TWO LEMMAS : one seemingly true and one seemingly false

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TWO LEMMAS :

one seemingly true and one seemingly false

Maria Chlouveraki

Universit´e de Versailles - St Quentin

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Lemma T

LetR be an integrally closed domain and letF be its field of fractions. Letpbe a prime ideal of R. Then

(R[x])_pR_[x]∩F[x] =Rp[x].

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Lemma F

LetR be an integrally closed domain andf(x) =P

iaixⁱ,g(x) =P

jbjx^j be two polynomials inR[x]. If there exists an elementc∈R such that all the coefficients off(x)g(x) belong tocR, then all the products aibj belong tocR.

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Fact

An integral domainR is integrally closed if and only if it is the intersection of a family of valuation rings contained in its field of fractions.

LetR be an integral domain and letF be its field of fractions. We say thatR is a valuation ringif there exists a totally ordered abelian group Γ and an application v :F →Γ∪ {∞}which satisfies the following properties:

(V1) v(xy) =v(x) +v(y) forx,y ∈F. (V2) v(x+y)>min(v(x),v(y)) for x,y∈F. (V3) v(0) =∞.

(V4) {x ∈F|v(x)>0}=R.

In particular, if v(x)6=v(y), Property(V2)givesv(x+y) =min(v(x),v(y)). Further, Property(V1)implies the following :

v(1) = 0.

v(1/x) =−v(x) forx ∈F^×.

Ifx,y∈R, thenx/y∈R if and only if v(x)>v(y).

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Fact

LetR be an integral domain and letF be its field of fractions. We say thatR is a valuation ringif there exists a totally ordered abelian group Γ and an application v :F →Γ∪ {∞} which satisfies the following properties:

(V1) v(xy) =v(x) +v(y) forx,y ∈F.

(V2) v(x+y)>min(v(x),v(y)) for x,y ∈F. (V3) v(0) =∞.

(V4) {x∈F|v(x)>0}=R.

In particular, if v(x)6=v(y), Property(V2)givesv(x+y) =min(v(x),v(y)). Further, Property(V1)implies the following :

v(1) = 0.

v(1/x) =−v(x) forx ∈F^×.

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Fact

(V4) {x∈F|v(x)>0}=R.

In particular, if v(x)6=v(y), Property(V2)givesv(x+y) =min(v(x),v(y)).

Further, Property(V1)implies the following : v(1) = 0.

v(1/x) =−v(x) forx ∈F^×.

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Fact

(V4) {x∈F|v(x)>0}=R.

Further, Property(V1)implies the following :

v(1) = 0.

v(1/x) =−v(x) forx ∈F^×.

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Fact

(V4) {x∈F|v(x)>0}=R.

v(1/x) =−v(x) forx ∈F^×.

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Fact

(V4) {x∈F|v(x)>0}=R.

v(1/x) =−v(x) forx∈F^×.

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Fact

(V4) {x∈F|v(x)>0}=R.

v(1/x) =−v(x) forx∈F^×.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

jbjx^j be two polynomials inR[x]. If there exists an elementc∈R such that all the coefficients off(x)g(x) belong tocR, then all the products a_ib_j belong tocR.

Proof: It is enough to prove the result in the case whereR is a valuation ring. Let κ:=min_i(v(a_i)) andλ:=min_j(v(b_j)). Thenκ+λ=min_i,j(v(a_ib_j)). We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Let a_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

We have that i1+j1<im+jn, ∀(m,n)6= (1,1).

Therefore, the coefficientci₁+j₁ ofxⁱ¹^+j¹ inf(x)g(x) is of the form c_i₁_+j₁ =a_i₁b_j₁+X

(terms with valuation> κ+λ). Now,v(P(terms with valuation> κ+λ))> κ+λ=v(a_i₁b_j₁), whence

v(ci₁+j₁) =min(v(ai₁bj₁),v(X

(terms with valuation> κ+λ))) =κ+λ. Now, since all the coefficients of f(x)g(x) are divisible byc, we have that v(ci1+j1)>v(c), as desired.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

Proof: It is enough to prove the result in the case whereR is a valuation ring.

Let κ:=min_i(v(a_i)) andλ:=min_j(v(b_j)). Thenκ+λ=min_i,j(v(a_ib_j)). We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Let a_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

Letκ:=min_i(v(a_i)) andλ:=min_j(v(b_j)).

Thenκ+λ=min_i,j(v(a_ib_j)). We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Let a_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

Letκ:=min_i(v(a_i)) andλ:=min_j(v(b_j)). Thenκ+λ=min_i,j(v(a_ib_j)).

We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Let a_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j.

Let a_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Leta_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ.

Respectively, letbj₁,bj₂, . . . ,bj_s withj1<j2< . . . <js be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

We will show thatκ+λ>v(c) and thusc divides all the productsa_ib_j. Leta_i₁,a_i₂, . . . ,a_i_r withi₁<i₂< . . . <i_r be all the elements among thea_i with valuation equal toκ. Respectively, letb_j₁,b_j₂, . . . ,b_j_s withj₁<j₂< . . . <j_s be all the elements among the bj with valuation equal toλ.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

We have that i1+j1<im+jn,∀(m,n)6= (1,1).

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

(terms with valuation> κ+λ).

Now,v(P(terms with valuation> κ+λ))> κ+λ=v(a_i₁b_j₁), whence v(ci₁+j₁) =min(v(ai₁bj₁),v(X

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

Now,v(P(terms with valuation> κ+λ))> κ+λ=v(a_i₁b_j₁),

whence v(ci₁+j₁) =min(v(ai₁bj₁),v(X

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

(terms with valuation> κ+λ))) =κ+λ.

Now, since all the coefficients of f(x)g(x) are divisible byc, we have that v(ci1+j1)>v(c), as desired.

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Lemma F (Bourbaki)

iaixⁱ,g(x) =P

(terms with valuation> κ+λ))) =κ+λ.

Now, since all the coefficients of f(x)g(x) are divisible byc, we have that v(ci₁+j₁)>v(c), as desired.

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

Proof: The inclusionR_p[x]⊆(R[x])_pR[x]∩F[x] is obvious. Now, letf(x) be an element of F[x]. Thenf(x) can be written in the formr(x)/ξ, wherer(x)∈R[x] andξ∈R. If, moreover,f(x) belongs to (R[x])_pR[x], then there exist

s(x),t(x)∈R[x] witht(x)∈/ pR[x] such thatf(x) =s(x)/t(x). Thus, we have f(x) =r(x)

ξ = s(x)

t(x).

All the coefficients of the product r(x)t(x) belong toξR. Due to Lemma F, if r(x) =P

ia_ixⁱ andt(x) =P

jb_jx^j, then all the productsa_ib_j belong toξR. Now, the fact that t(x)∈/pR[x] implies there exists j₀such thatb_j₀ ∈/ p. Since a_ib_j₀∈ξR, for alli, we deduce thatb_j₀f(x) = (b_j₀r(x))/ξ∈R[x] and so all the coefficients off(x) belong toRp.

This result generalises to polynomial rings and Laurent polynomial rings in multiple variables.

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

Proof: The inclusionR_p[x]⊆(R[x])_pR[x]∩F[x] is obvious.

Now, letf(x) be an element of F[x]. Thenf(x) can be written in the formr(x)/ξ, wherer(x)∈R[x] andξ∈R. If, moreover,f(x) belongs to (R[x])_pR[x], then there exist

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

Proof: The inclusionR_p[x]⊆(R[x])_pR[x]∩F[x] is obvious. Now, letf(x) be an element of F[x]. Thenf(x) can be written in the formr(x)/ξ, wherer(x)∈R[x] andξ∈R.

If, moreover,f(x) belongs to (R[x])_pR[x], then there exist

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

s(x),t(x)∈R[x] witht(x)∈/ pR[x] such thatf(x) =s(x)/t(x).

Thus, we have f(x) =r(x)

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x). All the coefficients of the product r(x)t(x) belong toξR.

Due to Lemma F, if r(x) =P

ia_ixⁱ andt(x) =P

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

jb_jx^j, then all the productsa_ib_j belong toξR.

Now, the fact that t(x)∈/pR[x] implies there exists j₀such thatb_j₀ ∈/ p. Since a_ib_j₀∈ξR, for alli, we deduce thatb_j₀f(x) = (b_j₀r(x))/ξ∈R[x] and so all the coefficients off(x) belong toRp.

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

Now, the fact that t(x)∈/pR[x] implies there existsj₀ such thatb_j₀ ∈/ p.

Since a_ib_j₀∈ξR, for alli, we deduce thatb_j₀f(x) = (b_j₀r(x))/ξ∈R[x] and so all the coefficients off(x) belong toRp.

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

Now, the fact that t(x)∈/pR[x] implies there existsj₀ such thatb_j₀ ∈/ p. Since a_ib_j₀∈ξR, for alli, we deduce thatb_j₀f(x) = (b_j₀r(x))/ξ∈R[x] and so all the coefficients off(x) belong toR_p.

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Lemma T (C.)

(R[x])_pR_[x]∩F[x] =R_p[x].

ξ = s(x)

t(x).

ia_ixⁱ andt(x) =P

Now, the fact that t(x)∈/pR[x] implies there existsj₀ such thatb_j₀ ∈/ p. Since a_ib_j₀∈ξR, for alli, we deduce thatb_j₀f(x) = (b_j₀r(x))/ξ∈R[x] and so all the coefficients off(x) belong toR_p.

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Lemma B (C.)

LetR be an integrally closed domain and let F be its field of fractions. Lets(x) andt(x) be two elements ofR[x] such thats(x)/t(x)∈F[x]. If one of the coefficients oft(x) is a unit inR, thens(x)/t(x)∈R[x].

Proof: Sinces(x)/t(x)∈F[x], there exist r(x)∈R[x] andξ∈R such that s(x)

t(x) = r(x) ξ .

ia_ixⁱ andt(x) =P

jb_jx^j, then all the productsa_ib_j belong toξR. By assumption, there existsj₀such that b_j₀ is a unit inR. Sincea_ib_j₀∈ξR for alli, we deduce that a_i∈ξR for alli, whencer(x)/ξ∈R[x].

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Lemma B (C.)

t(x) = r(x) ξ .

ia_ixⁱ andt(x) =P

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Lemma B (C.)

t(x) = r(x) ξ .

All the coefficients of the product r(x)t(x) belong toξR.

Due to Lemma F, if r(x) =P

ia_ixⁱ andt(x) =P

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Lemma B (C.)

t(x) = r(x) ξ .

iaixⁱ andt(x) =P

jbjx^j, then all the productsaibj belong toξR.

By assumption, there existsj₀such that b_j₀ is a unit inR. Sincea_ib_j₀∈ξR for alli, we deduce that a_i∈ξR for alli, whencer(x)/ξ∈R[x].

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Lemma B (C.)

t(x) = r(x) ξ .

iaixⁱ andt(x) =P

jbjx^j, then all the productsaibj belong toξR. By assumption, there existsj₀such that b_j₀ is a unit inR.

Sincea_ib_j₀∈ξR for alli, we deduce that a_i∈ξR for alli, whencer(x)/ξ∈R[x].

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Lemma B (C.)

t(x) = r(x) ξ .

iaixⁱ andt(x) =P

jbjx^j, then all the productsaibj belong toξR. By assumption, there existsj₀such that b_j₀ is a unit inR. Sincea_ib_j₀∈ξR for alli, we deduce that a_i∈ξR for alli, whencer(x)/ξ∈R[x].

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Lemma B (C.)

t(x) = r(x) ξ .

iaixⁱ andt(x) =P

jbjx^j, then all the productsaibj belong toξR. By assumption, there existsj₀such that b_j₀ is a unit inR. Sincea_ib_j₀∈ξR for alli, we deduce that a_i∈ξR for alli, whencer(x)/ξ∈R[x].

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Preview

Letλ, µ∈C^× anda,b,c∈Nwithgcd(a,b,c) = 1. The polynomial

λx^ay^b+µz^c is irreducible inC[x,y,z], becauseλx+µis irreducible inC[x].