text{Hom}_k(k, k)$ is a rank $1$ free $k$-module, which is isomorphic to $k$.

Therefore $$HC^{2n}(k) = \text{Hom}_k(k, k) \simeq k$$ as required.

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  • $\begingroup$ $\text{Hom}_k(k, k) \cong k$ because a $k$-linear map $k \to k$ is uniquely and freely determined by what it does to the basis vector $1 \in k$. $\endgroup$ Oct 8 '20 at 0:56
  • $\begingroup$ @QiaochuYuan yeah I actually make that same argument in one of my edits! I put this as my answer because I thought it was a little more succinct. $\endgroup$
    – SeraPhim
    Oct 8 '20 at 10:42

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