If A is a graded algebra, $A=\oplus A_i$ and $A_0=\mathbb{K}$ then $\mathbb{K}$ is a A-module with multiplication $a*k=ak$ for $a\in A_0$ and $a*k=0$ for $a\in \bar{A}=\oplus_{n\geq1}A_i$. My question is if the following is a free resolution of $\mathbb{K}$ $$...\to0\to \bar{A}\langle e_1\rangle\to_{f} A\langle e_1\rangle \to_{g} \mathbb{K}\to 0$$ where $f(e_1)=e_1$ and $g(e_1)=1$? If it is, can I use this to find $Tor_i^{A}(\mathbb{K},\mathbb{K})$? Sorry if this makes little sense it is new to me.
Free resolution of k as an A-module
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modules
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0No way! Btw, why is $\overline Ae_1$ free $A$-module? – 2017-02-08
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0Ok I see now that it's not, I thought $e_1$ was a basis for $\bar{A}e_1$ but it's not even in $\bar{A}e_1$.. Thanks for your comment! – 2017-02-09