2. PreliminariesIn this paper the ring means a commutative ring with unit.

Definition 2.1 A subring of a field is called a valuation ring of, if for every, , either or.

Definition 2.2 Let be a totally ordered abelian group. A valuation on with values in is a mapping satisfying:

i);

ii).

Definition 2.3 Let be field. A discrete valuation on is a valuation which is surjective.

Definition 2.4 A fractionary ideal of is an -submodule of such that, for some,.

Definition 2.5 A fractionary ideal is called invertible, if there exists another fractionary ideal such that.

Proposition 2.1 Let be a local domain. Every non zero fractionary ideal of is invertible if and only if is DVR (see [3] ).

Theorem 2.1 Let be a Noetherian local domain with unique maximal ideal and the quotient field of. The following conditions are equivalent.

i) is a discrete valuation ring;

ii) is a principal ideal domain;

iii) is principal;

iv) is internally closed and every non-zero prime ideal of is maximal;

v) Every non-zero ideal of is power of (see [3] ).

Definition 2.6 Let be a ring together with a family of subgroups of if satisfying the following conditions:

i);

ii) for all;

iii) for all;

Then we say has a filtration.

Definition 2.7 Let be a ring together with a family of subgroups of if satisfying the following conditions:

i);

ii) for all;

iii) for all;

Then we say has a strong filtration.

Example 2.1 Let be an ideal of, then is a filtration that is called adic filtration ring.

Definition 2.8 Let be a filtered ring. A filtered -module is an -module together with family of subgroup of satisfying:

1.;

2. for all;

3. for all.

Then we say has a filtration.

Definition 2.9 A map is called a homomorphism of filtered modules, if: i) is -module an homomorphism and ii) for all.

Definition 2.10 A graded ring is a ring, which can expressed as a direct sum of subgroup i.e. such that for all

Definition 2.11 Let be a graded ring. An -module is called a graded -module, if can be expressed as a direct sum of subgroups i.e. such that for all.

Definition 2.12 Let and be graded modules over a graded ring. A map is called homomorphism of graded modules if: i) is -module an homomorphism and ii) for all.

Definition 2.13 Let be a filtered ring with filtration. Let, and. Then has a natural multiplication induced from given

where. This makes in to a graded ring. This ring is called the associated graded ring of.

Definition 2.14 Let be a filtered -module over a filtered ring with filtration and respectively. Let, and. Then has a natural

-module structure given by, where.

3. Filtered Ring Derived from Discrete Valuation Ring and Its PropertiesIn this section we proved that, if is a discrete valuation ring, then is a filtered ring. And we prove some properties for.

Let be a field which be a domain and a discrete valuation ring (DVR) for. The map is valuation of.

Lemma 3.1 By above definition, the set is an ideal of.

Proof. (see [3] )

Theorem 3.1 If is a discrete valuation ring with valuation. Then is a filtered ring with filtration defined by

where.

Proof. By definition of valuation ring, it is obvious that. For the second condition for filtration ring we have, So we have.

For the third condition, we have for every and without losing generality. Since and are ideals of so

is an ideal of.

Now let then for and.

Thus

Consequently we have hence. Therefore is a filtered ring.

Proposition 3.1 Let be a local domain. If every non-zero fractionary ideal of invertible, then is filtered ring.

Proof. By proposition 2.1 is DVR then by theorem 3.1 is filtered ring.

Proposition 3.2 Let be a Noetherian local domain with unique maximal ideal and the quotient field of. Then is filtered ring if one of following conditions is held i) is a principal ideal domain;

ii) is principal;

iii) is integrally closed and every non-zero prime ideal of is maximal.

Proof. It follows from theorem (3.1) and theorem (2.1).

Definition 3.1 Let be a ring, and let be a totally ordered cancellative semigroup having identity. A function is a filtration if, and for alli), and ii), then is called a filtration.

For this filtration we have 1) the set of ideals;

2);

3);

4).

Lemma 3.2 Let be a filtration and let. Then:

i);

ii);

iii);

iv) if, then and.

Proof. See lemma 3.3 of [7] .

Proposition 3.3 If be a discrete valuation ring, then there exists a totally ordered cancellative semigroup, and such that:

i);

ii);

iii);

iv) if, then, and.

Proof. By theorem 3.1 there exists a filtration for, then by lemma 2.1 we have the all above conditions.

Proposition 3.4 Let be a filtered ring, , filtered -modules, and homomorphism of filtered -modules. If the induced map is injective, then is injective provided. (see [3] )

Corollary 3.1 Let be a valuation ring, , filtered -modules, and homomorphism of filtered -modules. If the induced map is injective, then is injective provided.

Proposition 3.5 If is a discrete valuation ring with valuation, Then is a strongly filtered ring with filtration defined by

where.

Proof. By theorem 3.1 is a filtered ring. Now we show for all. Since so

Consequently, and. Therefore.

Proposition 3.6 Let be a discrete valuation ring, and. If and, then

is smallest prime ideal in which contains, and is largest prime ideal in

which does not contains.

Proof. By proposition 3.5 is strongly filtered ring, then by proposition 4.2. of [7] -[9] we have If and, then is smallest prime ideal in such that contains, and is the largest prime ideal in such that does not contains.

Remark 3.1 Given a strong filtration on a ring, we say that a prime in is branched in, if cannot be written as union of prime ideals in such that properly contained in.

Corollary 3.2 Let be a discrete valuation ring and. Then a prime ideal in is branched in if and only if for some.

Proof. By proposition 3.5 is strongly filtered ring, then by proposition 4.5. of [7] a prime ideal in is branched in, if and only if, for some.