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I’d like to find a proof that the localization of a principal ideal domain on a prime ideal is a discrete valuation ring.
Apparently this is a well-known result (according to some question on math exchange, such as this one), but I haven’t found a proof of it online.
Edit: The definition of DVR I have is the following: a DVR is a PID with exactly one non-zero maximal ideal.
Recall that a local ring is a ring with a unique maximal ideal, and that a field is a ring where $0$ is maximal. So a discrete valuation ring is equivalently a local ring which is also a PID, but which is not a field.
Also recall that a local ring can equivalently be defined as a ring where $0 \neq 1$ and for all $x$, either $x$ or $1-x$ is a unit.
Lemma 1: the localisation of any ring at a prime ideal is a local ring.
Proof: Consider a prime ideal $p \subseteq R$. Suppose $0 = 1$ in $R_p$. Then there exists some $x \notin p$ with $0 = x$. But then $0 \notin p$, which contradicts that $p$ is an ideal.
Now consider some $\frac{a}{b} \in R_p$. Then either $a \notin p$ or $b - a \notin p$, since if both were in $p$, we’d have $a + (b - a) = b \in p$. So either $a$ is invertible in $R_p$, in which case $\frac{a}{b}$ is, or $b - a$ is invertible in $R_p$, in which case $1 - \frac{a}{b} = \frac{b - a}{b}$ is. So $R_p$ is indeed a local ring. $\square$
Lemma 2: Let $R$ be a PID. Than any nonzero localisation of $R$ is also a PID.
Proof: any nonzero localisation of an integral domain is an integral domain. Now consider an ideal $I \in S^{-1} R$. Then consider $I \cap R$, which is an ideal of $R$, hence principal. Then write $I \cap R = pR$. I claim $I = (p)$. For clearly $p \in I$. And if $\frac{a}{b} \in I$, then $a \in I \cap R$, so we can write $a = pc$ and thus $\frac{a}{b} = p \frac{c}{b}$. $\square$
Combining these two lemmas, we see that localising on a prime ideal gives us a local ring which is also a PID. However, if our original ring $R$ is a field, then localising the field at a maximal ideal will give us the same field, whose maximal ideal is zero. So we must add a condition to your claim, which is that the prime ideal in question be nonzero. In that case, write the ideal as $(p)$. Then $p$ is not invertible in $R_p$, nor is it zero. So $R_p$ is not a field, and thus its unique maximal ideal is nonzero.
Your PID (integral domain where each ideal is principal) is $A$ and $(c)$ is a non-zero prime ideal. We need to prove that any element $a\in A-0$ is of the form $a=c^n b$ with $b\not \in (c)$.
We set $v_c(a)=n$ and this discrete valuation extends naturally to $A_{(c)}$ as $v_c(a/s)=v_c(a)$ for any $s\in A-(c)$, making $A_{(c)}$ a DVR.
So why is $a=c^nb$? Let $n$ be the largest integer such that $a/c^n\in A$. If $n$ is finite then $b=a/c^n\not \in (c)$ and $a=c^nb$ as wanted. If $n$ is infinite then the ideal $I=\sum_{n\ge 0} \frac{a}{c^n} A$ is principal, equal to $(d)$ with $d\in I$ ie. $d=\sum_{j=1}^J r_j \frac{a}{c^{n_j}},r_j\in A$ and $J$ finite. Let $m=\max n_j$. We get that $(d)\subset (\frac{a}{c^m})$ contradicting that $\frac{a}{c^{m+1}}\in (d)$.
