Existence and Uniqueness of Algebraic Closure

Existence and Uniqueness of Algebraic Closure: Artin’sProof

Ayan Sengupta

March 15, 2015

Ayan Sengupta March 15, 2015 1 / 16

Recall

A relation ≤R on a set S is called partial ordering if

a ≤R b and b ≤R a =⇒ a = b ∀a, b ∈ S

a ≤R b and b ≤R c =⇒ a ≤R c ∀a, b, c ∈ S

A subset T of a partially ordered set S is called Totally ordered if∀s, t ∈ T s ≤R t or, t ≤R s.An upper bound of a totally ordered subset T of a partially ordered set(S ,≤R) is an element s ∈ S such that t ≤R s ∀t ∈ T .A maximal element of (S ,≤R) is an element s ∈ S such that s ≤R s

′=⇒

s = s′.


Recall


a ≤R b and b ≤R a =⇒ a = b ∀a, b ∈ S


A subset T of a partially ordered set S is called Totally ordered if∀s, t ∈ T s ≤R t or, t ≤R s.

An upper bound of a totally ordered subset T of a partially ordered set(S ,≤R) is an element s ∈ S such that t ≤R s ∀t ∈ T .A maximal element of (S ,≤R) is an element s ∈ S such that s ≤R s

′=⇒

s = s′.


Recall


a ≤R b and b ≤R a =⇒ a = b ∀a, b ∈ S


A subset T of a partially ordered set S is called Totally ordered if∀s, t ∈ T s ≤R t or, t ≤R s.An upper bound of a totally ordered subset T of a partially ordered set(S ,≤R) is an element s ∈ S such that t ≤R s ∀t ∈ T .

A maximal element of (S ,≤R) is an element s ∈ S such that s ≤R s′

=⇒s = s

′.


Recall


a ≤R b and b ≤R a =⇒ a = b ∀a, b ∈ S


A subset T of a partially ordered set S is called Totally ordered if∀s, t ∈ T s ≤R t or, t ≤R s.An upper bound of a totally ordered subset T of a partially ordered set(S ,≤R) is an element s ∈ S such that t ≤R s ∀t ∈ T .A maximal element of (S ,≤R) is an element s ∈ S such that s ≤R s

′=⇒

s = s′.


Zorn’s Lemma

Zorn’s LemmaIf every totally ordered subset of (S ,≤R) has an upper bound, then S hasa maximal element.


Zorn’s Lemma

Corollary 1Every commutative ring R has a maximal ideal M.Proof : Straight forward!Take the partially ordered set to be S , the set of all proper ideals of Rwith the ordering ⊆.Take any chain of ideals (totally ordered subset) of R -I0 ⊆ I1 ⊆ I2....It can easily verified that I = ∪Ii 6=< 1 > is a proper ideal of R and everyIi ⊆ I .Hence, I is an upper bound of the chain in S .Hence, Zorn’s lemmaimplies that S has a maximal element i.e. a maximal ideal in R.


Some Definitions

Definition : Algebraically Closed Fieldevery f (x) ∈ F [x ] of degree ≥ 1 has a root in F .Quick fact :To show a field algebraically closed it is sufficient to show that everymonic polynomial over it has a root in it.

Definition : Algebraic ClosureE ⊇ F is said to be algebraic closure of F if E is algebraic over F and E isalgebraically closed.e.g. - C is the algebraic closure of R.


Some Definitions

Definition : Algebraically Closed Fieldevery f (x) ∈ F [x ] of degree ≥ 1 has a root in F .Quick fact :To show a field algebraically closed it is sufficient to show that everymonic polynomial over it has a root in it.Definition : Algebraic ClosureE ⊇ F is said to be algebraic closure of F if E is algebraic over F and E isalgebraically closed.e.g. - C is the algebraic closure of R.


Theorem (Proposition 1)

Every field F has an algebraic closure which is unique upto isomorphism.

Proof : This proof was constructed by Emil Artin.


Existence of Algebraic Closure

S = f (x) ∈ F [x ]| f is monic . For each element f (x) in S we assign anindeterminant xf .Consider the ring F [S ], the polynomial ring with indeterminants xf .Consider the ideal I =< f (xf ) >f ∈S in F [S ].

Claim : I 6=< 1 >.Suppose I =< 1 >. Then

r∑i=1

gi .fki (xfki ) = 1 (1)

for some indexing ki and gi ∈ F [S ]. Now, we have r polynomials fki fori = 1, 2, ...r . Take F0 = F [S ]. If fk1 is irreducible in F0 then takeF1 = F0/ < fk1 >.Else F1 = F0.In this way we can generate Fr such that every polynomial fk1 , fk2 , ...fkr hasa root in it.



S = f (x) ∈ F [x ]| f is monic . For each element f (x) in S we assign anindeterminant xf .Consider the ring F [S ], the polynomial ring with indeterminants xf .Consider the ideal I =< f (xf ) >f ∈S in F [S ].Claim : I 6=< 1 >.

Suppose I =< 1 >. Then

r∑i=1

gi .fki (xfki ) = 1 (1)




S = f (x) ∈ F [x ]| f is monic . For each element f (x) in S we assign anindeterminant xf .Consider the ring F [S ], the polynomial ring with indeterminants xf .Consider the ideal I =< f (xf ) >f ∈S in F [S ].Claim : I 6=< 1 >.Suppose I =< 1 >. Then

r∑i=1

gi .fki (xfki ) = 1 (1)




Precisely, if the image of xfki in Fi−1[S ]/ < fki > is ¯xfki , then

f ( ¯xfki ) = ¯f (xfki ) = 0 in Fi .In Fr , if we substitute xfki by ¯xfki in equation (1) then we get

1 =r∑

i=1

gi .fki ( ¯xfki ) = 0 (2)

contradiction.Hence our claim is verified.Now, as I 6= 1 we can conclude from the above corollary that somemaximal ideal m of F [S ] contains I .Now we take E0 = F [S ] and E1 = E0/m.In E1 every f (x) ∈ S has a root. Moreover, in E1 every f (x) ∈ F [x ] has aroot. (Precisely, roots are xf )





1 =r∑

i=1

gi .fki ( ¯xfki ) = 0 (2)

contradiction.Hence our claim is verified.Now, as I 6= 1 we can conclude from the above corollary that somemaximal ideal m of F [S ] contains I .

Now we take E0 = F [S ] and E1 = E0/m.In E1 every f (x) ∈ S has a root. Moreover, in E1 every f (x) ∈ F [x ] has aroot. (Precisely, roots are xf )





1 =r∑

i=1

gi .fki ( ¯xfki ) = 0 (2)

contradiction.Hence our claim is verified.Now, as I 6= 1 we can conclude from the above corollary that somemaximal ideal m of F [S ] contains I .Now we take E0 = F [S ] and E1 = E0/m.

In E1 every f (x) ∈ S has a root. Moreover, in E1 every f (x) ∈ F [x ] has aroot. (Precisely, roots are xf )





1 =r∑

i=1

gi .fki ( ¯xfki ) = 0 (2)

contradiction.Hence our claim is verified.Now, as I 6= 1 we can conclude from the above corollary that somemaximal ideal m of F [S ] contains I .Now we take E0 = F [S ] and E1 = E0/m.In E1 every f (x) ∈ S has a root. Moreover, in E1 every f (x) ∈ F [x ] has aroot. (Precisely, roots are xf )


Inductively, we construct E0 ⊆ E1 ⊆ E2....We take E = ∪Ei . Then E is a field extension of F .

E is also algebraically closed.Consider E

′= u ∈ E |u is algebraic over F. We will show that E

′is an

algebraic extension of F (which is clear from our construction) and it isalgebraically closed.


Inductively, we construct E0 ⊆ E1 ⊆ E2....We take E = ∪Ei . Then E is a field extension of F .E is also algebraically closed.

Consider E′

= u ∈ E |u is algebraic over F. We will show that E′

is analgebraic extension of F (which is clear from our construction) and it isalgebraically closed.


Inductively, we construct E0 ⊆ E1 ⊆ E2....We take E = ∪Ei . Then E is a field extension of F .E is also algebraically closed.Consider E

′= u ∈ E |u is algebraic over F.

We will show that E′

is analgebraic extension of F (which is clear from our construction) and it isalgebraically closed.


Inductively, we construct E0 ⊆ E1 ⊆ E2....We take E = ∪Ei . Then E is a field extension of F .E is also algebraically closed.Consider E

′= u ∈ E |u is algebraic over F. We will show that E

′is an

algebraic extension of F (which is clear from our construction) and it isalgebraically closed.


Uniqueness of Algebraic Closure

Theorem

If Ω is an algebraically closed field containing F , and let E be an algebraicextension of F . Then there exists an F -homomorphism from E → Ω.Moreover, if E is algebraically closed then this homomorphism is anisomorphism.

Proof :Consider the inclusion map σ : F → Ω

Now, if E = F [α1, α2, ...αn] then, we can construct our F -homomorphismσ as follows:for each αi take the minimal polynomial over F . Let α

′i is a root of the

minimal polynomial in Ω then σ(αi ) = α′i

It is clear that this procedure will give us a homomorphism σ : E → Ωsuch that σ|F = σ.



Theorem


Proof :Consider the inclusion map σ : F → ΩNow, if E = F [α1, α2, ...αn] then, we can construct our F -homomorphismσ as follows:for each αi take the minimal polynomial over F . Let α






Theorem


Proof :Consider the inclusion map σ : F → ΩNow, if E = F [α1, α2, ...αn] then, we can construct our F -homomorphismσ as follows:for each αi take the minimal polynomial over F . Let α






For general case,Let S be the set of all fields M, F ⊆ M ⊆ E along with the correspondingF -homomorphism (M, σM) and let the partial ordering be(M, σM) ≤R (N, σN) if M ⊆ N and σN |M = σM .Now take a totally ordered subset T of S .M

′= ∪M∈TM is a field containing F and we define the corresponding

F -homomorphism, σM′ as follows:

if a ∈ M′

is in Mi , then σM′ (a) = σMi(a). Hence, (M

′, σM′ ) is an upper

bound of T .Hence, Zorn’s lemma implies that S has a maximal element (Mmax , σMmax ).Claim : Mmax = E .

Proof :



For general case,Let S be the set of all fields M, F ⊆ M ⊆ E along with the correspondingF -homomorphism (M, σM) and let the partial ordering be(M, σM) ≤R (N, σN) if M ⊆ N and σN |M = σM .Now take a totally ordered subset T of S .M

′= ∪M∈TM is a field containing F and we define the corresponding

F -homomorphism, σM′ as follows:

if a ∈ M′

is in Mi , then σM′ (a) = σMi(a). Hence, (M

′, σM′ ) is an upper

bound of T .Hence, Zorn’s lemma implies that S has a maximal element (Mmax , σMmax ).Claim : Mmax = E .Proof :



If E is algebraically closed, then σ : E → Ω is an isomorphism.


Any Questions ?


References

J.S.Milne (2012)

Fields and Galois Theory

Patrick Morandi (2004)

Artin’s Construction of an Algebraic Closure


Thank you


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Existence and Uniqueness of Algebraic Closure