Let R be a subring of the field F. R is a valuation ring if it contains either x, or 1/x, or both, for every x in F.

Since R is an integral domain it has a well defined fraction field; call this field K. We know that K is a subfield of F. If F is larger than K, then F contains an x such that x and 1/x are not in R. Therefore F is precisely the fraction field of R.

The integers (within the rationals) don't cut it, because Z contains neither 3/2 nor 2/3. However, R could be all the fractions that do not have a factor of p in the denominator, for some prime p. R is a localization, Zp, and R contains either x or 1/x. R doesn't contain 1/p, for instance, but it contains p, so we're ok.

A ring extension of a valuation ring, within its fraction field, is still a valuation ring, since it still contains x or 1/x.

F … 1/x
R … x

Let x and y be nonunits in a valuation ring R. If xz is a unit then x would be too, so x times anything in R remains a nonunit. If x+y is also a nonunit we have an ideal.

Since x and y are not units, 1/x and 1/y are not in R. Without loss of generality, assume x/y is in R. Hence 1 + x/y = (x+y)/y is in R. If 1/(x+y) is in R then multiply by (x+y)/y to put 1/y in R. This is a contradiction, hence x+y is not invertible, and the nonunits form an ideal in R.

Call this ideal M, since it is maximal. After all, the only things not in M are units. Note that M contains every other proper ideal. Thus M is the only maximal ideal of R, and R is a local ring.

R is integrally closed in its fraction field F. Let x, an element of F-R, be integral over R. Thus x satisfies a monic polynomial p(x). Write xn in terms of smaller powers of x, then divide through by xn-1. Now x is a linear combination of powers of 1/x, with coefficients in R. Remember that 1/x is in R, thus x is in R, which is a contradiction. Thus R is integrally closed in its fraction field.

If R is noetherian, the converse is also true. A noetherian local ring that is integrally closed is a valuation ring. However the proof will have to wait until we are well in to dedekind domains.

The ideals of an integral domain R are linearly ordered by containment iff R is a valuation ring.

Suppose two ideals in the valuation ring R do not exhibit containment in either direction. This means there exist x (in the first ideal) and y (in the second ideal) such that x does not generate y and y does not generate x. Yet either x/y or y/x lies in R. This is a contradiction. Therefore, given any two ideals H and J, either H contains J or J contains H. The ideals can be arranged in a line, via containment. We already know R is local, so the maximal ideal M sits at the top of this chain.

Conversely, assume ideals are linearly ordered in an integral domain R. Is R a valuation ring? Let F be the fraction field of R. Let x/y be a fraction in F. Either x*R contains y*R, or y*R contains x*R. Either x generates y or y generates x. Either x/y is in R or y/x is in R. This holds for every x/y in F, hence R is a valuation ring.

As a corollary, the quotient of a valuation ring is a valuation ring. Mod out by a prime ideal P, so the result is an integral domain. Two ideals in the image that do not contain each other pull back to ideals in R that do not contain each other. This is a contradiction, hence the ideals of R/P are linearly ordered, and R/P is a valuation ring.

Also, S inverse of R is a valuation ring. Since R embeds in S inverse of R, we are merely producing a larger ring inside F. This ring extension certainly contains x or 1/x, because R does.

Let R and S be local rings in the field F. S dominates R if S contains R, and the maximal ideal of S contains the maximal ideal of R. This latter condition is equivalent to saying the maximal ideal of S intersected with R gives the maximal ideal of R. In other words, the nonunits of S, intersect R, produce the nonunits of R - just as the units of S, intersect R, produce the units of R.

Verify that dominance is a partial ordering.

Let C be an ascending chain of dominant local rings, and let U be their union. Two nonunits in U are nonunits in some ring R in the chain. Let z be their sum, which is in R, and in U. If z is a unit in U then 1/z appears in some ring S in the chain. Thus z is a unit in S. Intersect S and R, and if z is common to both, and a nonunit in R, it should be a nonunit in S. This is a contradiction, hence z is a nonunit in U. The sum of nonunits remains a nonunit.

Similarly we can multiply a nonunit in U by anything else in U and obtain a nonunit, so the nonunits form an ideal, and U is a local ring.

Intersect U with any ring R in the chain, and the nonunits remain nonunits. If z and 1/z are units in U, they are brought in by some ring S, where they are still units, and z remains a unit in every ring above S, or below S where z still appears. Therefore U dominates the entire chain. U is a local ring that dominates C.

Every ascending chain of local rings is bounded by a local ring, i.e. the union, that dominates the entire chain. By zorn's lemma, there is always a maximal local ring in F, that dominates any given local ring R, though this maximal ring may be F itself. Such is the case when F is the union of all finite fields of characteristic p. If R is a ring inside F, and x is an element of R, then Z/p adjoin x creates a finite field in R, every x is invertible, and R is a field. All the rings in F are fields, which are local rings. Keep expanding the field until R = F.

The idea of dominance will be expanded, in the next section, to homomorphisms.

Let F be a field that contains various subrings, and let C be an algebraically closed field. Let R and S be rings in F, such that R is a subring of S, and let g and h be ring homomorphisms from R and S into C. If g and h agree on R, then h dominates g. In other words, h(S) is an extension of g(R).

Given an ascending chain of dominant homomorphisms, let U be the union of the rings, and let V(U) be the union of the homomorphisms. Verify that U is a ring, and V(U) is a ring homomorphism.

Since V(U) dominates everything in the chain, every chain is bounded. Apply zorn's lemma, and any homomorphism g(R) is contained in a maximal homomorphism V(U), such that V agrees with g when restricted to R.

So - what does all this have to do with valuation rings? Well, the domain of a maximal homomorphism is in fact a valuation ring. Let's see why this is so.

If g(R) is trivial, mapping everything to 0, then g(1) = 0. Now g (extended) maps any ring to 0, including all of F. Obviously F is a valuation ring, so let's set this trivial case aside and assume g(1) = y. With y2 = y, y is 0 or 1, hence g(1) = 1. That is the default assumption in this book anyways: a ring homomorphism should map 1 to 1.

Choose r and g so that g(R) is maximal with respect to dominance. Let g(R) have kernel M. The image is an integral domain, thus M is prime. Let E be the fraction field of R, which lives in the larger field F. R embeds in E, so E is a ring extension of R.

Suppose there is some nonunit x in R-M. Let S be the set of fractions whose numerators are drawn from R, and denominators are powers of x. The powers of x never intersect M, M being prime, so M remains a proper prime ideal in S. S is a subring of E, and since x is a nonunit, S is strictly larger than R. Extend the homomorphism g to all of S via g(a/b) = g(a)/g(b). Since g(x) is nonzero, this is well defined. This new homomorphism dominates g(R), and that is a contradiction. Therefore the nonunit x does not exist. Everything outside of M is a unit, and R is a local ring, with maximal ideal M. The image g(R), which is isomorphic to R/M, is a subfield of C.

Let x be anything in F-R, and let y = 1/x. Let R[x] be the ring produced by adjoining x to R. In other words, the polynomials of x with coefficients in R. This is a subring of F.

Abuse the notation just a bit, and let M[x] be the polynomials of x with coefficients in M. Verify that M[x] is an ideal in R[x].

The grand assumption, which we will soon contradict, is that R[x] = M[x], and R[y] = M[y]. An equivalent formulation of this condition says some polynomial p(x) = 1, and some other polynomial q(y) = 1, where all coefficients lie in M.

There may be many polynomials in M[x] and M[y] that equal 1. Select p(x) from among these, so that its degree is minimal, and similarly for q(y). Assume without loss of generality that p(x) has the greater degree.

Let q(y) have a constant term t, which could be 0. Now 1-t = q(y)-t, which has all positive powers of y. Remember that 1-t is not in M, hence it is a unit.

Let p(x) have degree n, and multiply q(y)-t by xn. If q(y) included the term vyn, this term is now v. Terms with lesser powers of y become terms with powers of x, having exponents between 1 and n-1. Therefore (1-t)xn is a linear combination of lesser powers of x. Multiply through by the inverse of 1-t, which is an element in R, and all coefficients remain in M. Now xn is equal to a linear combination of lesser powers of x, with coefficients taken from M. Substitute in p(x) to find a new polynomial, equal to one, with a smaller degree. This is a contradiction. Therefore either M[x] ⊂R[x], or M[y] ⊂R[y].

Ok, we're almost home. Suppose R is not a valuation ring inside F. Then there is some x, with inverse y, such that R contains neither x nor y. The above tells us M[x] is a proper ideal inside R[x]. (Reverse the roles of x and y if necessary.) Let R′ be R[x], and let M′ be any maximal ideal in R′ that contains M[x]. Intersect M′ with R and you get at least M. Any additional element u would be a unit in R, which is a unit in R′, and is not part of M′. Therefore M′ ∩ R = M.

Let z represent a coset of M in R. z also represents a coset of M′ in R′. The previous paragraph shows this map is well defined. Each coset of M in R determines a unique coset of M′ in R′. This map respects addition and multiplication, hence the field R/M embeds in the field R′/M′.

The larger field is in fact an extension of the smaller; simply adjoin x. Since the larger field is in fact a field, x cannot be transcendental. It must be algebraic over R/M.

We have extended the ring R to R′; now we are ready to extend its homomorphism g. Let g map all of M′ to 0. Of course g(R) is unchanged. Let g(x) be the corresponding algebraic element in C, which exists, because C is a closed field. This new homomorphism dominates g(R), which is a contradiction, since g(R) is maximal with respect to dominance. Therefore R is a valuation ring.

A ring R is a valuation ring for the field F iff R is a maximal dominant local ring inside F.

Take the forward direction first. Let R be a valuation ring, hence it is a local ring, and suppose there is some other ring S that dominates it. Recall the definition of dominance. The ring S contains the ring R, and W, the maximal ideal of S, contains M, the maximal ideal of R. In other words, W ∩ R = M. Let x be any element in S-R. This means y, the inverse of x, lies in R. Both x and y belong to S, and they are inverses, hence they are not contained in W. This means y does not belong to M. That means y is a unit in R, and it has an inverse z in R. Within the larger ring S, x and z are both inverses of y; yet the inverse must be unique. Therefore x = z, and x lies in R, which is a contradiction. Our valuation ring is maximal with respect to dominance.

Now for the converse. Let R be a local ring in F that is not dominated by any other local rings. Let M be its maximal ideal. Let g(R) be the ring homomorphism implemented by R/M. The result is a field; call it K. Let C be the algebraic closure of K, hence g maps R into C.

Suppose there is a larger ring S, with a homomorphism h(S) into C, such that h dominates g. In other words, h and g agree on R. Let S be as large as possible, inside F, such that h(S) dominates g(R). In other words, h(S) is a maximal, dominant ring homomorphism. In the last section we proved such a homomorphism exists, and that S is a valuation ring. In particular, S is a local ring, with a maximal ideal W. If z is in M, then g(z) = 0, and h(z) = 0, z is not a unit in S, and z is in W. Therefore S dominates R. Yet R is maximal with respect to dominance, so this is a contradiction.

There is no dominating homomorphism h(S), thus g(R) is already maximal with respect to dominance. Again, call upon the previous theorem to assert R is a valuation ring. That completes the proof; R is a valuation ring inside F iff R is a maximal dominant local subring of F.

Let R be a valuation ring in its fraction field F. Recall that F* is the group of units in F, and similarly for R*. The latter is a subgroup of the former. The quotient group G = F*/R* is the valuation group associated with R. The map from F* onto G is a valuation of F. There may be several valuations. Each valuation ring R inside F implies a different subgroup R*, which produces a different homomorphism, and a different valuation on F.

For a given ring R and valuation group G, take any two elements of G and pull them back to x and y in F. Either x/y or y/x is in R. Assume x/y lies in R. Multiply x or y by any unit, anything in R*, and x/y still lies in R. It doesn't matter which x and y we select. It only depends on the elements of G. For any x and y in G, either x/y is in R, or y/x is in R, or both.

Use this relationship to build a partial ordering on G. Write y ≤ x if x/y is in R.

Clearly x ≤ x; the relation is reflexive. And if x ≤ y and y ≤ x then x/y and y/x are both in R, x/y is a unit in R, x/y is the identity element of G, and x and y represent the same element in G. If two elements are ≤ each other they are equal in G.

Finally show transitivity. Assume z ≤ y ≤ x, whence x/y and y/z are in R. This implies x/z is in R, and z ≤ x. We have a partial ordering.

Since R is a valuation ring, x/y or y/x is in R. Every pair of elements is comparable, and the partial ordering becomes a linear ordering. All the elements of G can be arranged in a line.

Assume y1 ≤ x1 and y2 ≤ x2. This means x1/y1 and x2/y2 are in R. Multiply them together and (x1x2)/(y1y2) is in R. In other words, y1y2 ≤ x1x2. Multiplication in F, which defines the operation in the group G, respects order.

What about strict inequality? Suppose y1 < x1 and y2 ≤ x2. We know, from the above, that y1y2 ≤ x1x2. Suppose they are equal. Thus (x1x2)/(y1y2) is a unit in R. Also, x1/y1 and x2/y2 are in R, and the former is a nonunit. So a nonunit times an element of R gives a unit; this is impossible. Therefore y1y2 < x1x2, and strict inequality is preserved.

Might x be a torsion element? Let x have a positive valuation in the group G. In other words, x is in R, and x is not a unit. Now the powers of x in R have valuations 1x 2x 3x 4x and so on. The subgroup generated by x produces a copy of the integers, and the order of these "integers" is compatible with the ordering on G. Remember that strict inequality is preserved, hence 0 < x implies x < 2x. (Add x to both sides.) The powers of x advance in valuation, and do not cycle back around to zero, x being a nonunit. Every nonzero element of G has infinite order, and G is torsion free.

If G is trivial then R is a trivial valuation ring. Such a ring is a field, nothing but units. If the ring is nontrivial, let x be a nonzero nonunit and note that the powers of x have increasing valuations, as described above. The powers of x also produce an infinite descending chain of ideals, else xn = xn-1v, and x is a unit. Therefore a valuation ring with a nontrivial maximal ideal cannot be artinian.

Here is an example of a valuation ring. Like most valuation rings, it is based on localization. Let p be any prime, and let R be the fractions whose denominators are not divisible by p. The fraction ring F is the rational numbers. Given a fraction x in F, write it in lowest terms and look at the exponent e on p. Remember that this exponent can be negative if p is in the denominator. Now x is in R iff e is nonnegative, and x is a unit in R iff e = 0.

When two fractions are multiplied together, the corresponding exponents on p are added. (You need the fundamental theorem of arithmetic to make this rigorous, but it's pretty clear.) When looking at the exponents of p, multiplication in F becomes addition in the integers. The kernel is precisely those fractions with e = 0, hence the quotient group, i.e. the valuation of F, is equal to Z, the group of integers under addition.

If x and y are fractions then y ≤ x iff x/y is in R, iff there are more powers of p in x than in y. As the fraction gets smaller, with more and more powers of p in the denominator, its valuation decreases. The fractions with negative valuation are the fractions that are not in R.

Let x and y be nonzero elements of the field F. The valuation of xy is v(x) + v(y); that comes from the definition of the valuation group. How about v(x+y)?

If x ≤ x + y, then 1+y/x belongs to R. Similarly, if y ≤ x + y than 1+x/y belongs to R. R contains either x/y or y/x, and it certainly contains 1, so either x or y is less than x+y. The valuation of the sum is at least as large as the lesser valuation.

Assume v(x) > v(y). This means 1 + x/y is in R, but not 1 + y/x. Hence y ≤ x+y, but not so for x.

Since R does not contain 1 + y/x it contains the inverse, x/(x+y). It also contains 1, hence R contains y/(x+y). Therefore x+y ≤ y. The sum is both ≤ and ≥ y, hence it equals y. If x and y have different valuations, their sum has the lesser valuation.

Recall the example of the last section, fractions without p in the denominator. The valuation is the exponent of p. Sure enough, p2 + p4 has exactly two powers of p in its factorization. The valuation of the sum is 2. When the operands have the same valuation the sum could have the same valuation or larger. Add p+p to get the same valuation, or p + (p-1)p to get a larger valuation.

Let v(0) be a point at infinity, higher than anything in G. The valuation of x + -x is now infinity, which is certainly higher than v(x). And the valuation of x + 0 is v(x), as it should be, since 0 has a higher valuation. The rules for the valuation of the sum or the product now apply across all of F.

In this section we will begin with the valuation group, and work backwards to produce the valuation ring. Here are the conditions that must hold.

  1. F is a field and G is an abelian group that is linearly ordered.

  2. The action of G respects order. Add something larger and you get something larger.

  3. A valuation homomorphism maps the nonzero elements of F onto G. Multiplication in F corresponds to addition in G.

  4. 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 these four properties, is a valuation group, with a corresponding valuation ring.

Recall that R is the preimage of the nonnegative members of G, union 0. This time let e be a positive element of G and let H be the preimage of the elements of G that lie at or above e. Throw in 0, and use the properties of the previous section to show H is an ideal. The same holds for the preimage of the elements of G that are strictly greater than e.

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. We already showed the ideals of a valuation ring are linearly ordered, so this is nothing new.

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 I called E above. The ideals of R and the initial segments of G correspond 1 for 1.

If an ideal H 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 H. Conversely, if x maps to the least element in the image of H, then everything in H has a higher valuation than x, everything in H is generated by x, and H is principal. An ideal is principal iff it has a least element in G.

If G is isomorphic to the integers, then the image of an ideal H has a least element, and H is principal. Therefore R is a pid.

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. 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.

A pid is noetherian; how about the converse? Suppose x is an element of G, and a cluster point from above. Subtract x across the board, so that 0 is a cluster point from above. Let the initial segment E contain 0, and infinitely many positive elements. Let E decrease, step by step, although E always includes infinitely many positive elements. The ideals that correspond to these decreasing initial segments form an infinite ascending chain, hence the ring is not noetherian. Turn this around, and a noetherian valuation ring produces a valuation group with no cluster points from above.

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. Therefore a noetherian valuation ring has a valuation group with no cluster points anywhere. This is called a discrete valuation group, and R is a discrete valuation ring, or dvr.

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 such that v(y) is not a multiple of v(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. The following conditions on a valuation ring R are equivalent.

  1. R is noetherian

  2. R is a dvr

  3. R is a pid

  4. valuation group = Z

If R is a dvr, its maximal ideal is principal, generated by some element t, and every ideal is a power of this maximal ideal, generated by a power of t. An example is the fractions of polynomials in t over a field K, that do not have t in the denominator, e.g. (t2+2) / (t+1). The power of t is the valuation, and the valuation group is the integers.

You may be wondering why it is called a discrete valuation ring. Since G is linearly ordered, it has a linear topology, wherein base sets are open intervals or open rays. A topology is discrete if every point is an open set. If G is discrete, then every point is an open interval, having a point above and a point below. G has no cluster points, and is equal to Z. R is a discrete valuation ring iff its valuation group is discrete as a topological space.

So far all the valuation groups we've encountered are discrete. In fact we've only seen examples of Z. But any ordered abelian group will do.

Let G be a linearly ordered abelian group, where the group operator respects order. That is, a ≤ b and x ≤ y implies a+x ≤ b+y.

If a < b, and x ≤ y, then b+x ≤ b+y, and subtracting b-a from the left, a+x < b+y. Addition in G respects strict inequality.

Let R be any integral domain and let R[G] be the groupring G over R. If you're not familiar with grouprings, I can explain it pretty quickly. Let R[G] consist of finite linear combinations of elements of G with coefficients taken from R. It's basically a ring of polynomials where the elements of G multiply as prescribed by the group. Hence the notation R[G], which is usually reserved for polynomials. The additive identity is 0, which indicates no group elements are used. The multiplicative identity is 1e, where e is the group identity. R embeds in R[G] via R*e. Nonabelian groups create interesting noncommutative rings, but in this case the group, and the ring, are commutative.

Let the valuation of a member of R[G] be the lowest group element represented, according to the linear order of G. The valuation of 0 is infinite, as usual.

Multiply two polynomials together, and the lowest group element in the first times the lowest group element in the second gives the lowest group element in the product. The coefficients on the two lowest group elements are multiplied to produce the coefficient on the lowest group element in the product, and since R is an integral domain, this coefficient is nonzero. G respects strict order, so the other pairwise products taken from the two polynomials are higher. Therefore R[G] is an integral domain. Also, the valuation of the product is the sum of the two valuations. The valuation function is a group homomorphism from the nonzero elements of R[G] onto G, extracting the least group element of x.

Let K be the fraction field of R[G], which can be represented as quotients of polynomials from R[G]. Note that K includes F, the fraction field of R.

Build a valuation homomorphism from the nonzero elements of K onto G. If an element of K is represented by a/b, the quotient of two polynomials in F[G], map this fraction to v(a) - v(b). If c/d represents the same fraction, then ad = bc, hence v(a) + v(d) = v(b) + v(c), and v(a) - v(b) = v(c) - v(d). The valuation function is well defined. Multiply two fractions together to show this is indeed a homomorphism.

Review the four criteria for a valuation group. The first three criteria are satisfied. We only need prove condition 4, the valuation of the sum.

Picture an arbitrary multiplier m, a member of K. If a and b are elements of K, and v(a+b) is at least as large as the lesser of v(a) and v(b), then the same relationship holds for m*a and m*b. This is because multiplication by m distributes over a+b, and multiplication by m shifts everything in G, preserving order.

If a and b are fractions, multiply through by the common denominator. This becomes our multiplier m above. Evaluate the valuation of the sum in R[G], then you can multiply through by 1/m to get back to the fractions in K.

Let a and b be finite linear combinations of elements of G. The least element in the sum is the least element in either a or b, or perhaps something larger if the least elements of a and b cancel each other out. Therefore v(a+b) is greater than or equal to the lesser of v(a) and v(b). This proves condition 4, and makes G a valuation group.

The valuation ring associated with G is the preimage of the nonnegative elements of G. These are the fractions in K where the denominator has a valuation no higher than the numerator. The units are the fractions where the numerator and denominator have the same valuation.

In conclusion, any ordered abelian group will do. There is no limit on cardinality. Let R be an integral domain, the integers for instance. Upgrade R to its fraction field, (you always have to do this anyways), build F[G], find its fraction field K, and G becomes the valuation group for K.

In an earlier section we placed a linear topology on the valuation group G. In this section I will place a topology on the field F. In fact F becomes a metric space. All we need do is define a valid metric.

This process assumes the valuation group G can be embedded in the reals. This is usually the case, since G is linearly ordered. It certainly holds when G = Z.

Let c be any real number between 0 and 1, and establish the following metric.

|x,y| = cr(v(x-y))

In other words, subtract x and y, find the valuation of the difference, map that to a real number, and raise c to that power. If x-y = 0, the valuation is infinite, and c, a number less than 1, raised to the infinite power, is driven to 0. Otherwise the metric is positive.

As you can see, |x,y| = 0 iff x = y. Since the valuation does not depend on the sign, -1 belonging to R, |x,y| = |y,x|. That leaves only the triangular inequality.

Let x y and z be elements of the field F. We want to show |x,z| ≤ |x,y| + |y,z|. Verify by hand that this is true when any two of the three variables are equal.

Note that z-x = z-y + y-x. As per the valuation of the sum: v(z-x) is at least as large as the lesser of v(z-y) and v(y-x). Since c is less than 1, larger valuations lead to smaller metrics. Thus the metric on the left is bounded by one of the metrics on the right. It is certainly bounded by the sum of the metrics on the right, and that proves the triangular inequality. F is a valid metric space.

In this space, every triangle is isosceles. If z-y and y-x have different valuations, then their sum, z-x, has the lesser of the two valuations. Two of the three lengths are always the same.

As usual, a circle is the locus of points a fixed distance from a given center. The unit circle is the elements of F with metric 1, having valuation 0. These are the units of R. The unit disk is all of R.

Now consider any circle with center c and radius t. Let p be a point inside the circle and let q be any point on the circle. Draw the triangle cpq. We know that the distance from c to p is less than the distance from c to q. That's what it means to be "inside" the circle. So cq has a smaller valuation. The valuation of the sum, from p to q, has to equal this lesser valuation. Thus the distance pq is the same as the distance cq. This holds for every q on the circle. The conclusion: every point inside a circle is at the center of the circle.

Return to our favorite example, the ring of fractions without p in the denominator. The valuation is the exponent on p, which embeds nicely into the reals. Hence the metric is well defined. Set c = ½ for illustration. The points that are ½ distance from the origin have a factor of p in the numerator. Points that are ¼ distance from the origin have a factor of p2 in the numerator, and so on. Points like p100, which use to be far away under the traditional distance metric, are just a hair's breath from the origin. Conversely, the point p-100 has been hurled light-years away. The rationals have definitely been rearranged, but the result is still a metric space. This is called the p-adic topology on the rationals.

F is a topological field, i.e. the operators of F are continuous. Let's start with addition. This makes sense in the p-adic topology. If you change x and y by adding high powers of p, their sum will differ from the original by a high power of p. Numbers close to x and y add up to something close to x+y. Let's prove this in general.

Let v be any valuation that is larger than the valuation of x or y. Add s to x and t to y, where s and t have valuation at least v. This gives x+y+(s+t). The valuation of s+t is at least v, so (x+s)+(y+t) is within ε of x+y, as long as s and t are less than ε.

There are some special cases when x or y is 0, or when x = -y; I'll let you deal with those.

Multiplication is also continuous. Consider the valuation of (x+s) × (y+t) - xy. This is at least the valuation of xt or the valuation of ys or the valuation of st. Since s is under our control, make sure its valuation is at least v - the valuation of y. Thus the valuation of ys is at least v. Do the same for t, and the valuation of xt is at least v. Finally, make sure s has a valuation at least v, and t has a valuation at least 0. Now st has a valuation at least v, and the same is true of the sum. When the factors differ by s and t respectively, the product is within ε of xy.

Again there are special cases when x or y or both are 0.

Next look at the inverse map 1/x. If x is changed by s, look at the difference between 1/x and 1/(x+s). This is s over x*(x+s). Select s so that its valuation is higher than x. The denominator has the same valuation as x2, which is twice the valuation of x. Add v to this, and make sure s has an even higher valuation. Now the valuation of s/x2 is at least v, and we are within ε of 1/x. Obviously this fails when x = 0.

Put this together and division is a continuous operator from F cross F into F, provided the divisor is nonzero.

If the valuation group of F embeds in the reals, then F is a metric space, as described in the previous section. Is F a complete metric space? If not, what does its completion look like? If you aren't familiar with cauchy sequences and complete metric spaces, you may want to start here.

The points in the completion of F are cauchy sequences. To be precise, a point is an equivalence class of cauchy sequences that converge to the same thing. If s and t are cauchy, and s-t converges to 0, then s and t represent the same point in the completion of F. This is a bit like 3.699999… and 3.700000… representing the same real number.

The original elements of F embed in the completion in the usual way. If x is in F then the sequence x,x,x,x,x… is in the completion of F.

The completion of F is a field, just as the reals form a field. There is a lot of stuff you have to do if you want to be rigorous. You need to show, for instance, that addition is well defined in the completion. Fix a high valuation corresponding to ε. For some n, and for j beyond n, sj-sn has at least this valuation, as does tj-tn, hence (sj+tj) - (sn+tn) has this same valuation or higher. The sum s+t is a convergent sequence, which is a point in the completion of F. But what if s is replaced with some other sequence that also converges to s? Does s+t still land in the same equivalence class? Add a sequence e to s and consider (s+e)+t. Here e approaches zero, so s and s+e represent the same point in the completion. By associativity, (s+e)+t = e+(s+t). The new result is really the same as s+t, adjusted by a sequence that approaches 0. Thus addition is well defined.

If c is a scalar and s a sequence, and e is a sequence that approaches 0, c*(s+e) = c*s+c*e. Since c*e drops to 0, multiplication by c is well defined. Set c = -1 and subtraction is well defined.

Step gently into multiplication by considering a sequence s and another sequence e that approaches 0. Move out beyond some large n, and let a = si and a+v = sj. Of course v has a high valuation, and the terms ei and ej also have a high valuation - as high as you like. Consider (a+v)ej - aei. Rewrite this as a(ej-ei) + vej. Since e is cauchy the first term drops to zero. Since e and v both have high valuation, the second term also approaches 0. Any sequence in the completion, multiplied by a sequence that approaches 0, produces another sequence that approaches 0. Symbolically, s*0 = 0.

Consider the product of two arbitrary sequences s*t. Add a sequence e to s, and the product sequence is shifted by e*t. If e approaches 0 then e*t approaches 0. All the sequences that represent s give the same result, namely s*t, and multiplication is well defined.

You can also prove that addition and multiplication are commutative and associative, and the latter distributes over the former, and 0 and 1 are the respective identities. In other words, the completion forms a ring. Furthermore, the original space embeds in the completion as a subring, and as a subspace with the induced topology. (I'll leave all this to you.)

Finally, assume the sequence s does not approach 0, and consider the term by term inverse of s. Since 1/x is continuous in F, 1/s is cauchy. Add the sequence e to s. The difference between 1/(s+e) and 1/s is e/(s*(s+e)). If s is convergent, yet does not approach 0, then it is, beyond some n, bounded away from 0. (This is a lemma that you may want to verify.) The valuation of the terms of s is, eventually, bounded below by some value b. Go out further, so that the valuation of the terms of e is always beyond b. Now the denominator is no worse then b2 in valuation, and the numerator keeps increasing. The metric approaches 0, and the sequence 1/(s+e) has the same limit as the sequence 1/s. Division is well defined, and the completion forms a field.

I ran through this pretty fast, but it's all very similar to theorems you've seen before, when completing the rationals to produce the reals.

Apply the above to build the p-adic numbers. Let p be a prime and let R be the ring of rational numbers that do not have p in the denominator. This is a valuation ring, and it produces the p-adic topology on R, and on F, the fraction field of R, which is in this case the rational numbers. The completion of this valuation space is, by definition, the p-adic numbers. In other words, a sequence of rationals that is cauchy, using the valuation metric, represents a p-adic number. Such can be written, canonically, as an infinite series of digits from 0 to p-1, just as a real number has an infinite decimal expansion. This was described in chapter 1, including the field operators + - * /.

In the last section I defined the p-adic numbers as the completion of the rationals Q, using the p-adic topology. The result, Q′, is both a field and a complete metric space, but many questions remain.

  1. Is Q′ larger than Q? Did we really get anything new? Or does every cauchy sequence converge to a rational number?

  2. If Q′ is in fact larger, is it algebraically closed? Does it at least contain the closure of Q? Is every algebraic element of Q contained in Q′?

We aren't going to characterize all the elements of Q′, but we can certainly answer questions 1 and 2 above.

When the linear metric was used, the completion of Q produced R, the real numbers. This brings in lots of new, algebraic elements, like the square root of 2, but R is not algebraically closed, since there is no square root of -1. If the reals are a guide, the answer to (1) will be yes, and the answer to (2) will be no. Indeed, this is the case.

Let q(x) be an irreducible polynomial with integer coefficients. To keep things simple, let the degree of q be small, say 5 or less, and let p be odd, and let the coefficients of q be nonzero mod p. Let s be a p-adic number that is a solution to q(x). Thus s is a cauchy sequence of rational numbers that produces 0 when substituted into q(x). Since s is a sequence, rather than a canonical series of digits, operations are performed term by term. In other words, x2 becomes a sequence whose ith term is si2. The result is a sequence whose ith term is q(si). s is a root of q(x) iff q(sn) approaches 0 as n approaches infinity, iff the valuation of q(sn) approaches infinity as n approaches infinity. We are looking for a sequence of rational numbers, such that q(sn) is divisible by ever higher powers of p.

Start the sequence by assuming there is an integer c such that q(c) is divisible by p. Thus c is a root of q(x) mod p. If q(c) = 0 then c is a root of q, and we are done. Otherwise let q(c) = gpj. Now c is the first term in our sequence, s0 = c, and the valuation of q(c) equals j.

Next, evaluate q(c+pj) mod pj+1. Multiply the linear coefficient by pj, and add this to gpj. Move to the squared term and use the binomial theorem. We already accounted for c2. Multiply the coefficient on x2 by 2cpj, and add this to the growing total. The square of pj will give a higher valuation, so no worries about that. Then bring in 3c2pj times the coefficient on the cubed term. Again, the higher powers of pj have higher valuations. Add 4c3pj times the coefficient on the fourth term, and so on. This gives the value of q(c+pj) mod pj+1.

Add up the linear coefficient, plus 2c times the coefficient on the squared term, plus 3c2 times the coefficient on the cubed term, and so on up to ncn-1 times the lead coefficient, and call this sum u. We are only interested in u mod p. By the selection of j, u is nonzero. Let w = -g/u. Evaluate q(c + wpj). Now w is multiplied by u, then added to g. This is 0 mod p, hence q(c + wpj) is divisible by pj+1. The valuation has increased.

Set s1 = c + wpj, and repeat the above procedure to find s2, s3, and so on. Thus s is a root of q(x). The solution mod p has been parlayed into a solution in the p-adic numbers.

The integer p2+1 has no square root in Q, but it always has a square root in Q′. Reduce mod p, and p2+1 becomes 1, and 1 always has a square root mod p, namely 1. Set s1 = 1 to start the sequence. The sum of the coefficients times their degree is 2, which is nonzero mod p, at least when p is odd. The inductive step is valid, and the process described above builds the entire sequence s.

Turning to p = 2, Q′ contains the cube root of 3, but Q does not. Again, 1 seeds the sequence, and the sum of the coefficients times degree is 3, which is 1 mod 2.

We have answered question (1) above. Q is not complete with respect to the p-adic topology. Its completion Q′ brings in additional elements, including quite a few algebraic elements that were not part of Q to begin with. But don't confuse the square root of 2 in the 7-adic topology with 1.4142135…, which is the square root of 2 in the linear topology. The numbers are both sequences, or series if you prefer, but they bear no relationship to one another, because the metrics are different. In fact the p-adic square root is a series of integers that grow larger and larger (in the traditional sense). However, these integers, when squared, do indeed approach 2 under the p-adic metric. This is so weird, maybe we should derive the first few terms.

Since 32 = 2 mod 7, set s0 = c = 3. Now q(3) = 7, or 1×7 if you prefer. The value of u is 6, so our first instance of w is -1/6, or 1. The next term in the sequence is therefore 3 + 1×7, or 10.

Evaluate q(10) and get 98, which is 2×72. Since -1/u is conveniently 1, just use the coefficient 2 as is. Therefore s2 = 10 + 2*49 = 108. Square this and subtract 2 to get 34×73, whence s3 = 108 + 6*343 = 2166. The next value of q is 1954×74, hence s4 = 2166 + 1×2401 = 4567. Continue this process to build sqrt(2).

Let's see if we can answer question (2). This time we are looking for a q(x) that has no solution in the p-adic numbers.

Assume there is no integer c that is a root of q(x) mod p. Further assume there is no solution for the mirror image of q. If q = x3 + 2x + 19, then the reverse of q is 19x3 + 2x2 + 1. Neither polynomial has a root mod p.

Suppose a/b is a solution in the p-adic numbers. Either a or b could be divisible by p, but not both. If a is divisible by p, then a/b has positive valuation, and a/b is also divisible by p. Reduce everything mod p, and q(0) = 0. Thus q has a root mod p, namely 0, which is a contradiction.

If a and b are units, both nonzero mod p, then a/b is a root of q, again a contradiction.

If b is divisible by p, multiply through by bn, and divide by an, and 0 is a root of the reverse polynomial of q. This is another contradiction, hence q has no solution in the p-adic numbers.

It is enough to know that q has no solution mod p, and the lead and constant coefficients of q are not divisible by p.

When p = 7, and q = x2 + 1, there is no integer c such that c2 = -1 mod 7, hence no square root of -1 mod 7 in the p-adic numbers. Q′ is not algebraically closed, and it doesn't even contain the algebraic closure of Q. The answer to (2) is no, as expected.

This generalizes to any prime p. Let w be an integer that is not a square mod p, and w does not have a square root in the p-adic numbers. When p = 2, there is no solution to x2 + x + 1 in the 2-adic numbers.

In summary, the p-adic numbers bring in some, but not all, of the algebraic elements of Q.

The connection between equations mod p and the p-adic numbers can be generalized to other rings, and is described formally by Hensel's lemmas, coming up in a later chapter.

Like any unbounded metric space, the p-adic space is not compact; however, it is locally compact. Any point x lies in a closed ball of radius r. To illustrate, let r = 1, the unit ball. Prove this is totally bounded, and it becomes compact.

Points with metric ≤ 1 have valuation ≥ 0, and are integers in the traditional sense, or completions of these integers. They are power series, not laurent series. The entire series of integers mod p lies to the right of the radix point. Fix an ε, which sets a valuation v. To illustrate, let v = 3. Any series starting with .2153… (in the 7-adic numbers), when subtracted from .2153, gives a difference whose valuation exceeds 3. The open ball of radius ε about .2153 includes all series that start with .2153. Thus 74 = 2401 open balls of radius ε cover the closed unit ball of radius 1. The same can be done for any radius and any ε, and that completes the proof.

What was done for the integers about a prime number p, can be done for any pid R about a prime element p. Localize about p, and RP is still a pid, but there is only one prime ideal, generated by p. All other ideals are powers of this ideal, generated by powers of p. If a/b is not in this ring then b is divisible by p, whence a is not divisible by p. Flip this around and b/a is in RP, hence RP is a valuation ring. Since it is also a pid, it is a dvr. Valuations correspond to ideals, and thus the valuation of x is the largest power of p that generates x. This implies a metric, and the fraction field F becomes a metric space. F has a completion, just as ZP gave birth to the p-adic numbers. Here is another example.

Let R = K[y], the polynomials in y over a field K. This is a pid. Let p(y) be an irreducible polynomial in R, a prime element in the pid. (The simplest case would be p(y) = y.) Let T be the fractions whose denominators are not divisible by p. This is a valuation ring inside the field F = K(y). T is a dvr, and defines a metric on the field F. Let F′ be the completion of F. Does this bring in any new algebraic elements? Are there algebraic elements that are not present in F′?

Apply the proofs in the previous section. Let q be a polynomial with coefficients in K[y], not divisible by p, and let q have the root c(y) mod p. If c joins the coefficients of q to build a polynomial u(y) that is not divisible by p, then q has a root in F′, which was (perhaps) not present in F. On the other hand, if q has no root c(y) mod p, then q has no roots in F′.

Many different sequences represent the same entity in the completion of a valuation ring. The same thing occurs when the rationals produce the reals. However, in that case we have settled on canonical, decimal representatives. Many sequences converge to π, but the most familiar is 3.14159265… This is shorthand for a series that builds a cauchy convergent sequence. Each decimal digit is added to the growing real number, and the sequence of partial sums becomes the convergent sequence. This is the canonical series, and all real numbers can be represented in this fashion.

Let R be a pid, and let p generate a maximal ideal in R, hence p acts as the base for the valuation. An entity in R has valuation v if it is divisible by pv. The same holds for an entity in F, although v could be negative.

Let s be a cauchy sequence, an element in the completion of F. Select any v, and beyond some sn, the difference between sn and sj has valuation v or higher. In other words, sj = sn mod pv. Beyond n, sj is constant mod pv. Later on, s becomes constant mod pv+1, then it becomes constant mod pv+2, and so on. At each step we add a power of p times something in K, where K is the quotient field R/p. Replace s with a series that carries the same information. It looks something like 3 + 6p + 4p2 + 0p3 + 5p4 + p5 + …

The series could start with some negative powers of p, perhaps 3/p2 + 4/p, but there are finitely many of these. The canonical series for any cauchy sequence is a laurent series, the sum of powers of p with coefficients in K.

In the case of the p-adic numbers, p is a specific prime, and the terms of the series are integers mod p times powers of p. When R = K[y], and p = y (y being an irreducible polynomial), the canonical series for an element in the completion of K(y) is a formal laurent series in y.

In the reals, there is some overlap, e.g. 0.399999… = 0.400000… But there is no overlap in the p-adic completion. Suppose two series s and t differ in the coefficient on pj. The difference, s-t, admits an instance of pj, and perhaps instances of higher powers of p. The valuation is always j. Since s-t does not approach 0, s and t represent different entities in the completion.

In many cases operations can be performed on these canonical series instead of the underlying cauchy sequences. The simplest ring is K[y], its completion consisting of laurent series. Consider the term by term sum of two laurent series, and look at the resulting partial sums. Each is the sum of the corresponding partial sums in the original series. Therefore entities in the completion can be added or subtracted using standard polynomial arithmetic. Just add two laurent series together; that's all.

The above is similar to adding two decimal numbers digit by digit, although there is no carry operation. The p-adic numbers include a carry operation which flows to the right. If p = 7, then 3p2 + 6p2 becomes 2p2 + p3. The p3 carries into the next term, as we add the series together.

Series are multiplied using the diagonal algorithm associated with polynomials. If both power series begin with a constant term, the coefficient on p3, in the product, is a0b3 + a1b2 + a2b1 + a3b0. In the case of the p-adic numbers, there may be a carry operation, which is folded into the computation of the coefficient on p4. The result is the same as that obtained by multiplying, in the limit, the sequences of partial sums. They converge to the same thing, or more accurately, their difference converges to 0.

Since multiplication behaves as expected under polynomial operations, and since division is well defined in a field, two series can be divided using synthetic division. The quotient times the divisor reproduces the dividend, so it must be right. Dividing laurent series is straightforward, at least in theory. When dividing p-adic numbers, you may need to borrow from the higher power - the inverse of the earlier carry operation.

In summary, the completion of F, with respect to its valuation p, is the laurent series based on p, with coefficients in K. In some cases, polynomial arithmetic, with possible carry operations, implements addition, subtraction, multiplication, and division in the completion of F.

The formal power series, starting with p0 or higher, is a ring within the field of formal laurent series. Call this ring R′. Given a laurent series w, either w or w inverse is a proper power series with no reciprocal terms. Thus R′ is a valuation ring. Note that p generates the maximal ideal in the local ring R′, and the quotient is once again K. Given a laurent series in F′, let the exponent on the first term indicate the valuation. Verify that this satisfies the properties of a valuation, hence the valuation group of F′ is Z, and R′ is a dvr, and a pid. Furthermore, the valuation of F, embedded in F′, is the same as the valuation within F′. In other words, the valuation of F has been extended to its completion F′.

Beyond this, one can extend the valuation up to a field extension. Let R be a valuation ring in a field F, and let E be a field extension of F. Now R may not be a valuation ring in E, but it is still a local ring, and it can be extended to a maximal local ring S that dominates R, and is a valuation ring for E. This defines a group homomorphism from E* onto a valuation group G. The image of F* becomes a subgroup of the larger group G. Thus the valuation of F can be extended to a valuation on E, which is consistent with F.

Let E be the algebraic closure of F, and extend the valuation as above. The result is rarely a dvr, even when R is a dvr. Consider once again the polynomials in K[y], localized about y. The algebraic closure includes the square root of y, and the fourth root of y, and the eighth root of y, and so on. If y has valuation 1, these roots have valuations ½, ¼, and so on. The valuation group cannot equal Z.

Put this all together and a dvr induces a valuation on its fraction field, which extends to a valuation on the completion of F as a metric space, which then extends to a valuation on the algebraic closure of the completion. We may not have a simple formula that computes the valuation, but a valuation exists nonetheless, and it is compatible with the valuation produced by R.