# Is every local ring a valuation ring?

Is every local ring a valuation ring?

I know the answer is no and the first example comes to my mind was as following (I started with smallest fields, as $\mathbb{Z}_2$ and $\mathbb{Z}_3$ are not interesting so I came to next possible one, means the field with 4 elements.

Let $F=\dfrac{\mathbb{Z}_2[T]}{\langle T^2+T+1\rangle}$ and $\alpha=\bar{T}\in F$, so $F=\{0,1,\alpha,\alpha+1\}$. The set $\{0,1\}$ forms a subring of $F$ which is isomorphic to $\mathbb{Z}_2$ so a field and a local ring but doesn't contain $\alpha$ or $\alpha+1$ while $\alpha^{-1}=\alpha+1$.

In fact finite fields have no nontrivial valuation ring because their multiplicative groups are cyclic. So every subring of a finite field which is local is a suitable example.

I would to know other kind of examples so if someone knows already other examples different than mine I will be pleased if he shares them here with me.

As this question has no unique answer I only upvote persons who are giving me new classes of examples.

By the answer of "user26857" we have the second type of examples. Let $A$ be a domain and $F=frac(A)$, its field of fractions. For every prime ideal of $A$ which can't be generated by less than 2 elements say $\mathfrak{p}$ we can make an example. Let $B$ be a generator for $\mathfrak{p}$ and $x,y\in B$ two elements that their greatest common divisor is not in $\mathfrak{p}$ (otherwise we can replace them both with their g.c.d.). Assume $d=\text{g.c.d.}(x,y)$ then $\exists x_1,y_1\in A\;:\;x=dx_1,y=dy_1$ and $x_1,y_1$ are prime to each other. Since $\mathfrak{p}$ is prime and $d\notin\mathfrak{p}$ we have $x_1,y_1\in\mathfrak{p}$. Now none of $\dfrac{x_1}{y_1}$ and $\dfrac{y_1}{x_1}$ are in $A_\mathfrak{p}$ so it is not a valuation ring of $F$ while it's local.\ Thanks to "user26857".

• But, $F$ isn't the field of fractions of $\{0,1\}\subseteq F$? – Yai0Phah Sep 11 '14 at 16:22
• What about $\mathbb C[X,Y]_{(X,Y)}$? (In a valuation ring the ideals are linearly ordered; in our example the ideals $(X)$ and $(Y)$ are not contained one in another.) Btw, fields are valuation rings. – user26857 Sep 11 '14 at 16:27
• @user26857 Yes, you're right I've seen this field is not a valuation 3 days ago in a lecture of Pro. Teissier but when I was thinking to this question as sometimes I like to start from smallest objects and then coming up I wasn't taking care! Very very thank you for this nice example. Please write it as an answer. – AmirHosein Sadeghimanesh Sep 11 '14 at 16:35
• @FrankScience You are right but I don't see why $F$ should be field of fraction of $\{0,1\}$! In fact I think you were thinking to this statement "If $V$ be a valuation ring of a field $F$ then $F=frac(V)$" but the examples which we are looking for are not valuation so their field of fraction can be different than $F$. – AmirHosein Sadeghimanesh Sep 11 '14 at 16:40
• Could you perhaps say why it is you want more examples? What is your motivation? Otherwise I think this question is too broad. – RghtHndSd Sep 11 '14 at 16:44

$\mathbb C[X,Y]_{(X,Y)}$ is a local ring which is not a valuation ring. (In a valuation ring the ideals are linearly ordered; in our example the ideals $(X)$ and $(Y)$ are not contained one in another.)
• What exactly is the expression $\mathbb{C}[X,Y]_{(X,Y)}$? Is it something like a localization (I suppose $(X,Y)$ is a prime, even a maximal, ideal)? – Diglett Jul 5 '19 at 2:23
Valuation rings are integral domains (by definition). Not every local ring is an integral domain (for example $\mathbb{Z}/p^2$).
• Yes you are right but in definition of a valuation ring I've seen so far like this "Let $F$ be a field. A subring $R\subseteq F$ is called a valuation ring of $F$ if for every nonzero element of $F$ say $x$ we have $x\in R$ or $x^{-1}\in R$." Thus by this definition we don't discuss about rings wich we can't sit them in a field. If you have seen another definition please introduce the source. But thank you too. – AmirHosein Sadeghimanesh Sep 11 '14 at 17:59