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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 10 Rik Sengupta December 18, 2009 1. Let p be a prime. Write down a normal basis for Q ( ζ p ) over Q . Now let n be any positive integer. Is the following statement true of false: { ζ a n } a ∈ ( Z /n Z ) × is always a normal basis for Q ( ζ n ) over Q ? Solution. Let α be a generator of ( Z /p Z ) × . So then, the operator σ : Q ( ζ p ) → Q ( ζ p ) defined by σ ( ζ p ) = ζ α p , σ  Q = id Q is a generator of Gal( Q ( ζ p ) / Q ) ’ ( Z /p Z ) × . So, the normal basis is given by n ζ p ,ζ α p ,ζ α 2 p ,...,ζ α p 2 p o = e,ζ p ,ζ 2 p ,...,ζ p 2 p . All the elements ζ α k p , for 0 ≤ k ≤ p 2, are distinct, because α is a generator of ( Z /p Z ) × , and so the fact that it is a basis is easily verified, because Q ( ζ ) ’ Q [ x ] / Irr( ζ p , Q ,x ) , with deg (Irr( ζ, Q ,x )) = p 1. The statement is not true for any general n ∈ N . In fact, for a counterexample, consider ζ 4 = i . Then, { ζ 4 ,ζ 3 4 } = { i, i } does not form a basis, since we have the equation i + ( i ) = 0 . 2 2. Find an element β in F 8 df = F 2 [ x ] / ( x 3 + x + 1) such that the three elements { β,β 2 ,β 4 } form a normal basis for F 8 over F 2 . Solution. We see almost immediately that the element β = x + 1 satisfies the condition. This is because 1 β 2 = x 2 + 1 β 4 = x 4 + 1 = x 2 + x + 1 . So, it is easy to see that span F 2 n x + 1 , x 2 + 1 , x 2 + x + 1 o = span F 2 n x 2 , x, x 2 + x + 1 o = span F 2 n 1 , x, x 2 o = F 8 . So, they form a basis. Furthermore, Gal( F 8 / F 2 ) = h Frob i , and so β 2 = Frob( β ) β 4 = Frob 2 ( β ) , and therefore this is a normal basis as well, as desired. 2 3. Use Hilbert’s Theorem 90 for Q ( i ) to solve the Pythagorean equation over the integers: Let x , y , z be nonzero integers satisfying x 2 + y 2 = z 2 , GCD ( x,y,z ) = 1 . Show that there are two coprime integers m,n , unique up to a sign, with x = m 2 n 2 , y = 2 mn, z = m 2 + n 2 . Solution. By Hilbert’s Theorem 90, we know, since Q ( i ) / Q is cyclic and generated by complex conjugation, for any α ∈ Q ( i ), we must have α · ¯ α = 1 ⇐⇒ a = ¯ β β for some β ∈ Q ( i ) × . Hence, 2 x 2 + y 2 = z 2 = ⇒ x + iy z x iy z = 1 = ⇒ x iy z = m ni m + ni , for some m , n ∈ Q ....
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 '09
 CLAUSSORENSEN
 Algebra, Prime number, Finite set, Hilbert, normal basis, 1 T rE/F, Frob

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