# Tagged Questions

For questions about principle ideal domains: rings without zero divisors where every ideal is principle.

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### Is $\mathbb{Q}$ a principal ideal domain?

Is $\mathbb{Q}$ a principal ideal domain? I know that $\mathbb{Z}$ is, but I'm not sure about $\mathbb{Q}$.
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### About the rank of submodules over PID

If $M$ is a free $R$-module of finite rank $n$ and $R$ is a PID, do proper submodules of $M$ have strictly less rank than $M$? I know that in this case, every submodule of $M$ is free and has ...
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### Does torsion-free implies flat over a “locally principal” domain?

Let $R$ be a commutative ring. An $R$-module $M$ over $R$ is said to be torsion free if for every $r\in R$ which is not zero divisor and for every $0\neq m\in M$, we have $r\cdot m\neq 0$. I know ...
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### Ideal class group of $\mathbb{Z}[ \sqrt{2} ]$

How does one compute the ideal class group for $\mathbb{Z}[\sqrt{2}]$? Motivation: I wish to prove that $\mathbb{Z}[\sqrt{2}]$ is a PID. I have seen proofs which use norm and go on to show that it ...
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### Why is $(2,x)$ non-principal in $\Bbb Z[x]$? [duplicate]

Why is $(2,x)$ non-principal in $\Bbb Z[x]$? Apparently this is the case, I just read it on wiki, as a counter example to $\Bbb Z[x]$ being a PID. What is $x$ here? I mean $2$ can surely generate ...
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### Proof about finitely generated torsion-free R-module M is free, where R is a PID

Proof about finitely generated torsion-free $R$-module $M$ is free, where $R$ is a PID. Can I prove by the following way? Proof by induction on $n$, where $M=\langle v_1,...,v_n\rangle$. If ...
### Principal ideal domain, $\forall x=(x_1,x_2)^t \in R^2~~\exists G \in SL_2(R) : Gx=(\gcd(x_1,x_2), 0)^t$
Let $R$ be a principal ideal domain. Prove that for every $x=(x_1,x_2)^t \in R^2$ exists a matrix $G \in SL_2(R)$ for which $Gx=(\gcd(x_1,x_2), 0)^t$. I think it's easy, but do not know how to ...