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For the zero polynomial we establish the following

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Unformatted text preview: For the zero polynomial, we establish the following conventions: its leading coefficient and constant term are defined to be 0 R , and its degree is defined to be “-∞ ”, where it is understood that for all integers x ∈ Z ,-∞ < x , and (-∞ )+ x = x +(-∞ ) =-∞ , and (-∞ )+(-∞ ) =-∞ . This notion of “negative infinity” should not be construed as a useful algebraic notion — it is simply a convenience of notation; for example, it allows us to succinctly state that for all a,b ∈ R [ T ], deg( ab ) ≤ deg( a ) + deg( b ), with equality holding if the leading coefficients of a and b are not both zero divisors. Theorem 5.5 Let D be an integral domain. Then 1. for all a,b ∈ D [ T ] , deg( ab ) = deg( a ) + deg( b ) ; 2. D [ T ] is an integral domain; 3. ( D [ T ]) * = D * . Proof. Exercise. 2 5.2.3 Division with remainder An extremely important property of polynomials is a division with remainder property, analogous to that for the integers: Theorem 5.6 (Division with Remainder Property) Let R be a ring. For a,b ∈ R [ T ] with lc( b ) ∈ R * , there exist unique q,r ∈ R [ T ] such that a = bq + r and deg( r ) < deg( b ) . Proof. Consider the set S of polynomials of the form a- xb with x ∈ R [ T ]. Let r = a- qb be an element of S of minimum degree. We must have deg( r ) < deg( b ), since otherwise, we would have r := r- (lc( r )lc( b )- 1 T deg( r )- deg( b ) ) · b ∈ S , and deg( r ) < deg( r ), contradicting the minimality of deg( r ). That proves the existence of r and q . For uniqueness, suppose that a = bq + r and a = bq + r , where deg( r ) < deg( b ) and deg( r ) < deg( b ). This implies r- r = b ( q- q ) . However, if q 6 = q , then deg( b ) > deg( r- r ) = deg( b ( q- q )) = deg( b ) + deg( q- q ) ≥ deg( b ) , which is impossible. Therefore, we must have q = q , and hence r = r . 2 If a = bq + r as in the above theorem, we define a rem b := r . Theorem 5.7 If K is field, then for a,b ∈ K [ T ] with b 6 = 0 K , there exist unique q,r ∈ K [ T ] such that a = bq + r and deg( r ) < deg( b ) . Proof. Clear. 2 39 Theorem 5.8 For a ring R and a ∈ R [ T ] and x ∈ R , a ( x ) = 0 R if and only if ( T- x ) divides a . Proof. Let us write a = ( T- x ) q + r , with q,r ∈ R [ T ] and deg( r ) < 1, which means that r ∈ R . Then we have a ( x ) = ( x- x ) q ( x ) + r = r . Thus, a ( x ) = 0 if and only if T- x divides a . 2 With R,a,x as in the above theorem, we say that x is a root of a if a ( x ) = 0 R . Theorem 5.9 Let D be an integral domain, and let a ∈ D [ T ] , with deg( a ) = k ≥ . Then a has at most k roots. Proof. We can prove this by induction. If k = 0, this means that a is a non-zero element of D , and so it clearly has no roots....
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For the zero polynomial we establish the following...

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