# Localization of a PID on a prime ideal is a DVR?

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.

• What's your definition of a DVR? Depending on which one you take, this can range from extremely short (i.e. not worth writing down for most authors) to a little more involved. (For instance, one of the equivalent definitions of a DVR is a PID with a unique nonzero prime, which immediately answers the question once you know a little about localization if you'll accept that definition.) Dec 4, 2022 at 17:22
• @KReiser I didn’t realize there were different definitions. I added mine to the question. Dec 4, 2022 at 19:09

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.

• Thank you! But where can I find a proof of this fact? ‘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’. Dec 6, 2022 at 16:41
• The existence of a maximal ideal implies $0 \neq 1$. Suppose the maximal ideal is unique. Then every element of the maximal ideal is not a unit. And if $x$ is not a unit, then $(x)$ is contained in some maximal ideal, namely the unique one. So the maximal ideal is the set of all non-units. It follows that non-units are closed under addition; thus, either $x$ or $1-x$ is a non-unit. Conversely, if $0 \neq 1$ and for all $x$, either $x$ or $1-x$ is a unit, we can show that the set of non-units is the unique maximal ideal. Dec 6, 2022 at 17:17

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 $$\frac{a}{c^{m+1}}\in (d)$$.