This is from some homological algebra notes. You are given that $R = k[x]/(x^3)$, $k$ is a field, and the complex $$C_.= \cdots R \to R \to R \to \cdots $$ where our differential map $\delta_n$ is multiplication by $x^2$. Then $\text{im}\,\delta_{n+1} = (x^2)$ and $\text{ker}\,\delta_n = (x)$, which contains $(x^2)$.
Then $\bf{H_n(C_.) = (x)/(x^2) \cong R/(x)}$.
The last line is where I'm stuck. I suspect there is a lot of abuse of notation, and I am trying to figure out how this is explicitly isomorphic. Firstly, when they say $(x)$ and $(x^2)$, do they mean $\overline{(x)}$ and $\overline{(x^2)}$, where each is an ideal of the quotient ring $R$? Similarly, shouldn't it be $R/\overline{(x)}$, where $\overline{(x)}$ is the ideal of $R$? I'm also having difficulty understanding the elements of $R/\overline{(x)}$. I believe that $R/\overline{(x)} = \{\overline{p(x)} +(x) \mid \overline{p(x)} \in R\} = \{(p(x)+(x^3))+(x)\}$.
Now I know that if I use the first isomorphism theorem, then I want some map $$\overline{(x)} \to R/\overline{(x)}$$ such that the kernel is $\overline{(x^2)}$ and the map is onto. But I'm confused about the zero element in $R/\overline{(x)}$.
My thought was to use the map that sends $\overline{p(x)}$ to $\overline{q(x)} + (x) = (q(x) +(x^3))+(x)$, where $p(x)=xq(x)$, i.e., the polynomial we get if we factor out an $x$. Since $(x^k) \subseteq (x^{k-1})$, if the smallest degree of $p(x)$ is greater than or equal to $2$, then $p(x) \in (x^2)$, which means that $q(x) \not \in (x^3)$, but $q(x) + (x^3) \in (x)$. If the smallest degree of $p(x)$ is 1, then if we factor out an $x$, $q(x)$ has a constant and so $q(x) + (x^3) \not \in (x)$. If this is the right thinking, then the kernel of my map is $\overline{(x^2)}$. I think it would also be onto because if $\overline{a(x)} \in R/\overline{(x)}$ then $\overline{a(x)} = \overline{f(x)} +(x) = (f(x) +(x^3)) +(x)$, where $f(x) \in k[x]$. If we assume that $f \neq 0$, then if we multiply $f(x)$ by $x$ then $xf(x) \in (x)$. So our map is onto.
This may be completely wrong. Any feedback would be much appreciated!!!
It is pretty standard when dealing with a ring like $k[x]/(x^3)$ to abuse notation and write $x$ for what is really the coset determined by $x$. Therefore I hope it is clear that $R/(x)$ is isomorphic to $k$. For example, the composite $k \to R \to R/(x)$ is an isomorphism. You can think of $R$ as consisting of polynomials of degree at most 2, and the map $R \to R/(x)$ is the map sending $x \mapsto 0$ — that is, it is the same as the map $R \to k$ which is obtained by evaluating a "polynomial" $a + bx + cx^2 \in R$ at $x=0$.
When I say "you can think of $R$ as consisting of polynomials of degree at most 2," this can be made precise: every element of $R$ is represented by a coset of the form $a + bx + cx^2 + (x^3)$, and this representation is unique: every "polynomial" $a + bx + cx^2$ determines an element of $R$, and different degree two polynomials determine different elements. So we can think of $R$ as consisting of elements of the form $a + bx + cx^2$, with multiplication defined by ordinary polynomial multiplication plus the rule that $x^3 = 0$. No need to actually write explicit cosets: just define $R = \{a + bx + cx^2 \,|\, a, b, c \in k\}$.
So then the ideal $(x) \subset R$ consists of elements $ax + bx^2$, and $(x^2)$ consists elements $bx^2$, so the quotient $(x) /(x^2)$ is just the equivalence class of elements $ax$ — again isomorphic to $k$.