Result: {v: \mathbb{Q}^* \rightarrow \mathbb Z} is a discrete valuation satisfying the following properties:

  • {v} is surjective.
  • {v(ab) = v(a) + v(b) \quad \text{for all} \quad a,b\in \mathbb Q^*}
  • {v(a+b) \geq \{ v(a), v(b) \} \quad \text{provided} \quad a+b \neq 0}

Then v=v_p for some prime p, given by, v_p\displaystyle\left(p^r \displaystyle\frac{a}{b} \right) = r for (a,p)=(b,p)=1.

Proof: It is a fact (cf. Dummit & Foote Ex 39 Sec. 7.4) that {R = \{ x \in \mathbb{Q}^* : v(x) \geq 0 \}} is a local ring with a unique maximal ideal {\mathfrak m} of elements of positive valuation. (Recollect that an element of {R} is a unit iff its valuation is zero.) Now, {v(1)=0 \Rightarrow 1 \in R \Rightarrow \mathbb Z \leq R}.
Claim: {\mathbb Z \cap \mathfrak m = (p)}. Clearly, {v} being surjective, {\mathbb Z \cap \mathfrak m \neq (0)} because otherwise, each nonzero integer would have valuation zero and so would every nonzero rational. If {\mathbb Z \cap \mathfrak m = (n)} then we factorize {n=ab} with {a,b} integers. Then {ab\in \mathfrak m \Rightarrow a\in \mathfrak m } or {b \in \mathfrak m}. Thus {\mathbb Z \cap \mathfrak m} must be a prime ideal. So the claim is justified.
Now given {\displaystyle\frac{a}{b}}, write {\displaystyle\frac{a}{b}=p^r \frac{a'}{b'}} with {(p,a')=(p,b')=1}. I claim that {a'} and {b'} are units. Since {(p,a')=1}, we can write {px+a'y=1}. If {a'} is not a unit then {a'\in \mathfrak m} and thus {1\in \mathfrak m}, a contradiction. A similar argument suggests that {b'} is also a unit. Hence,
\displaystyle v\displaystyle\left(\frac{a}{b}\right)=rv(p)+v(a')-v(b')=rv(p)

Now {v} surjective implies, there exists {\displaystyle\frac{a}{b}=p^r \frac{a'}{b'}} such that {v\displaystyle\left(\displaystyle\frac{a}{b}\right)=rv(p)=1}. This leaves two possibilities, namely, {v(p)=\pm 1}. But {v(p)=-1} gives an easy contradiction: {-1=v(p)\geq \min\{v(1),v(p-1)\} = 0}. Thus the only possibility is that {v(p)=1} and thus, {v=v_p}. \blacksquare