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Let $R=k[x,y]$, $\mathfrak{a}=(x,y)$ be an ideal of $R$, how to compute $\text{Tor}_i^R(k,\mathfrak{a}^n)$ for all $i\ge 0$ and $n\ge 1$? Here $\mathfrak{a}^n$ denotes the ideal obtained by taking product of $\mathfrak{a}$ with itself, $n$ times.

Thanks for your help.

Ryze
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1 Answers1

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The Koszul complex gives a free resolution of $k$: $$0\to R\to R^2\to R\to k\to0$$ where the map $R^2\to R$ is given by $(f,g)\mapsto xf+yg$ and $R\to R^2$ is given by $h\mapsto (y,-x)$. So $\textrm{Tor}_*^R(k,M)$ is the homology of the complex $$0\to M\to M^2\to M\to0$$ where the maps are $m\mapsto(ym,xm)$ and $(m_1,m_2)\mapsto xm_1+ym_2$. Now you just have to set $M=\mathfrak a^n$.

Angina Seng
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  • Thanks for your help! – Ryze Apr 06 '20 at 13:00
  • I found that “$\text{Tor}_0^R(k,\mathfrak{a}^n)=\mathfrak{a}^n/{xm_1+ym_2|m_1,m_2\in \mathfrak{a}^n}$, $\text{Tor}_1^R(k,\mathfrak{a}^n)={(m_1,m_2)|xm_1+ym_2=0, m_1, m_2\in\mathfrak{a}^n}$, $\text{Tor}_i^R(k,\mathfrak{a}^n)=0, i\ge 2$. Could you please tell me how to simplify them? – Ryze Apr 08 '20 at 09:33
  • What you have these is $\text{Tor}_0$ is $a^n/a^{n+1}$. Indeed $\text{Tor}_0(k,M)=k\otimes_RM=M/aM$. – Angina Seng Apr 08 '20 at 09:35
  • Also $\text{Tor}_1$ is what you've written but factored out by the image of $M\to M^2$. I think it comes out as zero, but you should check it.... – Angina Seng Apr 08 '20 at 09:38
  • Yes, I forgot to type the image, sorry. I think in the last line of your answer, $m\mapsto (ym,xm)$ should be $m\mapsto (ym,-xm)$. Right? – Ryze Apr 08 '20 at 09:56