Abstract Algebra, Lecture 14 Jan Snellman Abstract Algebra, Lecture 14 Field Extensions

Abstract Algebra, Lecture 14 Jan Snellman Abstract Algebra, Lecture 14 Field extensions

General ﬁeld 1 extensions Jan Snellman Simple extensions 1Matematiska Institutionen Zeroes of Link¨opingsUniversitet polynomials

Link¨oping,fall 2019

Lecture notes availabe at course homepage http://courses.mai.liu.se/GU/TATA55/ Abstract Algebra, Lecture 14

Jan Snellman Summary

General field extensions 1 General field extensions Iterated simple extensions Simple extensions Degree, dimension Zeroes of polynomials Algebraic extensions 3 Zeroes of polynomials 2 Simple extensions Zeroes and multiplicities Classification of simple Splitting field extensions Algebraic closure Abstract Algebra, Lecture 14

Jan Snellman Summary

Jan Snellman Definition Suppose that E, F are fields, and that E is a subring of F . We write E ≤ F and say that E is a subfield of F , and that F is an overfield of E. General field The inclusion map i : E F is called a field extension (or equivalently, extensions the pair E ≤ F ). Degree, dimension Algebraic extensions → Simple extensions Example Zeroes of polynomials • Any field is an overfield of its prime subfield • Q ≤ R ≤ C • C ≤ C(x) ≤ C(x)(y) • Z2[x] Z2 ≤ (x2+x+1) Abstract Algebra, Lecture 14

Jan Snellman Definition Let E ≤ F be a field extension. Then F is a vector space over E. The dimension is denoted by [F : E], and refered to as the degree of the General field extension. If this dimension is finite, then the extension is said to be finite extensions dimensional. Degree, dimension Algebraic extensions Example Simple extensions Zeroes of polynomials • [C : R] = 2, so R ≤ C is a finite dimensional extension of degree 2. • [R : Q] = , so this extension is infinite dimensional. It is a theorem (as long as you accept the axiom of choice) that any vector space has a basis.∞ In the first example, we can take {1, i}, in the second, we need set-theory yoga to produce a Hamel basis. Abstract Algebra, Lecture 14

Jan Snellman Theorem (Tower thm) If K ≤ L ≤ M, then [M : K] = [M : L][L : M].

Proof General ﬁeld extensions Obvious if any extension involved is inﬁnite, so suppose [M : L] = m < , Degree, dimension Algebraic extensions [L : K] = n < . Then M has an L-basis Simple extensions ∞ u1,..., um, Zeroes of ∞ polynomials and L has a K-basis v1,..., vn. Claim: ui vj , 1 ≤ i ≤ m, 1 ≤ j ≤ n is a K-basis for M. Abstract Algebra, Lecture 14

Jan Snellman Proof (of claim) Spanning: take w ∈ M. Then

General ﬁeld m extensions Degree, dimension w = ci ui , ci ∈ L Algebraic extensions i=1 X Simple extensions m  n  Zeroes of =  dij vj  ui , dij ∈ K polynomials i=1 j=1 X X = dij vj ui i,j X Abstract Algebra, Lecture 14 Proof (of claim) Jan Snellman K-linear independence: suppose that

dij vj ui = 0. i,j General ﬁeld X extensions Degree, dimension Then Algebraic extensions m  n  Simple extensions  dij vj  ui = 0, i=1 j=1 Zeroes of X X polynomials so since the ui ’s are L-linearly independent, all coeﬃcients

n dij vj = 0. j=1 X

But the vj ’s are K-linearly independent, so all dij ’s are zero. Abstract Algebra, Lecture 14

Jan Snellman

Example

General ﬁeld [x] extensions Z2 [y] Z2[x] (x2+x+1) Degree, dimension Z2 ≤ ≤ Algebraic extensions (x2 + x + 1) (y 3 + y + 1) Simple extensions has degree 3 ∗ 2 = 6, and a basis consists of Zeroes of polynomials 1, x, y, xy, y 2, xy 2.

This ﬁnite ﬁeld thus has 26 = 64 elements. Abstract Algebra, Lecture 14

Jan Snellman Deﬁnition If E ≤ F , u ∈ F is algebraic over E if there is a non-zero polynomial f (x) ∈ E[x] having u has a zero, i.e., General ﬁeld extensions n Degree, dimension i Algebraic extensions f (x) = ai x , ai ∈ E, i=0 Simple extensions X Zeroes of and polynomials n i f (u) = ai u = 0 ∈ F . i=0 X The smallest degree of a polynomial that works is the degree of u over E. If u is not algebraic over E, then it is transcendental over E. Abstract Algebra, Lecture 14

Jan Snellman Example √ s = 2 + 1 ∈ R is algebraic of degree 2 over Q, since it satisﬁes

General ﬁeld 2 extensions (s − 1) − 2 = 0, Degree, dimension Algebraic extensions but no non-trivial algebraic relation of lower degree. Simple extensions On the other hand, the number Zeroes of polynomials 10−j! ∞ j=1 X is transcendental over Q, as proved by Liouville. Abstract Algebra, Lecture 14 Deﬁnition Jan Snellman The extension E ≤ F is algebraic if every u ∈ F is algebraic over E.

Example √ √ Let E = , and let F = a + b 2 a, b ∈ . Put u = a + b 2. General ﬁeld Q Q extensions • F is a ﬁeld, since Degree, dimension √ Algebraic extensions √ −1 1 a − b 2 a −b Simple extensions u = √ = = + 2. a + b 2 a2 + 2b2 a2 + 2b2 a2 + 2b2 Zeroes of polynomials 2 2 Note that a + 2b 6= 0 when (0, 0) 6= (a, b) ∈ Q × Q. • E ≤ F • E ≤ F is algebraic, with every element of F algebraic over Q with degree at most 2, since

(u − a)2 − 2b2 = 0. Abstract Algebra, Lecture 14

Jan Snellman Theorem If [F : E] = n < then E ≤ F is algebraic.

General ﬁeld Proof. ∞ extensions Degree, dimension Take u ∈ F , and consider Algebraic extensions 2 n Simple extensions 1, u, u ,..., u ∈ F Zeroes of polynomials These n + 1 vectors must be linearly dependent over E, which means that there are ci ∈ E, not all zero, such that

n c01 + c1u + ··· + cnu = 0

Thus, u is algebraic over E. Abstract Algebra, Lecture 14 Example Jan Snellman There are algebraic extensions that are not finite-dimensional. For instance, let E = , and let F be the smallest subfield of that contains √ Q R all p√for all primes√ p. Then all elements of F are algebraic; for instance, General field if u = 2 + 7 3 then extensions 12 Degree, dimension √ 3 ∗ 49 49 Algebraic extensions (u − 2)2 = = 122 48 Simple extensions √ 2 49 Zeroes of u − 2 2u + 2 − = 0 polynomials 48 √ 49 47 2 2u = u2 + (2 − ) = u2 + 48 48 47 8u = (u2 + )2 48 √ √ √ But the set 2, 3, 5,... is infinite and Q-linearly independent, so [F : E] = .

∞ Abstract Algebra, Lecture 14 Jan Snellman Definition Let E ≤ F be a field extension, and let u ∈ F . We denote by E(u) the smallest subfield of F containing E and u, in other words General field \ extensions E(u) = K E≤K≤F Simple extensions u∈K Classification of simple extensions Iterated simple Picture! extensions We can also describe it as Zeroes of polynomials p(u) E(u) = p(x), q(x) ∈ E[x], q(x) 6= z.p q(u)

We call E(u) a simple extension, and u a primitive element of the extension. Abstract Algebra, Lecture 14

Jan Snellman Example √ Q( 2) consists of all rational expressions like

√ √ 2 √ n a0 + a1 2 + a2 2 + ··· + an 2 General field √ √ 2 √ m , extensions b0 + b1 2 + b2 2 + ··· + bm 2 Simple extensions but this actually simplifies to just all Classification of simple extensions √ Iterated simple extensions a0 + a1 2.

Zeroes of −j! polynomials On the other hand, put u = j=1 10 , then all expressions ∞ P 2 n a0 + a1u + a2u + ··· + anu 2 m , b0 + b1u + b2u + ··· + bmu

that are not identical, are different. So Q ≤ Q(u) is infinite dimensional. Abstract Algebra, Lecture 14 Theorem Jan Snellman Let E ≤ F be a field extension, and let u ∈ F .

1 If u is algebraic over E, of degree n, then E ≤ E(u) is algebraic, and

E[x] General ﬁeld E(u) ' , extensions p(x) Simple extensions where the minimal polynomial p(x) Classiﬁcation of simple extensions 1 is irreducible Iterated simple extensions 2 has degree n Zeroes of 3 is the unique (up to association) non-zero polynomial of smallest polynomials degree such that p(u) = 0

2 If u is transcendental over E, then E ≤ E(u) is transcendental, and inﬁnite dimensional, and

E(u) ' E(x),

the ﬁeld of rational functions with coeﬃcients in E. Abstract Algebra, Lecture 14 Proof Jan Snellman • Consider

φ : E[x] F General field φ(f (x)) = f (u) extensions → Simple extensions • The image is a subring of F , and is contained in E(u). In fact, it is Classification of simple extensions E[u], the smallest subring containing u. Iterated simple extensions • Let I = ker φ. Zeroes of • • If I 6= (0), then I = (p(x) for a polynomial, which is (up to polynomials association) the unique polynomial of smallest degree in I . • Of course p(u) = 0; every pol in I has u as a zero, by definition. • By the first iso thm, E[x]/I ' E[u] ⊆ E(u) ⊆ F . • So E[u], a subring of a field, is a domain; so I is a prime ideal; so p(x) is irreducible; so I is maximal; so E[x]/I is a field; so E[u] is already a field; so E[u] = E(u). Abstract Algebra, Lecture 14

Jan Snellman Proof, cont

• • If I = (0), then φ is injective. • Then it factors through the splitting field E(x) of E[x]. That is, it extends to General field extensions φ^ : E(x) F Simple extensions f (x) f (u) Classification of simple φ^( ) = extensions g(x) → g(u) Iterated simple extensions • It is injective, by general nonsense Zeroes of polynomials • The image is precisely E(u), the simple extension • So E(x) ' E(u).

This explains “if not identical, then different”; two rational expressions in the transcendental u are equal iff they coincide as rational functions. Abstract Algebra, Lecture 14 Example Jan Snellman √ Let Q ≤ C 3 2 + i = u. What is Q(u)? We see that √ 2 General field u = 2 − 1 + 2i extensions √ 2i = u2 − 1 Simple extensions 2 2 Classification of simple −2 = (u − 1) extensions 4 2 Iterated simple u − 2u + 3 = 0 extensions Zeroes of Since f (x) = x4 − 2x2 + 3 ∈ [x] is irreducible, it is the minimal polynomials Q polynomial of U, and

[x] (u) ' Q Q (x4 − 2x2 + 3)

We have that [Q(u): Q] = 4. Abstract Algebra, Lecture 14

Jan Snellman

Deﬁnition

Let E ≤ F , and let u1,..., ur ∈ F . We deﬁne E(u1,..., ur ) either as

General ﬁeld • The smallest extension of E inside F which contains u1,..., ur , i.e., extensions \ Simple extensions K, Classiﬁcation of simple extensions E≤K≤F Iterated simple u1,...,ur ∈K extensions Zeroes of • or as the iterated extension polynomials E(u1)(u2) ··· (ur ) Abstract Algebra, Lecture 14

Jan Snellman Example √ √ √ √ Consider√ Q√( 2)( 3). We have that Q( 2) has a Q-basis 1, 2 , and√ that 3 6∈ 3. In fact, x2 − 3 is irreducible both over and over ( 3), General field √ √ Q Q extensions so it is the minimal polynomial of√ 3 over√ Q( 2). The tower theorem, Simple extensions and its proof, then yields that Q( 2)( 3) has a Q-basis Classification of simple extensions √ √ √ √ Iterated simple 1, 2, 3, 2 3. extensions √ √ √ √ Zeroes of Now consider u = 2 + 3 ∈ Q( 2, 3). Obviously, polynomials √ √ Q ≤ Q(u) ≤ Q( 2, 3), where the first inclusion√ √ is proper. By the tower theorem again, [Q(u): Q]√is a divisor of [Q( 2, 3): Q]√ = 4,√ so it is either 2 or 4. But u 6∈ Q( 2), so it is 4; and Q(u) = Q( 2, 3). Abstract Algebra, Lecture 14 Jan Snellman Theorem The extension E ≤ F is finite dimensional iff there are a finite number of elements u1,..., ur ∈ F , algebraic over E, such that F = E(u1,..., ur ). General field extensions Proof. Simple extensions If [F : E] = n < then there is a basis u1,..., un ∈ F . These basis Classification of simple extensions elements are elgebraic over E. Iterated simple extensions If there are such∞ algebraic elements, then clearly uj is algebraic over Zeroes of E(u1,..., uj−1), and [E(u1,..., uj−1, uj ): E(u1,..., uj−1)] ≤ [E(uj ): E], polynomials so by the tower theorem,

[F : E] = [E(u1,..., ur ): E] ≤ [E(u1): E] ··· [E(ur ): E] < .

∞ Abstract Algebra, Lecture 14

Jan Snellman

Theorem (Primitive element thm) General field extensions Let E ≤ F be a finite dimensional extension, and suppose that either char Simple extensions (E) = 0 (that is, Q ≤ E) or that E is finite. Then there exists a Classification of simple primitive element u ∈ F for the extension: F = E(u). extensions Iterated simple extensions Proof. Zeroes of polynomials The proofs are in Svensson, maybe also in Judson. Abstract Algebra, Lecture 14

Jan Snellman Example Consider the iterated simple extension General field extensions Z2[x] E[y] Z2 ≤ E ≤ F , E = 2 , F = 3 Simple extensions (x + x + 1) (y + y + 1) Classification of simple extensions Clearly Iterated simple span 2 2 extensions F = Z2(x, y) = Q(1, x, y, xy, y , xy ). Zeroes of polynomials Let’s find a primitive element! Put v = x + y. Then [1, v, v 2, v 3, v 4, v 5] Abstract Algebra, Lecture 14

Jan Snellman Example is 1, x + y, y 2 + x + 1, xy 2 + xy, y 2 + x + y, xy 2 + y 2 + x + y

General field We take the coordinate vectors of these (w.r.t. our prefered basis) and put extensions them in a matrix. Then these 6 powers span F iff they are linearly Simple extensions independent iff the matrix is invertible iff it has determinant 1 in Z2. The Classification of simple extensions matrix is Iterated simple  1 0 1 0 0 0  extensions  0 1 1 0 1 1  Zeroes of    0 1 0 0 1 1  polynomials      0 0 0 1 0 0     0 0 1 0 1 1  0 0 0 1 0 1

and it has determinant 1. So Z2(x + y) = Z2(x, y) = F . Abstract Algebra, Lecture 14 Definition Jan Snellman Let E ≤ F , and let n i f (x) = ai x ∈ E[x] i=0 X General field Then u ∈ F is a zero of f (x) if extensions Simple extensions n f (u) = a ui = 0 ∈ F Zeroes of i polynomials i=0 Zeroes and X multiplicities It is a simple zero if Splitting field Algebraic closure (x − u)|f (x) but (x − u)2 6 |f (x),

and more generally, a zero of multiplicity r if

(x − u)r |f (x) but (x − u)r+1 6 |f (x). Abstract Algebra, Lecture 14

Jan Snellman Theorem (Kronecker) n i Let f (x) = i=0 ai x ∈ E[x] be non-constant. Then f (x) has a zero somewhere. General ﬁeld P extensions Proof. Simple extensions Let f (x) = g(x)h(x) ∈ E(x), with g(x) irreducible. Put Zeroes of polynomials E[x] Zeroes and F = . multiplicities (g(x)) Splitting ﬁeld Algebraic closure Then E ≤ F , and x ∈ F is a zero of f (x).

This might look like some dubious sleight-of-hand, but it is completely on the up-and-up! Abstract Algebra, Lecture 14 Example Jan Snellman 2 The polynomial f (x) = x + x + 1 ∈ Z2[x] is irreducible, hence has no linear factor, hence no zero (in Z2). In [x] F = Z2 , General field (x2 + x + 1) extensions Simple extensions the elements are Zeroes of 0, 1, x, x + 1, polynomials Zeroes and with the relation multiplicities 2 Splitting field x = x + 1. Algebraic closure Now f (x) = x2 + x + 1, viewed as a polynomial with coefficients in F , has two zeroes:

f (x) = x2 + x + 1 = 0 f (x + 1) = (x + 1)2 + (x + 1) + 1 = x2 + 1 + x + 1 + 1 = 0 Abstract Algebra, Lecture 14

Jan Snellman Deﬁnition Let E ≤ F . The polynomial

General ﬁeld n i extensions f (x) = ai x ∈ E[x] i=0 Simple extensions X Zeroes of polynomials is said to split inside the extension F if there are distinct zeroes Zeroes and u1,..., ur ∈ F , and multiplicities bj ∈ Z+, such that multiplicities Splitting ﬁeld r Algebraic closure n bj f (x) = a (x − uj ) j=1 Y Abstract Algebra, Lecture 14

Jan Snellman Example 2 The√ polynomial f (x) = x + 2 ∈ Q[x] is irreducible over Q but splits over Q( 2, i), since General ﬁeld √ √ √ extensions 2 x + 2 = (x − i 2)(x + i 2) ∈ Q( 2, i)[x]. Simple extensions Zeroes of polynomials Example Zeroes and multiplicities 2 Splitting ﬁeld The polynomial f (x) = x + x + 1 ∈ Z2[x] splits over the “Kronecker Algebraic closure extension” we studied earlier, as

x2 + x + 1 = (x + x)(x + x + 1). Abstract Algebra, Lecture 14

Jan Snellman Definition The polynomial n i f (x) = ai x ∈ E[x] General field i=0 extensions X has F as its splitting field if Simple extensions 1 E ≤ F , Zeroes of polynomials 2 f (x) splits in F [x], Zeroes and multiplicities 3 Write Splitting field r n bj Algebraic closure f (x) = a (x − uj ) j=1 Y Then F = E(u1,..., ur )

So, we adjoin zeroes of f (x), but nothing unnecessary. Abstract Algebra, Lecture 14 Theorem Jan Snellman The polynomial n i f (x) = ai x ∈ E[x] i=0 General field X extensions has a splitting field F , and this splitting field is unique up to a rigid Simple extensions isomorphism:

Zeroes of φ polynomials F L Zeroes and multiplicities i j Splitting ﬁeld id Algebraic closure E E

The degree [F : E] ≤ n! and if f (x) is irreducible then [F : E] ≥ n.

Proof. Read your textbook! Abstract Algebra, Lecture 14

Jan Snellman Example (The most famous splitting field example there is!) 3 Let f (x) = x − 2 ∈ Q[x]. What is its splitting field? Put √ General field α = 3 2 extensions √ 2πi 1 2 Simple extensions β = exp( ) = − + i 3 2 3 Zeroes of polynomials Zeroes and which have minimal defining polynomial relations (over Q) multiplicities Splitting field 3 Algebraic closure α − 2 = 0 β3 − 1 = β2 + β + 1 = 0 β − 1 Abstract Algebra, Lecture 14

Jan Snellman Example (Cont)

Then, in Q(α, β), f (x) splits as

x3 − 2 = (x − α)(x − αβ)(x − αβ2) General field extensions So the splitting field is contained in Q(α, β), and of course contains the Simple extensions zeroes Zeroes of α, αβ, αβ2. polynomials Zeroes and αβ multiplicities But then it also contains β , so it is actually equal to Q(α, β). Splitting field Since x3 − 2 ∈ [x] is irreducible, [ (α): ] = 3. We have that Algebraic closure Q Q Q (x − αβ)(x − αβ2) = x2 + x + 1

is irreducible over Q(α), so [Q(α)(β): Q(α)] = 2; the tower theorem now reveals that [Q(α, β): Q] = 3 ∗ 2 = 6. Abstract Algebra, Lecture 14

Jan Snellman Example (Cont) Note: General field • α ∈ R, so Q(α) ∈ R extensions 3 • α has minimal polynomial x − 2 over Q Simple extensions • This minimal polynomial factors over (α) as Zeroes of Q polynomials 3 2 2 2 3 2 2 Zeroes and x − 2 = (x − α)(x − α(β + β )x + α β ) = (x − α)(x + αx + α ), multiplicities Splitting field Algebraic closure where the latter factor is irreducible • In particular, we have not found the splitting field after adjoining α to Q. Abstract Algebra, Lecture 14

Jan Snellman

Example General field Let f (x) = x3 + x + 1 ∈ [x]. It is irreducible. In F [y] = Z2[x] [y], we extensions Z2 (x3+x+1) have that Simple extensions Zeroes of y 3 + y + 1 = (y + x) ∗ (y + x2) ∗ (y + x2 + x) polynomials Zeroes and multiplicities So, F is the splitting field, since [F : Z2] = 3. Splitting field Algebraic closure We found the splitting field after adjoining just one zero! Abstract Algebra, Lecture 14

Jan Snellman

Example General field 4 extensions Let f (x) = x + 4 ∈ Q[x]. Then Simple extensions f (x) = x4 + 4 = (x2 + 2x + 2)(x2 − 2x + 2), Zeroes of polynomials Zeroes and By Eisenstein’s criteria, these two factors are irreducible. Further analysis multiplicities Splitting field reveals that the splitting field has degree two over Q. Algebraic closure Abstract Algebra, Lecture 14

Jan Snellman Example p Let p be a prime number, and let f (x) = x − 1 ∈ Q[x]. Then General ﬁeld extensions f (x) = xp − 1 = (x − 1)(xp−1 + xp−2 + ··· + x1), Simple extensions Zeroes of and the latter factor (call it g(x)) can be shown to be irreducible. The polynomials zeroes of g(x) are Zeroes and multiplicities 2πi Splitting ﬁeld ξk 1 ≤ k ≤ p − 1 , ξ = exp( ) Algebraic closure p

and the splitting ﬁeld is Q(ξ), which has degree p − 1 over Q. Abstract Algebra, Lecture 14

Jan Snellman

General field extensions Example Simple extensions p Let p be a prime number, and let f (x) = x − 2 ∈ Q[x]. Then the Zeroes of splitting field of f (x) is an extension of degree p(p − 1). polynomials Zeroes and multiplicities Splitting field Algebraic closure Abstract Algebra, Lecture 14

Jan Snellman

Lemma Let E ≤ F be a field extension. The set of elements of F that are General field extensions algebraic over E forms a field, which is an aglebraic extension over E. Simple extensions Example Zeroes of polynomials Let ≤ . The set of complex numbers that are algebraic over is Zeroes and Q C Q multiplicities called the field of algebraic numbers. By definition, any zero of a rational Splitting field Algebraic closure polynomial is an algebraic number. More surprisingly, every zero of a polynomial with algebraic number coefficients — is an algebraic number! Abstract Algebra, Lecture 14

Jan Snellman

Definition The field K is an algebraic closure of K if General field extensions 1 K ≤ K

Simple extensions 2 the extension is algebraic Zeroes of polynomials 3 every f (x) ∈ K[x] splits over K[x]. Zeroes and multiplicities Splitting ﬁeld Example Algebraic closure Q, the ﬁeld of algebraic numbers, is an algebraic closure of Q. Abstract Algebra, Lecture 14

Jan Snellman Definition The field E is algebraically closed if every non-constant f (x) ∈ E[x] has a zero in E. General field extensions Lemma Simple extensions If E is algebraically closed, and f (x) ∈ E[x], then f (x) splits in E. Zeroes of polynomials Proof. Zeroes and multiplicities Since f (x) has a zero u ∈ E, it has a factor (x − u) ∈ E[x]. Split it off; Splitting field Algebraic closure the remaining factor also has a zero, and so on.

Theorem (Fundamental theorem of algebra)

The complex ﬁeld C is algebraically closed. Abstract Algebra, Lecture 14 Lemma Jan Snellman Let K be an algebraic closure of K. Then K is algebraically closed (so equal to its closure).

Proof. General ﬁeld extensions 1 Take a polynomial f (x) ∈ K, and pick a zero u (somewhere). Simple extensions 2 Then K ≤ K(u) is algebraic. Zeroes of polynomials 3 Furthermore K ≤ K is algebraic. Zeroes and multiplicities 4 This means that K ≤ K(u) is algebraic. Splitting ﬁeld Algebraic closure 5 In particular, u is algebraic over K.

6 But then it belongs to K.

7 Hence, all zeroes of polynomials in K[x] remain in K.

8 So this ﬁeld is algebraically closed. Abstract Algebra, Lecture 14

Jan Snellman

Theorem Let E be a field. Then there exists a unique (up to rigid isomorphism) General field algebraic closure E, i.e. extensions Simple extensions 1 E ≤ E, and this extension is algebraic Zeroes of 2 Any polynomial with coefficients in E have a zero in E, polynomials 3 Zeroes and Any polynomial with coefficients in E have a zero in E, multiplicities Splitting field Algebraic closure Proof. Needs set theory yoga. Abstract Algebra, Lecture 14

Jan Snellman

Example

Q ≤ Q ≤ C ≤ C(x). The ﬁeld of algebraic numbers is General ﬁeld extensions 1 the algebraic closure of Q, Simple extensions 2 algebraically closed,

Zeroes of 3 the set of complex numbers that are algebraic over . polynomials Q Zeroes and The complex field algebraically closed, and its own algebraic closure. It multiplicities C Splitting field is still properly contained in the field of rational functions with complex Algebraic closure coefficients — this latter field is not algebraically closed! I belive he algebraic closure is the field of P-x series.