By Flegg H.G.

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Extra resources for Boolean algebra

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2) B is its only conjugate if and only if G B is a normal subgroup of Gal(E/F). Proof. 8. Returning to the situation of the Fundamental Theorem of Galois Theory, let E be a finite Galois extension of F, and let B be an intermediate field between E and F. 12. Let Q B = {σ0 : B → B an isomorphism | F ⊆ B ⊆ E, σ0 | F = id}. Let : Q B → {left cosets of G B in G} be defined as follows: Let σ0 ∈ Q B . 3. Then set (σ0 ) = [σ ] ∈ G/G B . 13. (1) is well defined. (2) is one-to-one and onto. (3) Gal(B/F) ⊆ Q B with equality if and only if G B is normal, or equivalent if and only if B/F is Galois.

And F = B. Now f (X ) is simply the polynomial f (X ) = (X − X 1 )(X − X 2 ) · · · (X − X d ), which obviously has splitting field E, as its roots are X 1 , . . , X d . 5. (1) Note in particular that we have constructed an explicit polynomial f (X ) of degree d whose Galois group is Sd . Explicit computation shows that d f (X ) = X d + (−1)i si X d−i . i=1 (2) Let us write f (X ) = X d + ad−1 X d−1 + · · · + a0 . Then we see from the above expression that the coefficients of f (X ) are, up to sign, the elementary symmetric functions of its roots, ad−i = (−1)i si , i = 1, .

That “some” but not “all” of the roots of f (X ) are √ 4 2). Then the polynomial in B. As an example of this, we may take B = Q( √ √ √ √ 4 4 4 4 X 4 − 2 is irreducible in Q[X ] with √ roots 2, √ i 2,− 2, and −i 2, and B 4 4 contains just two of these four roots ( 2 and − 2). We shall soon see that any two splitting fields of f (X ) ∈ F[X ] are isomorphic. Granting that, for the moment, we have the following result. 3. Let f (X ) ∈ F[X ] be a nonconstant polynomial, with deg f (X ) = d. Let E be a splitting field of f (X ).