Quotient ring and PID Is $ \ \mathbb{Q}[x] / ((x-1)^{2}) $  a Prinicipal Ideal domain and Euclidean domain ? $$ $$ I have tried in the this way- Since $ \mathbb{Q}[x] /(x-1)^{2} \sim \mathbb{Q} [x]/(x) $ . So is a PID.  I am not sure about that . Any help is appreaciating
 A: $\mathbb{Q}[x] / ((x-1)^{2})$ is not a PID because it is not even a domain.
Nevertheless, all its ideals are principal because they correspond to the ideals of $\ \mathbb{Q}[x]$ that contain $((x-1)^{2})$.
Therefore, $\mathbb{Q}[x] / ((x-1)^{2})$ is a principal ideal ring.
A: All PID's and Euclidean domains are also integral domains$^\dagger$.  So if $\mathbb{Q}[x]/\langle (x-1)^2 \rangle$ is an example of these, then we should not be able to find two nonzero elements that multiply to zero.
But the polynomial that generates the ideal that we're modding out by is reducible!  With this in mind, can you find two nonzero elements that will multiply to give zero in the quotient ring?

If we did have $\mathbb{Q}[x]/ \langle (x-1)^2 \rangle \cong \mathbb{Q}[x]/\langle x \rangle$, then the latter is a field and you'd be right (all fields are Euclidean domains & PIDs).  Unfortunately, this cannot be true since $F[x]/ \langle f(x) \rangle$ is a field $\iff f(x)$ is irreducible over $f$ (click here for further discussion).  In this scenario, $x$ is irreducible over $\mathbb{Q}$, but $(x-1)^2$ is not.

$\dagger$ You can find a useful set of inclusions at the bottom of the introductory paragraphs here.  Try proving some of them.  :)
