ac.commutative algebra - Proving that two local PIDs, one inside the other, with the same field of fractions are equal.

I believe that the OP meant to include the condition that each of the local PIDs is not a field. In this case the result is true, and as several people have said, is a rather standard exercise.

At this moment it seems to me that if we get asked a rather standard question that has not been asked on MO before, it would be nice to use it as an opportunity to explain something a little deeper / slightly less standard related to the question. In this regard, let me mention a generalization:

A local PID $R$ (which is not a field!) with fraction field $K$ is precisely a discrete valuation ring, i.e., is the valuation ring $R = {x in K | |x| leq 1 }$ of a norm $| |: K rightarrow mathbb{R}^{geq 0}$ such that $|K^{times}|$ is a discrete subgroup of $mathbb{R}^{times}$. Now for nontrivial norms $| |_1$, $| |_2$ (Archimedean or not) on a field $K$, there is the following result:

Theorem: The following are equivalent:
(i) There exists $alpha > 0$ such that $| |_2 = | |_1^{alpha}$.
(ii) For all $x in K$, $|x|_1 < 1 implies |x|_2 < 1$.
(iii) For all $x in K$, $|x|_1 leq 1 implies |x|_2 leq 1$.

(See e.g. http://math.uga.edu/~pete/8410Chapter1.pdf, p. 4, for a proof.)

Now the implication (iii) $implies$ (i) shows that there can be no proper containments among DVRs with the same fraction field. The same holds for all rank one valuation ring because, by definition, a rank one valuation ring is one whose value group is a subgroup of $mathbb{R}$; therefore the data of a rank one valuation is equivalent to that of a non-Archimedean norm. (Note that if $| |$ is a non-Archimedean norm, then $v = - log | |$ is a valuation, and conversely if $v$ is a rank one valuation, then $| | = e^{-v}$ is a non-Archimedean norm.) It is not true for valuation rings of higher rank.