- F is a field and G is an abelian group that is linearly ordered.
- The action of G respects order. Add something larger and you get something larger.
- A valuation homomorphism maps the nonzero elements of F onto G. Multiplication in F corresponds to addition in G.
- The valuation of the sum is at least as large as the lesser valuation.

Every valuation ring produces a valuation group with these properties. In this section we are interested in the converse.

Let R be the preimage of the elements of G that are nonnegative, union 0. Use property 4 to show R is closed under addition. Use property 2 to show R is closed under multiplication. Property 3 maps the multiplicative identity to the additive identity, hence R contains 1. Since R inherits its properties from F, it is an integral domain.

Let x be an element that is not in R. Now v(x) + v(1/x) must equal 0. This means v(1/x) is positive, and 1/x is in R. Therefore R is a valuation ring.

Assume v(x) = 0, and write xy = 1 in F. Since v(1) = 0, v(y) is also 0. So x and y are both in R, and x is a unit. Conversely, if x and y are both in R then two nonnegative valuations sum to 0, and v(x) and v(y) are 0. The kernel of the homomorphism is precisely the units of R.

If we had started with R, we would use R* as the kernel, and build the same homomorphism on F, and create the same valuation group G. Therefore any group homomorphism, satisfying our four properties, is a valuation group, with a corresponding valuation ring.

Now let's be a bit more general. Let E be a subset of G that includes 0, and is closed under <. In other words, if x < y and y ∈ E then x ∈ E. In this case E is called an initial segment of G. Let H be the preimage of the elements of G that are beyond E, and H becomes an ideal.

It's pretty easy to show that the initial segments of G are linearly ordered. We're basically biting off larger and larger pieces of G. The preimages of the regions above E are also linearly ordered, forming a descending chain of ideals.

What about the converse? Let H be an ideal and consider its image in G. If x is in H and y > x, then x times something in R gives y. Thus y is also in H, and the image is closed under >. Its complement is an initial segment, which we called E above.

The ideals of R and the initial segments of G correspond 1-1. But don't be fooled; ideals do not always correspond to elements of G. Consider the maximal ideal M. If M is principal, generated by x, then the image of x in G caps the initial segment E, and starts the region of G that corresponds to M. Conversely, if x maps to the least element in the image of M, then everything in M has a higher valuation than x, everything in M is divisible by x, and M is principal. Of course M could be the union of an ascending chain of ideals, whence the image of M in G has no least element.

This generalizes to any ideal. The ideal H is principal iff its image in G has a least element.

Conversely, assume R is a pid.
The maximal ideal M maps to the positive elements of G,
and since M is principal, it has a generator x,
which attains the least positive valuation.
Powers of x map to the integers, a subgroup **Z** in G.

Suppose there is some y between nx and (n+1)x.
Subtract nx and find an element of G between 0 and x, which is impossible.
If there is more to G, it lies beyond **Z**.

Let E be the initial segment that includes **Z**,
and let H be the preimage of the region above E.
Now H is principal, generated by some y.
Map y into G and find a cluster point from below.
This means every point in G is a cluster point from below, including x.
This is a contradiction, hence there is nothing beyond **Z**.
The proper ideals of R are the powers of x, period.

In summary, R is a pid iff G = **Z**.

Remember our example, the fractions without p in the denominator?
This has valuation group equal to **Z**, according to the exponent on p.
This must be a pid, and it is.
The maximal ideal is generated by the integer p,
and everything outside this ideal is a unit.

If x is a cluster point from below, then subtract x, and 0 is afflicted with the same malady. Everything in G has an inverse, so 0 is also a cluster point from above, and R is not noetherian. Put this all together, and a noetherian valuation ring has a valuation group with no cluster points. This is called a discrete valuation group, and R is a discrete valuation ring, or dvr. (We will explore this topic in the next section.)

Let R be a noetherian valuation ring,
whence G has no cluster points.
Let x have the smallest positive valuation in G.
If y is an element of G that is not generated by x,
and y lies between nx and (n+1)x, subtract nx to find an element with valuation between 0 and x.
This is a contradiction,
hence y exceeds all powers of x.
Now consider the initial segments capped by: y, y-x, y-2x, y-3x, etc.
Take the preimage of the complement of each of these segments,
thus producing an infinite ascending chain of ideals in R.
This contradicts the fact that R is noetherian.
There is nothing above **Z**.
Therefore a noetherian valuation ring has valuation group **Z**, and is a pid.
Since a pid is noetherian,
the following conditions on a valuation ring R are equivalent.

- R is noetherian
- R is a pid
- valuation group =
**Z**