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I found that A003172 in OEIS lists integers $n$ such that $ \mathbb Q[\sqrt n]$ is an UFD; but I think for all positive square-free integer $n$, $\mathbb Q[\sqrt n]$ should be a field, and all fields are UFDs, right?

I think there might be something wrong in my understanding.

Bill Dubuque
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Lake Oliver
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    the issue is actually whether the ring of integers in the field is a UFD – J. W. Tanner Nov 01 '21 at 21:14
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    The "title" of the sequence is confusing, but if you look at the linked-to sequence, A003649, that one states it correctly: it is the fields $\mathbb{Q}(\sqrt{n})$ with class number $1$, which is equivalent to the ring of integers being a UFD. – Arturo Magidin Nov 01 '21 at 21:38
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    To elaborate: a widespread convention in number theory is that divisibility theory problems in an algebraic number field implicitly refer to divisibility theory in its subring of (algebraic) integers . – Bill Dubuque Nov 01 '21 at 21:38

2 Answers2

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The reference to $\text{A}003649$ in the comments to $\text{A}003172$ clarifies the matter in the same way the comments do: we are talking about the ring of integers $\mathcal O(\mathbb Q(\sqrt n))$ of $\mathbb Q(\sqrt n)$ and not the field extension itself. Of course, you are right that $\mathbb Q[\sqrt n]=\mathbb Q(\sqrt n)$ is always a field, so in particular a UFD.

The referenced sequence lists the class numbers of real quadratic fields, i.e. $\mathbb Q(\sqrt n)$ with $n>0$ (and $n$ squarefree to really get different fields for different values of $n$). The class number is an arithmetical quantity attached to a field, or more precisely to its ring of integers, which measures the failure of unique factorization. It is equal to $1$ if and only if the ring of integers is a UFD. In the case of quadratic fields one can show that

$$ \mathcal O(\mathbb Q(\sqrt n))=\begin{cases}\mathbb Z\left[\frac{1+\sqrt n}2\right]&,n\equiv1\mod 4\\\mathbb Z[\sqrt n]&,n\not\equiv1\mod4\end{cases} $$

and these are the rings which are UFDs exactly for the $n$ listed on $\text{A}003649$, i.e. the $n$ for which the class numbers is $1$ as listed on $\text{A}003172$.

The title question is answered by Rene Schipperus. In general, a good way of investigating if a ring is a UFD is to find different factorizations and show that all factors are irreducible but not associated. For such quadratic rings $\mathbb Z[\sqrt n]$ a good tool for is the norm function:

$$ \mathrm N\colon\mathbb Z[\sqrt n]\to\mathbb Z,a+b\sqrt n\mapsto a^2-nb^2 $$

For example, an element is a unit iff its norm is $\{\pm1\}$ and if its norm is a prime, it is irreducible. You can find more about this in the answers to this question.

mrtaurho
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It is easy to see that $6$ has two different factorizations in $\mathbb{Z}[\sqrt{10}]$, which is the ring of integers of the field, $\mathbb{Q}[\sqrt{10}]$.

$$6=(\sqrt{10}+2)(\sqrt{10}-2)=2\cdot 3$$ thus there are two different factorizations. Of course one also has to show that $2$ and $3$ do not factor, but this is obvious as

$$x^2\equiv 2 \bmod 10$$ $$x^2\equiv 3 \bmod 10$$ have no solution.

KCd
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Rene Schipperus
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    This does not answer the OP's question (please read the question body - not only the title). This answer is also a dupe (of many). – Bill Dubuque Nov 02 '21 at 00:09
  • Note to readers: I updated the title to be more specific (originally it was: Why $,Q[sqrt(10)],$ is not an UFD? (A003172) – Bill Dubuque Nov 02 '21 at 00:22