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".
Best Answer
$\mathbb C[X,Y]_{(X,Y)}$ 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.)
Btw, fields are valuation rings.