For an algebra $A$ we can define its Hochschild cohomology $HH^{\bullet}(A,A)$ as in this wikipedia page.

Now let $A=k[x]$ be the polynomial ring where $k$ is a field. It is well-known that $HH^{0}(A,A)\cong k[x]$ and $HH^1(A,A)\cong k[x]$ and $HH^i(A,A)=0$ for $i\geq 2$. But all the computations I can find is through the Koszul resolution, for example in Weibel's An introduction to Homological algebra Chapter 9.

If we use the original Hochschild complex, then it is still easy to see that $HH^{0}(A,A)\cong A$ and $HH^1(A,A)\cong A$ but it is not so straightforward to me that $HH^i(A,A)=0$ for $i\geq 2$.

My question is: can we compute $HH^{\bullet}(A,A)$ through the Hochschild complex? Is there any reference?

  • 3
    $\begingroup$ It's not hard to prove that $HH^2$ with coefficients in any bimodule is zero because k[x] is a free algebra so any square-zero extension splits. Thus k[x] has projective dimension 1 as a bimodule and so all its higher cohomology vanishes for any coefficients. This argument works for free algebras as well and with a little extra work for path algebras of quivers $\endgroup$ Feb 17, 2022 at 4:46
  • $\begingroup$ But why do you want to carry out the computation with the Hochschild complex? $\endgroup$
    – Pedro
    Feb 17, 2022 at 7:10
  • $\begingroup$ @PedroTamaroff I am trying to deform $k[x]$ to a curved dg-algebra with trivial differential but non-trivial curvature and then compute its Hochschild cohomology. In this case I'm afraid that the Koszul resolution doesn't work so we have to stick to the original definition. Hence I am curious how to compute the Hochschild cohomology of $k[x]$ through the original definition. $\endgroup$ Feb 17, 2022 at 14:32

1 Answer 1


I am not sure how helpful this may be, but it is too long for a comment.

There may be no reference where people carry out the computation using the usual Hochschild complex because no one wants to use that complex, in general, to compute the Hochschild cohomology groups of an algebra, when better resolutions exist.

For free algebras (or more generally quiver path algebras with no relations over a semi-simple base) there is a canonical bimodule resolution of $A=TV$ of the form (starting at degree $-1$)

$$0\to A\otimes V\otimes A\to A\otimes A\to A \to 0$$

where the leftmost map is $d(x\otimes v\otimes y) = xv\otimes y - x\otimes vy$ and the second map is the product.

Let me call this complex $L$. Note that there is an obvious inclusion $\iota:L\to B$ where $B$ is the double sided bar construction of $A$, and this inclusion is a chain map. In the other direction, there is a map $\pi : B\to L$ that is the identity in degree $-1$ and $0$, on $A^{\otimes 3}$ acts as follows:

$$ \pi(f\otimes v_1\cdots v_n\otimes g) = \sum fv_1\cdots v_j\otimes v_{j+1}\otimes v_{j+2}\cdots v_n g\in A\otimes V\otimes A, $$ while it is necessarily zero in the other remaining degrees. You can check that $\pi$ is a chain map, and $\pi \iota=1$. It is a bit more tricky to check that $\iota\pi$ is chain homotopic to $1$. In the end, what you get is that $\hom_{A^e}(B,A)$ and $\hom_{A^e}(L,A)$ are chain homotopy equivalent.

Of course $\hom_{A^e}(L,A)$ is naturally isomorphic to $$0\to A\to \hom(V,A) \to 0$$ where the only non-zero map sends $a\in A$ to $V\to A$ such that $v\longmapsto [a,v]$, giving in your case $\mathrm{HH}^0(A) = \mathrm{HH}^1(A) = A$ as $V$ is one dimensional. What you'd have to do now is figure out how to lift these obvious cycles to representative cycles in $\hom_{A^e}(B,A)$, but you can do this precomposing with the chain map $\pi$.


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