Student: Yu Cheng (Jade) Math 612 Final Presentation Draft April - PDF document

Student: Yu Cheng (Jade) Math 612 Final Presentation Draft April 27, 2011 Problem: Show that Φ n ( x ) is irreducible over Q . Proof: First we want to show that Φ n ( x ) ∈ Z [ x ] . This is proved in class by induction. The root of unity ζ n is an algebraic integer since there exists a monic polynomial, x n − 1 , such that ζ is a root. Equivalently, the minimal polynomial m ζ n ( x ) ∈ Q [ x ] is in Z [ x ] . We claim that Φ n ( x ) = m ζ n ( x ) . By definition, m ζ n ( x ) is monic and irreducible over Q , so Φ n ( n ) is irreducible over Q . We can express m ζ n ( x ) as the following, where a 1 , · · · , a r ∈ Q are the roots. m ζ n ( x ) = ( x − a 1 ) · ( x − a 2 ) · · · · · ( x − a r ) . According to its definition Φ n ( x ) can be expressed as the following, where b 1 , · · · , b s ∈ Q are the roots and s = ϕ ( x ) . ϕ is the Euler’s totient function, the number of positive integers less than or equal to n that are co-prime to n . � ( x − ζ a Φ n ( x ) = n ) gcd ( a,n )=1 1 ≤ a<n = ( x − b 1 ) · ( x − b 2 ) · · · · · ( x − b s ) . To prove the claim, Φ n ( x ) = m ζ n ( x ) ∈ Q , we want to show that all roots of m ζ n ( x ) are also the roots for Φ n ( x ) , and vice versa. Since m ζ n ( x ) is irreducible over Q , we just need to show all roots for Φ n ( x ) are also roots for m ζ n ( x ) . Because if there are other roots in m ζ n ( x ) that are not in Φ n ( x ) , it indicates Φ n ( x ) ∈ Q is a factor of m ζ n ( x ) . This is a conflict. All roots for Φ n ( x ) are in the form ζ p n where p is a positive integer co-prime with n . � ( x − ζ a Φ n ( x ) = n ) gcd ( a,n )=1 1 ≤ a<n ( x − ζ p 1 n ) · ( x − ζ p 2 n ) · · · · · ( x − ζ p s = n ) . 1

So the problem is converted to proving an arbitrary ζ p n is a root for m ζ n ( x ) . We will prove this by contradiction. Let’s assume that ζ p n is not a root for m ζ n ( x ) . Since ζ p n is a root in x n − 1 we have the following relation. x n − 1 = m ζ n ( x ) · g ( x ) ⇒ g ( ζ p n ) = 0 . We can consider ζ n as a root for polynomial g ( x p ) . Since m ζ n ( x ) is the minimal polyno- mial of ζ n , m ζ n ( x ) has to be a factor in g ( x p ) . g ( x p ) = m ζ n ( x ) · h ( x ) . Let’s take the polynomials on both sides and mod p . g ′ ( x p ) = m ′ ζ n ( x ) · h ′ ( x ) g ′ ( x p ) , m ′ ζ n ( x ) , h ′ ( x ) ∈ Q p [ x ] . According to proposition 35 in Dummit & Foote, if a field F has a characteristic p , then for any a, b ∈ F we have the following. a p + b p ( a + b ) p = ( ab ) p . a p b p = Hence, we derive that g ′ ( x p ) = [ g ′ ( x )] p . g ′ ( x p ) c 0 + c 1 x p + c 2 ( x p ) 2 + c 3 ( x p ) 3 + · · · = x 2 � p + c 3 x 3 � p + · · · c 0 + c 1 x p + c 2 � � = c 0 + c 1 x + c 2 x 2 + c 3 x 3 · · · � p � = [ g ′ ( x )] p . = Plug this in the earlier equation. [ g ′ ( x )] p = m ′ ζ n ( x ) · h ′ ( x ) . Since Q p is a UFD, there is only one way to factorize a polynomial in Q p . Therefore, m ′ ζ n ( x ) and g ′ ( x ) have to share at least one common factor I ( x ) ∈ Q p [ x ] . Recall that we have x n − 1 = m ζ n ( x ) · g ( x ) . We can mod p on both sides of this equation as well. ζ n ( x ) · g ′ ( x ) ( x n − 1) mod p m ′ = [ I ( x )] 2 · J ( x ) = I ( x ) , J ( x ) Q p [ x ] . ∈ 2

This indicates that ( x n − 1) mod p has duplicate roots in Q p . Furthermore, x n − 1 has dupli- cate roots in Q p since ( x n − 1) mod p is a factor in x n − 1 . Now, let’s evaluate the derivative polynomial of x n − 1 . D x ( x n − 1) nx n − 1 = According to proposition 33 in Dummit & Foote, a polynomial f ( x ) has a multiple root α if and only if α is also a root of D x f ( x ) . But x n − 1 does not share any common factor with nx n − 1 for p being relatively prime to n . So, we’ve derived a contradiction. Namely, x n − 1 cannot have duplicated roots in Q p . Therefore, ζ p n has to be a root in m ζ n ( x ) rather than a root in g ( x ) , for x n − 1 = m ζ n ( x ) · g ( x ) . At this point, we’ve shown all roots in Φ n ( x ) are also roots in m ζ n ( x ) , and hence Φ n ( x ) = m ζ n ( x ) . Since m ζ n ( x ) is irreducible over Q , Φ n ( x ) is irreducible over Q . 3