# Calculating homology objects in this chain complex

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.

## 2 Answers

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 $$xR$$?

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 $$bx +(x^2) \mapsto b$$.