If R is a valuation ring it is integrally closed and local. Because R is noetherian, it is also a pid, and a dvr with valuation group Z. The ideals are precisely the powers of the maximal ideal, and the maximal ideal is the only prime ideal.
In this section we will prove the converse. If R is local and integrally closed, with one prime ideal, it is a valuation ring. And since R is noetherian, our valuation ring becomes a pid, and a dvr with valuation group Z. So the two characterizations of a noetherian ring are equivalent: valuation ring ⇔ integrally closed and one prime ideal.
This is a theorem about valuation rings, and you'd expect to find it under valuation rings, but the proof involves invertible fractional ideals, which is why it is here. In fact, valuation rings and dedekind domains are interrelated, and it's hard to talk about one without talking about the other.
Ok, here we go.
Let R be integrally closed with one prime ideal M. I'm calling it M because R has to have a maximal ideal, which is prime, so that is the one prime ideal.
Let F be the fraction field of R.
Let H be a nonzero fractional ideal of R, and let S be the set of elements in F that drive H into itself. In other words, x ∈ S iff x*H ⊆H. Verify that S is a fractional ideal in F.
Since 1*H = H, S contains 1, and S contains R.
Since H is a fractional ideal, let dH lie in R. Here d is the "common denominator" of H. Let y be an element of H, and let x be an element of S. Since xy is in H, dxy is in R. Put another way, dy times x lies in R, for every x in S. Multiply dy by the denominator of y and call the result z. Now z drives S into R. Therefore S is a fractional ideal.
Multiplication by z implements an R module isomorphism, hence S is isomorphic to an ideal of R. Remember that S contains R, and zr carries a submodule of S onto an ideal of R, the principal ideal generated by z. Select an x in S and do the same for the submodule in S generated by 1 and x. This becomes an ideal in R, generated by z and zx. A larger submodule, generated by 1, x, and x2, becomes a larger ideal. Continue this process through the powers of x. If each xn is new, not spanned by the lower powers of x, then we are building an infinite ascending chain of ideals in R. Since R is noetherian, this is impossible. Therefore some power of x is spanned by the lower powers of x. This makes x integral over R, and since R is integrally closed, x is already in R. Therefore S = R. The elements of F that drive H into itself, for any ideal H, belong to R, and that's the end of it.
Next, let J be an ideal, and consider J′, the elements of F that drive J into R. Of course J′ includes R. Let S be the set of ideals J in R such that J′ properly contains R. The "inverse" of J includes fractions outside of R.
If x is a nonzero element of M, the principal ideal generated by x belongs to S. The inverse of this ideal is generated by 1/x, which lies outside of R. All the proper principal ideals are contained in S.
Start with the ideal generated by x, and find a larger ideal in S, then a larger one, then a larger one, and so on, until the process stops, which it must, since R is noetherian. Let J be the maximal ideal in S containing x. Note that J is not all of R, since R is not in S at all. We want to show J is prime. Since there is but one prime ideal, that will prove J = M.
Let a*b lie in J, and let c be a fraction in J′ that is outside of R. Let a and J generate a larger ideal K. Note that cb drives K into R. Since J is maximal, K is not an ideal in S. That means K′ includes nothing outside of R, and cb is in R. Therefore c drives the ideal generated by b and J into R. The ideal generated by b and J belongs to S, and contains J, and is no larger than J, hence J contains b. This makes J a prime ideal, and since there is but one prime ideal, J = M. The maximal ideal M belongs to S, and M′ includes fractions outside of R.
We are going to use these facts to show M is invertible. We know that MM′ is an ideal of R. If M is not invertible than this product becomes a proper ideal. Now R is a local ring, so MM′ is contained in M. We showed that precisely R drives an ideal into itself, so R drives M into M, and M′ is contained in R. Yet we just showed that M′ properly contains R. This is a contradiction, so we must have MM′ = R, and M is an invertible ideal.
Since M is invertible every prime ideal is invertible, and that makes R a dedekind domain by an earlier theorem. Every ideal is a power of M, ideals are linearly ordered, and R is a valuation ring.
That completes the proof. When R is noetherian, declaring it a valuation ring is the same as saying it is integrally closed with one prime ideal. In either case it is a pid, and a dvr with valuation group Z.