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Problem: Let $R = k[x]/(x^2) $ where $k$ is a field and consider the chain complex $$C : \qquad 0 \xrightarrow{d_2} R \xrightarrow{d_1} R \xrightarrow{d_0} 0 $$ where $d_1 : R \rightarrow R: f \mapsto xf. $ Calculate the homology objects of the complex $C$.

Attempt: I calculated $$ H_0 (C) = \frac{ \ker(d_0)}{ \text{Im}(d_1)} = \frac{R}{xR} $$ where $xR := \left\{ xf + (x^2) \mid f \in k[x] \right\}. $

Furthermore, we have $$ H_1 (C) = \frac{ \ker(d_1)}{ \text{Im} (d_2)}. $$ Now $\text{Im}(d_2) = \left\{0 \right\}$ and $\ker(d_1) = \left\{ f + (x^2) \mid xf \in (x^2) \right\}$. So that means $f$ cannot have a constant term, and it must be a polynomial in $x^{k}$ with $k$ odd, since $xf$ must be divisible by $x^2$. Then $$ H_1(C) \cong \ker(d_1) = \left\{ f + (x^2) \mid f = \sum_{i=1, i = odd}^n a_i x^{i} \right\}. $$

Is this correct reasoning? Thank you in advance.

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Yes and no. Your presentation of $H_0(C)$ is correct, but that of $H_1(C)$ is overly complicated: Use that $k[x]$ is a unique factorization domain: For any $f ∈ k[x]$, $$xf ∈ (x^2) \iff x^2 \mid fx \iff x \mid f \iff f ∈ (x).$$ Hence $\ker d_1 = \{f + (x^2);~f ∈ (x)\} = xR = (x)/(x^2)$. You don’t need the monomials within the representing polynomials $f$ to be odd here.

So the homology groups are $R/xR$ and $xR$.

I also think that you are supposed to further calculate $H_0$ and $H_1$ here. For example, you can use an isomorphism theorem to show $$R/xR = \frac{k[x]/(x^2)}{(x)/(x^2)} \cong \frac{k[x]}{(x)} = k,$$ as $R$-modules. Can you do something similar with $xR$?

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Your reasoning is correct, but your answers are not in the most simplified form, I would say.

The kernel in $H_1$ is $(x)/(x^2)$, since every element of $R$ has an expression of the form $a + bx + (x^2)$, with $a$ and $b$ in $k$, and the only elements of $R$ that are annihilated by $x$ (i.e. in the kernel of $d_1$) are those of the form $bx + (x^2)$. Moreover, $(x)/(x^2) \cong k$, the isomorphism this time being given by $bx +(x^2) \mapsto b$.

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