Subring of PID is also a PID?

abstract-algebra ring-theory

Answer: no. $\mathbb{Q}[X]$ is a PID, whereas $\mathbb{Z}[X]$ is not. See this

Related

Proof of the power rule for logarithms
Why $P(A) \cup P(B)$ is not equivalent to $P(A \cup B)$
$x_n$ is the $n$'th positive solution to $x=\tan(x)$. Find $\lim_{n\to\infty}\left(x_n-x_{n-1}\right)$
Linear independence of vectors that form non-square matrix
Compute $\int_0^\infty \left(\frac{1}{x}\right)^{\ln(x)}\,\mathrm{d}x$ [closed]
about partial derivatives of implicit functions
Why $(a+d)(b+c)$ equals $-16$?
Derivation of the termwise Hadamard product of two generating functions.
Proving that $a,n$ and $b, n$ relatively prime implies $ab,n$ relatively prime
If $f:D\to D’$ is analytic and $u: D'\to R$ is harmonic then the composition of $u$ and $f$ is harmonic in $D$
Proof, is $\forall x : (P(x) \lor Q(x)) \Leftrightarrow \forall x : P(x) \lor \forall x : Q(x)$?
If $(F:E)<\infty$, is it always true that $\operatorname{Aut}(F/E)\leq(F:E)?$