HW 10 - MAT 322: Algebra with Galois Theory Professor Claus...

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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 ( p ) over Q . Now let n be any positive integer. Is the following statement true of false: { a n } a ( Z /n Z ) is always a normal basis for Q ( n ) over Q ? Solution. Let be a generator of ( Z /p Z ) . So then, the operator : Q ( p ) Q ( p ) defined by ( p ) = p , | Q = id Q is a generator of Gal( Q ( p ) / Q ) ( Z /p Z ) . So, the normal basis is given by n p , p , 2 p ,..., p- 2 p o = e, p , 2 p ,..., p- 2 p . All the elements k p , for 0 k p- 2, are distinct, because is a generator of ( Z /p Z ) , and so the fact that it is a basis is easily verified, because Q ( ) Q [ x ] / Irr( p , Q ,x ) , with deg (Irr( , Q ,x )) = p- 1. The statement is not true for any general n N . In fact, for a counterexample, consider 4 = i . Then, { 4 , 3 4 } = { i,- i } does not form a basis, since we have the equation i + (- i ) = 0 . 2 2. Find an element in F 8 df = F 2 [ x ] / ( x 3 + x + 1) such that the three elements { , 2 , 4 } form a normal basis for F 8 over F 2 . Solution. We see almost immediately that the element = x + 1 satisfies the condition. This is because 1 2 = x 2 + 1 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 2 = Frob( ) 4 = Frob 2 ( ) , and therefore this is a normal basis as well, as desired. 2 3. Use Hilberts 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 Hilberts Theorem 90, we know, since Q ( i ) / Q is cyclic and generated by complex conjugation, for any Q ( i ), we must have = 1 a = for some 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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HW 10 - MAT 322: Algebra with Galois Theory Professor Claus...

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