By I. N. Herstein

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I=1 Therefore, setting γi := sn χi tn if i < n otherwise we have n−1 1 = sn gn + tn h n = sn χi Hi gn + tn h n i=1 n−1 n (sn χi ) (Hi gn ) + tn h n = = i=1 γi h i . i=1 42 Intermezzo: Chinese Remainder Theorems Let us denote ci := γi h i (mod m), Ci := c j , ∀i, j=i and ci , C I := c I := i∈I ci , ∀I ⊂ {1, . . , n}, i∈I so that c I + C I = 1, ∀I. 2. We have cj ≡ 1(mod m i ) 0(mod m i ) if i = j otherwise Proof Since h j ≡ 0(mod m i ) if j = i, we have c j ≡ 0(mod m i ) and ci = 1 − γ j h j ≡ 1(mod m i ).

Proof Part 1 is obvious and implies deg(h p ) ≤ deg(gcd( f p , g p )). The assumption of Part 2 implies that lc(h) ≡ 0 (mod p) so that deg(h) = deg(h p ) ≤ deg(gcd( f p , g p )). 5. If lc( f ) ≡ 0 ≡ lc(g)(mod p), then there exists R ∈ Z such that p does not divide R ⇒ h p = gcd( f p , g p ). 6 will show that, given f , g ∈ Z[X ], there is R ∈ Z such that the following are equivalent R ≡ 0 (mod p); gcd( f p , g p ) = 1. Therefore we only have to apply this result to f := f / h and g := g/ h since gcd( f p , g p ) = h p gcd( f p , g p ).

4 They also knew that up to a linear transformation1 , cubic equations could be easily reduced to the form X 3 + p X + q = 0. The formula giving the solutions of this equations was discovered by Tartaglia and later divulged by Cardano. The formula is: 3 −q + 2 q 2 2 + p 3 3 + 3 −q − 2 q 2 2 + p 3 3 . 1. Let us consider the equation X 3 + 3X − 14 = 0. It is easy to verify that it has a single (real2 ) root – the function is increasing, since 1 In general, if f (X ) ∈ k[X ] is a polynomial, solving the equation f (x) = 0 and solving the equation lc( f )−1 f (X ) = 0 are the same, so we can assume that we have been given a monic polynomial to solve, let us say: n f (X ) = an−i X i = X n + a1 X n−1 + · · · + an−1 X + an .

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