By James Roeder

Shaped in California in Dec of '42 and built with P-39s. multiple 12 months later, the gang used to be thrown into strive against flying P-51 Mustangs opposed to the Luftwaffe. The heritage & strive against operations from its formation to the tip of the battle in Europe. Over one hundred forty pictures, eight pages colour profiles, sixty four pages.

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Now let na(X) = YY^^i(X — at). Since a e G permutes {a/}, a{na{X)) = na(X), In other words, every a e G fixes each coefficient of na{X). , that na(X) e ¥[X]. 7. But ma{X) is irreducible in F[Xlsoma(X)=na(X). H. Let F be a Galois extension of F and let a G E. Then (F(a)/F) is equal to the number of conjugates of a in E. 12. D The following theorem is a key result. 14. Let E be a finite extension ofF. The following are equivalent: (1)E is a Galois extension ofF. (2) E is a normal and separable extension ofF.

Xd) have degree k. , Xd-\, 0) is a synmietric polynomial in Z i , . . , Xd-\ so by induction / ( Z i , . . 12). Then / ( Z i , . . ,5^_i) = / i ( X i , . . , Xj) is a symmetric polynomial in X i , . . , X^ with / i ( X i , . . , Xj_i, 0) = 0. Hence / i ( Z i , . . , Xd) is divisible by Xd in D [ X i , . . , Xj], and hence, since this polynomial is symmetric, it is divisible by the product X\.. Xd as well. Hence / i ( X i , . . , Xd) = (Xi • • • Xd) fiiXu • . , X^) with / 2 ( X i , .

X^^^ = ( 2^ <^a(0, X 2 , . . , Xd)Xcj{y)''' x^^^ JX2 • • • Xj. Now we apply the inductive hypothesis. , Xd) over the subfield of synunetric functions in X 2 , . . , X^, so in particular they are linearly independent. , Xd) = 0. , Xd) is a polynomial that vanishes when we set Xi = 0, so it must have Zi as a factor, contradicting the definition of H[, Hence we see that H[ = 0, and so H[^ = Hi. , Xd) is divisible by Xi for every a e Hi. , Xd) € F is a symmetric polynomial in Z i , . . , Xd) is also divisible by X 2 , .