9.2. Splitting Fields
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“bookmt” — 2006/8/8 — 12:58 — page 422 — #434
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9. FIELD EXTENSIONS – SECOND LOOK 9.2. Splitting Fields
Now, we turn our point of view around regarding algebraic extensions.
Given a polynomial f .x/ 2 KŒx, we produce extension fields K L in which f has a root, or in which f has a complete set of roots.
Proposition 7.3.6 tells us that if f .x/ is irreducible in KŒx and ˛ is a root of f .x/ in some extension field, then the field generated by K and the root ˛ is isomorphic to KŒx=.f .x//; but KŒx=.f .x// is a field, so we might as well choose our extension field to be KŒx=.f .x//! What should ˛ be then? It will have to be the image in KŒx=.f .x// of x, namely, Œx D x C .f .x//. (We are using Œg.x/ to denote the image of g.x/ in KŒx=.f .x//.) Indeed, Œx is a root of f .x/ in KŒx=.f .x//, since f .Œx/ D Œf .x/ D 0 in KŒx=.f .x//.
Proposition 9.2.1. Let K be a field and let f .x/ be a monic irreducible element of KŒx. Then (a) There is an extension field L and an element ˛ 2 L such that f .˛/ D 0.
(b) For any such extension field and any such ˛, f .x/ is the minimal polynomial for ˛.
(c) If L and L0 are extension fields of K containing elements ˛ and ˛0 satisfying f .˛/ D 0 and f .˛0/ D 0, then there is an isomorphism W K.˛/ ! K.˛0/ such that .k/ D k for all k 2 K and .˛/ D ˛0.
Proof. The existence of the field extension containing a root of f .x/ was already shown. The minimal polynomial for ˛ divides f and, therefore, equals f , since f is monic irreducible. For the third part, we have K.˛/ Š KŒx=.f .x// Š K.˛0/, by isomorphisms that leave K pointwise fixed.
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We will need a technical variation of part (c) of the proposition. Recall from Corollary 6.2.8 that if M and M 0 are fields and W M ! M 0 is a field isomorphism, then extends to an isomorphism of rings M Œx !
M 0Œx by .P mi xi / D P .mi /xi .
Corollary 9.2.2. Let K and K0 be fields, let W K ! K0 be a field isomorphism, and let f .x/ be an irreducible element of KŒx. Suppose that ˛ is a root of f .x/ in an extension L K and that ˛0 is a root of .f .x// in an extension field L0 K0. Then there is an isomorphism W K.˛/ ! K0.˛0/ such that .k/ D .k/ for all k 2 K and .˛/ D ˛0.
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“bookmt” — 2006/8/8 — 12:58 — page 423 — #435
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9.2. SPLITTING FIELDS
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Proof. The ring isomorphism W KŒx ! K0Œx induces a field isomorphism Q W KŒx=.f / ! K0Œx=..f //, satisfying Q.k C .f // D .k/ C . .f // for k 2 K. Now, use K.˛/ Š KŒx=.f / Š K0Œx=..f // Š K0.˛0/.
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Now, consider a monic polynomial f .x/ 2 KŒx of degree d > 1, not necessarily irreducible. Factor f .x/ into irreducible factors. If any of these factors have degree greater than 1, then choose such a factor and adjoin a root ˛1 of this factor to K, as previously. Now, regard f .x/ as a polynomial over the field K.˛1/, and write it as a product of irreducible factors in K.˛1/Œx. If any of these factors has degree greater than 1, then choose such a factor and adjoin a root ˛2 of this factor to K.˛1/. After repeating this procedure at most d times, we obtain a field in which f .x/ factors into linear factors. Of course, a proper proof goes by induction; see Exercise 9.2.1.
Definition 9.2.3. A splitting field for a polynomial f .x/ 2 KŒx is an extension field L such that f .x/ factors into linear factors over L, and L is generated by K and the roots of f .x/ in L.
If a polynomial p.x/ 2 KŒx factors into linear factors over a field M K, and if f˛1; : : : ˛r g are the distinct roots of p.x/ in M , then K.˛1; : : : ; ˛r /, the subfield of M generated by K and f˛1; : : : ˛r g, is a splitting field for p.x/. For polynomials over Q, for example, it is unnecessary to refer to Exercise 9.2.1 for the existence of splitting fields; a rational polynomial splits into linear factors over C, so a splitting field is obtained by adjoining the complex roots of the polynomial to Q.
One consequence of the existence of splitting fields is the existence of many finite fields. See Exercise 9.2.3.
The following result says that a splitting field is unique up to isomor phism.
Proposition 9.2.4. Let K L and Q K Q L be field extensions, let W K ! Q
K be a field isomorphism, and let p.x/ 2 KŒx and Q p.x/ 2 Q
KŒx be polynomials with Q p.x/ D .p.x//. Suppose L is a splitting field for p.x/ and Q L is a splitting field for Q p.x/. Then there is a field isomorphism W L ! Q L such that .k/ D .k/ for all k 2 K.
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“bookmt” — 2006/8/8 — 12:58 — page 424 — #436
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9. FIELD EXTENSIONS – SECOND LOOK
L p p p p p p p p p p p p p p p p pqqqqqqq QL
qqq
q
q q
q q q
q q
q qqq qqq
K; p.x/
qqqq Q q K ; Q p.x/
Proof. The idea is to use Proposition 9.2.2 and induction on dimK.L/.
If dimK.L/ D 1, there is nothing to do: The polynomial p.x/ factors into linear factors over K, so also Q p.x/ factors into linear factors over Q K, K D L, Q K D Q L, and is the required isomorphism.
We make the following induction assumption: Suppose K M L and Q K Q
M Q L are intermediate field extensions, Q W M ! Q
M is a field isomorphism extending , and dimM L