Computing projective resolutions over quotients of polynomial rings

Question

I'd like to find projective resolutions for $k$ considered as an $R$-module, where $k$ is a field and $R$ as follows:

(i) $R=\frac{k[x]}{(x^n)}$;

(ii) $R=k[x,y]$

(iii) $R=\frac{k[x,y]}{(x^n,y^m)}$;

(iv) $ R=\frac{k[x,y]}{(x^2,y^2,xy)}$.

I try to find free resolutions, because I don't know any obvious projective non-free modules over these rings:

For (i), I've found: $$ \ldots \rightarrow R \xrightarrow{\cdot x} R \xrightarrow{\cdot x^{n-1}} R \xrightarrow{\cdot x} R \rightarrow k \rightarrow 0$$

For (ii), I've found: $$ 0 \xrightarrow{\cdot x} R \xrightarrow{1\mapsto(y,-x)} R\oplus R \xrightarrow{(1,0)\mapsto x, \ (0,1) \mapsto y} R \rightarrow k \rightarrow 0$$

For (iii), I think it should end similarly to (ii), but I'm not sure what comes before: $$ ? \xrightarrow{} R \xrightarrow{1\mapsto(y,-x)} R\oplus R \xrightarrow{(1,0)\mapsto x, \ (0,1) \mapsto y} R \rightarrow k \rightarrow 0$$

Are these correct?

Thanks in advance.

Answer 1

For (i), your resolution works. Let me call it $P(n) \to \mathbb k$. Your resolution for (ii) works too, it is called the Koszul resolution. Now note that if the corresponding algebra is $A(n)$, then $A(n)\otimes A(m)$ is the algebra in (iii). Since tensoring over $\mathbb k$ is exact, it follows that $P(n)\otimes P(m)$ is a resolution of the trival module in this case. This is a complex that has in degree $d$ the sum $P(n)_i\otimes P(n)_j$ where $i+j=d$.

Finally, let me do (iv) explicitly. This algebra $A$ has dimension 3, spanned by 1,$x$ and $y$, and we can begin the resolution by $A^2\to A\to \mathbb k$ where $f:A^2\to A$ covers $K=\ker(A\to \mathbb k) = (x,y)$ by sending $(p,q)$ to $xp+yq$. The kernel of this is easy to work out: since in this algebra $xp = xp(0,0)$ and $y q= y q(0,0)$, this map has kernel the pairs $(p,q)$ that both vanish at zero. Thus the kernel of $f$ is $K\oplus K$, which we can cover by $f\oplus f$. Repeating, we arrive at a minimal free resolution

$$ \cdots \to A^8\to A^4\to A^2\to A.$$

This is consistent with the fact that we can present $A$ by two free generators $x,y$ and monomial relations $x^2,xy,yx,y^2$, which generate $2^{n+1}$ Anick $n$-chains for each $n\in\mathbb N_0$.

Answer 2

I will give part solution to your question, this is what I can do. For (i), and (ii) the resolution you gave is right. More general, you can get the free resolution of $k[x_1,...,x_n]$ by tensoring $n$-copies $k[x_1,...,x_n]\xrightarrow {x_i}k[x_1,...,x_n]$ then we can get an acyclic complex which is called Koszul complex, and it is a resolution of $k$.

For (iv) $k[x,y]/{(x,y)^2}$ is a finite dimensional local algebra and radical square zero, so any kernel of a projective cover is semisimple. There is only one simple module $k$. If we get the short exact sequence $0\rightarrow k\coprod k\cong (x,y)/{(x,y)^2}\xrightarrow {inc}k[x,y]/{(x,y)^2}\rightarrow k\rightarrow 0$. So the projective dimension of $k$ is infinite.

For (iii), giving a concrete projective resolution is difficult I think. I am expecting someone can give one. Denote $A=k[x,y]/{(x^n,y^m)}$ is finite dimensional local algebra. Next I will give a proof that $A$ is self-injective and projective dimension of $k$ is $\infty$. Remark that $k[x,y]/{(x,y)^2}$ is not self-injective, this can be check directly by Baer's criterion.

In general, suppose $B$ is finite dimensional local commutative k-algebra, where k is a field. If $\mathrm{soc}(B)$ is simple, then $B$ is self-injective.

Proof: $\mathrm{soc}(B)\rightarrow B$ is essential extension. Select a injective envelope of $\mathrm{soc}(B)$, and denote it by $I(\mathrm{soc}(B))$, so there is a natural induced map $\alpha :B\rightarrow I(\mathrm{soc}(B))$. This is a monomorphism since essential extension. Remark $D=\mathrm{Hom}_k(-,k):B-\mathrm{mod}\rightarrow B-\mathrm{mod}$ is equivalence between finite generated modules. So $D(I(\mathrm{soc}(B))$ and $B$ are both projective covers of the unique simple module of $B$. Hence they have the same length. So $\alpha $ is isomorphism.

The global dimension of a self-injective algebra is $0$ or $\infty$.

So we know projective dimension of $k$ as $A$ is infinite since there is only one simple module $A-\mathrm{mod}$ and $A$ is finite dimensional algebra.