Question : What are the valuation rings of $\mathbb{Q}$?
I have started reading some Algebraic Geometry from Daniel Bump's book. I don't really have much background in commutative algebra (hence the choice of this particular book as it's more self contained in that sense). This is Exercise 2.2 from the book.
The definition given for valuation ring $R$ of a field $F$ is that $R$ is a subring of $F$ such that $x\in F\setminus\{0\}$ implies $x\in R$ or $x^{-1}\in R$. I also know that valuation rings are local.
It is easy to see that subrings of the form $\{a/b\in \mathbb{Q}\; |\; p\nmid b\}$ (for $p$ primes) are valuation rings of $\mathbb{Q}$. So the question boils down to whether rings like these exhaust all valuation rings of $\mathbb{Q}$? My guess is Yes they do. Is there a direct(elementary) way to show this?
The answer is yes, because the valuation rings in a field are also characterised by the property that they are maximal subrings among local subrings for the domination relation (i.e. the maximal ideal of the smaller ring is contained in the maximal ideal of the larger ring).
Some more details: any subring $V\subset\mathbf Q$ contains $\mathbf Z$, and if $\mathfrak m$ is the maximal ideal of $V$, $\mathfrak m\cap\mathbf Z $ is generated by a prime $p$ and clearly $\mathbf Z_{(p)}$ is dominated by $V$. As it is also a valuation ring, by maximality, we have $\mathbf Z_{(p)}=V$.