{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# For the zero polynomial we establish the following

This preview shows pages 44–46. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page44 / 74

For the zero polynomial we establish the following...

This preview shows document pages 44 - 46. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online