$R=\mathbb{C}[x]/\langle x^2\rangle $ is a PID.

Let $I$ be any ideal of $R$. To show $I=\langle \overline{f(x)}\rangle$ for some $\overline{f(x)}\in R$. Let $\overline{g(x)}\in I \implies g(x)\in \mathbb{C}[x].$ So, there exist unique $h(x)$, $a$ and $b$ such that \begin{align*} g(x)=h(x)x^2+ax+b\implies \overline{g(x)}=a\overline{x}+b \end{align*} Then how to conclude that it is principal. And what are the prime ideals of the ring $R$?


$R$ is not a PID because it is not even a domain: $\bar x \bar x = 0$.

$R$ is a principal ideal ring because it is the homomorphic image of a principal ideal ring.

The prime ideals of $R$ correspond to the prime ideals of $\mathbb{C}[x]$ that contain $\langle x^2\rangle$. Therefore, there is only one non obvious prime ideal in $R$: the one corresponding to $\langle x\rangle$.


As pointed out by lhf in his answer, $R = \Bbb C[x]/\langle x^2 \rangle$ is not a PID, since, denoting the coset $x + \langle x^2 \rangle \ne 0$ by $\tilde x$ etc., we have

$\tilde x^2 = 0 \tag 1$

in $R$; so $R$ is not an integral domain. Nevertheless, it is possible to completely classify all the ideals of $R$: they are $\{0\}$, $R$ (both improper), and $\langle \tilde x \rangle$, the principal ideal generated by $\tilde x$:

If $I$ is a proper ideal of $R$, then $0 \ne \tilde a \notin I$ for any $0 \ne a \in \Bbb C$; otherwise, we would have

$\tilde 1 = \tilde a^{-1} \tilde a \in I, \tag 2$

and thus $I$ cannot be proper, since it contains every $r \in R$: $r = r \tilde 1 \in I$. Now if $I \ne \{0\}$, there is some $0 \ne \tilde a + \tilde b \tilde x \in I$ with $a, b \in \Bbb C$; but then

$\tilde a^2 = (\tilde a - \tilde b \tilde x)(\tilde a + \tilde b \tilde x) \in I; \tag 3$

but we have seen we cannot have $0 \ne \tilde a^2 \in I$ if $I$ is proper; thus $\tilde a = 0$ and every element of $I$ must be of the form $\tilde b \tilde x$, so

$I = \langle \tilde x \rangle; \tag 4$

this works out fine, since for $\tilde c + \tilde d \tilde x \in R$,

$(\tilde c + \tilde d \tilde x)\tilde b \tilde x = \tilde c \tilde b \tilde x + \tilde d \tilde b \tilde x^2 = \tilde c \tilde b \tilde x \in I; \tag 5$

since the only proper ideal of $R$ is the principal ideal $\langle \tilde x \rangle$ (which is in fact maximal, since $R/\langle \tilde x \rangle \cong \Bbb C$, a field), we might call $R$ a principal ideal ring, or PIR for short.


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