I was wandering about the site, looking for interesting questions, and I stumbled upon an answer that I can't quite undertand. Unfortunately its creator isn't active on the site anymore so I cannot ask him.
He states that the localization of $k[x]/(x^2)$ at prime ideal $(x)$ leaves the dimension (as $k$-vector space) unchanged, but on the other hand that doing the same operation in $k[x]/(x^2-x)$ gives a dimension one module. I am struggling to see why this is true.
I think that $k[x]/(x^2)$ and $k[x]/(x^2-x)$ are both isomorphic as $k[x]$-modules, because both of them are isomorphic to $\{a_0+a_1x\}$. Please correct me if I am wrong in this statement. Now I move onto the localization part. For example, we have that $x-1\notin (x)$, so I would say that an element such as $1/(x-1)$ is in the localization of $k[x]/(x^2)$ at $(x)$ but not on $k[x]/(x^2)$ itself. So I struggle to see why the dimension is still the same in this case. In the case of $k[x]/(x^2-x)$ I get a similar reasoning. What am I doing wrong?
EDIT I would appreciate if someone could tell me how these localizations look like.
First of all, $k[x]/(x^2)$ and $k[x]/(x^2-x)$ are not isomorphic as modules. They both have dimension $2$ as vector spaces, and, as you say, they both are generated by the cosets of $1$ and $x$.
But the module structure also includes multiplication by elements of $k[x]$, and this makes the modules not isomorphic. As @Dustan Levenstein notes in the comment, on $k[x]/(x^2)$ multiplication by $x$ is represented by the matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$, while on $k[x]/(x^2-x)$ multiplication by $x$ is given by a matrix $\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$, and these maps do not transform into each other under basis change.
Now to localization. The element $\frac{1}{x - 1}$ is indeed in the localization $k[x]/(x^2)_{(x)}$, but remember that two fractions can be equal. In this case we have $\frac{1}{x - 1} = \frac{-1 - x}{1}$, because $$(x - 1) (-1-x) = (1 - x)(1+x) = 1 - x^2 = 1\cdot 1\text{ in $k[x]/(x^2)$}$$ In general, $$\frac{a + bx}{c + dx} = \frac{(a + bx)(c^{-1} - c^{-2}d\cdot x)}{1},$$ so every element of the localization $k[x]/(x^2)_{(x)}$ can be transformed into an element of the form $\frac{m}{1}$. Thus, localization does not give us anything new, and we have $k[x]/(x^2) \cong k[x]/(x^2)_{(x)}$ via $m \mapsto \frac{m}{1}$.
For the localization $k[x]/(x^2-x)_{(x)}$ in addition to that we also need to worry about annihilators. Remember that by definition $\frac{m}{a} = \frac{n}{b}$ if $c(mb - na) = 0$ for some $c$ in the multiplicative set with respect to which we localize (in our case, $c \notin (x)$). In contrast to the previous case, we have $(x - 1)x = 0$ in $k[x]/(x^2-x)$, and this means $\frac{x}{1} = \frac{0}{1}$ and more generally $\frac{a + bx}{c + dx} = \frac{a}{c} = \frac{ac^{-1}}{1}$, so the localization is isomorphic to $k$.