Differential Galois Theory - download pdf or read online

By M. van der Put, M. Singer

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If R is a ring, then R[X], the set of all polynomials in X with coefficients in R, is also a ring under ordinary polynomial addition and multiplication, as is R[X1 , . . , Xn ], the set of polynomials in n variables Xi , 1 ≤ i ≤ n, with coefficients in R. Formally, the polynomial A(X) = a0 + a1 X + · · · + an X n is simply the sequence (a0 , . . , an ); the symbol X is a placeholder. The product of two polynomials A(X) and B(X) is a polynomial whose X k -coefficient is a0 bk + a1 bk−1 + · · · + ak b0 .

Proof. , r is a constant. Apply the evaluation homomorphism X → a to show that r = f (a). 3 CHAPTER 2. RING FUNDAMENTALS Theorem If R is an integral domain, then a nonzero polynomial f in R[X] of degree n has at most n roots in R, counting multiplicity. Proof. 2), possibly applied several times, we have f (X) = q1 (X)(X − a1 )n1 , where q1 (a1 ) = 0 and the degree of q1 is n − n1 . If a2 is another root of f , then 0 = f (a2 ) = q1 (a2 )(a2 − a1 )n1 . e. a2 is a root of q1 (X). Repeating the argument, we have q1 (X) = q2 (X)(X −a2 )n2 , where q2 (a2 ) = 0 and deg q2 = n−n1 −n2 .

2), the quotient becomes constant, and we have f (X) = c(X − a1 )n1 . . (X − ak )nk where c ∈ R and n1 + · · · + nk = n. Since R is an integral domain, the only possible roots of f are a1 , . . , ak . 4 Example Let R = Z8 , which is not an integral domain. The polynomial f (X) = X 3 has four roots in R, namely 0, 2, 4 and 6. 5 In Problems 1-4, we review the Euclidean algorithm. Let a and b be positive integers, with a > b. Divide a by b to obtain a = bq1 + r1 with 0 ≤ r1 < b, then divide b by r1 to get b = r1 q2 + r2 with 0 ≤ r2 < r1 , and continue in this fashion until the process terminates: r 1 = r 2 q 3 + r 3 , 0 ≤ r 3 < r2 , ..

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Differential Galois Theory by M. van der Put, M. Singer


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