Abstract Algebra, Lecture 12 Jan Snellman Abstract Algebra, Lecture 12 Polynomial rings Polynomial rings Jan Snellman1 Coefficients in a domain 1Matematiska Institutionen Coefficients in a Link¨opingsUniversitet field Link¨oping,fall 2019 Lecture notes availabe at course homepage http://courses.mai.liu.se/GU/TATA55/ Abstract Algebra, Lecture 12 Jan Snellman Summary Division algorithm Polynomial rings 1 Polynomial rings K[x] is a PID Coefficients in a Zero-divisors, nilpotents, units GCD domain Degree Zeroes of polynomials and Coefficients in a Evaluation field linear factors 2 Coefficients in a domain Prime and maximal ideals in Divisibility K[x] Polynomial rings in several Unique factorization variables Ideal calculus in K[x] 3 Coefficients in a field Quotients Abstract Algebra, Lecture 12 Jan Snellman Summary Division algorithm Polynomial rings 1 Polynomial rings K[x] is a PID Coefficients in a Zero-divisors, nilpotents, units GCD domain Degree Zeroes of polynomials and Coefficients in a Evaluation field linear factors 2 Coefficients in a domain Prime and maximal ideals in Divisibility K[x] Polynomial rings in several Unique factorization variables Ideal calculus in K[x] 3 Coefficients in a field Quotients Abstract Algebra, Lecture 12 Jan Snellman Summary Division algorithm Polynomial rings 1 Polynomial rings K[x] is a PID Coefficients in a Zero-divisors, nilpotents, units GCD domain Degree Zeroes of polynomials and Coefficients in a Evaluation field linear factors 2 Coefficients in a domain Prime and maximal ideals in Divisibility K[x] Polynomial rings in several Unique factorization variables Ideal calculus in K[x] 3 Coefficients in a field Quotients Abstract Algebra, Lecture 12 Definition Jan Snellman Let L be a unitary, commutative ring. The polynomial ring L[x] is the set of all maps c : N L whose support Supp (c) = f n 2 c(n) 6= 0 g Polynomial rings ! N Zero-divisors, nilpotents, units is finite. Two such maps are added component-wise, and multiplied using Degree Evaluation Cauchy convolution Coefficients in a n domain (c ∗ d)(n) = c(i)d(n - i) Coefficients in a i=0 field X This makes L[x] into a commutative, unitary ring; via the injection L 3 ` 7 (N 3 n 7 ` 2 L) one can regard L as a subring.! ! Abstract Algebra, Lecture 12 Jan Snellman One usually uses the indeterminate x as a \placeholder" for the coefficients, so the map c : N L is displayed as Polynomial rings j Zero-divisors, f (!x) = c(j)x ; nilpotents, units 1 j=0 Degree X Evaluation j Coefficients in a where x can be thought of as the indicator function on fjg. domain The \Cauchy convolution" is then explained by the rule Coefficients in a field xi ∗ xj = xi+j and distributivity. Abstract Algebra, Lecture 12 Jan Snellman Example Polynomial rings Let L = Z, and let c : N L be given by c(0) = 2, c(1) = -3,c(2) = 1, Zero-divisors, and c(n) = 0 for n > 2. nilpotents, units Degree Let d : N L be given by!d(0) = -1, d(1) = 0,d(2) = 5, and d(n) = 0 Evaluation for n > 2. Coefficients in a domain The corresponding! polynomials, and their product, are Coefficients in a 2 2 2 3 4 field (2 - 3x + 1x ) ∗ (-1 + 0x + 5x ) = -2 + 3x + 9x - 15x + 5 ∗ x Abstract Algebra, Lecture 12 Jan Snellman Theorem L[x] is an integral domain iff L is. Proof. Polynomial rings Zero-divisors, If ab = 0 in L, then the same holds in L[x]. nilpotents, units Degree If Evaluation n m Coefficients in a 0 = (a0 + a1x + ··· + anx )(b0 + b1x + ··· + bmx ) domain n+m = a0b0 + (a0b1 + a1b0)x + ··· + anbmx Coefficients in a field then anbm = 0. Example Z[x] is a domain, Z6[x] is not. Abstract Algebra, Lecture 12 Jan Snellman Theorem n f = a0 + a1x ··· + anx 2 L[x] is invertible iff a0 is a unit in L and Polynomial rings a1;:::; an are nilpotent in L. Zero-divisors, nilpotents, units Degree Proof Evaluation We will make use of the fact (proved later) that Coefficients in a domain • nilpotent + nilpotent = nilpotent Coefficients in a • unit + nilpotent = unit field n If a0 unit, a1;:::; an nilpotent, then r = a1x + ··· + anx nilpotent, so f = a + r unit. Abstract Algebra, Lecture 12 Jan Snellman Proof (cont) n Conversely, if f = a0 + a1x + ··· + anx is a a unit, then there exists m g = b0 + b1x + ··· + bmx with fg = 1, thus a0b0 = 1, so a0 and b0 are Polynomial rings units. We need to prove that a1;:::; an are nilpotent. Zero-divisors, Since fg = 1, except for the constant coefficient, the coefficients of fg are nilpotents, units Degree zero, in particular Evaluation Coefficients in a anbm = 0 domain an-1bm + anbm-1 = 0 Coefficients in a field an-2bm + an-1bm-1 + anbm-2 = 0 . a0bn + a1bn-1 + ··· + anb0 = 0 Abstract Algebra, Lecture 12 Jan Snellman Proof (cont) Multiply the eqn an-1bm + anbm-1 = 0 Polynomial rings 2 Zero-divisors, by an to conclude that anbm-1 = 0. nilpotents, units Degree Multiply the eqn Evaluation Coefficients in a an-2bm + an-1bm-1 + anbm-2 = 0 domain 2 2 3 Coefficients in a by an to conclude (using anbm-1 = 0) that anbm-2 = 0. field m+1 m+1 Continue until you reach an b0 = 0. Since b0 is a unit, an = 0, so an is nilpotent. n Now f - anx is a unit, so repeat the previous procedure to conclude that an-1 is nilpotent, and so on. Abstract Algebra, Lecture 12 Jan Snellman Lemma If x; y are nilpotent, then so is x + y. Polynomial rings Zero-divisors, nilpotents, units Proof. Degree n n Evaluation Suppose that x = y = 0. Then Coefficients in a 2n domain 2n (x + y)2n = xj y 2n-j ; Coefficients in a j j=0 field X and either j or 2n - j are ≥ n. Abstract Algebra, Lecture 12 Lemma Jan Snellman If x unit, r nilpotent, then x - y (and x + y) is a unit. Proof. Polynomial rings Assume xx-1 = 1, y n = 0. Then Zero-divisors, nilpotents, units n-1 n Degree (1 - y)(1 + y + ··· + y ) = 1 - y = 1; Evaluation Coefficients in a so x-1(x - y) = x-1(1 - yx-1) has inverse domain Coefficients in a x-1(1 + (yx-1) + ··· + (yx-1)n-1); field hence (x - y) has inverse 1 + (yx-1) + ··· + (yx-1)n-1: Abstract Algebra, Lecture 12 Jan Snellman Definition n If f = a0 + a1x + ··· + anx 2 L[x], with an 6= 0, then the degree of f is deg f = max(Supp (f )) = n: Polynomial rings Zero-divisors, nilpotents, units The degree of the zero polynomial is - . Degree We define the Evaluation • leading term as a xn, 1 Coefficients in a n domain • leading monomial as xn, Coefficients in a • field leading coefficient as an Example 4 5 5 If Z[x] = 1 - 3x + 17x then the degree is 5, the l.t. is 17x , the l.c. is 17, and the l.m. is x5. Abstract Algebra, Lecture 12 Theorem Jan Snellman Let f ; g 2 L[x]. If L is a domain then • deg(f + g) ≤ • deg(fg) = deg(f ) + deg(g, max(deg(f ); deg(g)), Polynomial rings • • lt lt lt Zero-divisors, deg(fg) ≤ deg(f ) + deg(g. (fg) = (f ) (g) nilpotents, units Degree Evaluation Example Coefficients in a domain In Z4[x], Coefficients in a field (2x2 + x + 1)2 = 4x4 + x2 + 1 + 4x3 + 4x2 + 2x = x2 + x + 1; so the degree of products can drop in the presence of zero-divisors. In Q[x], (2x2 + x + 1) + (-2x2 - x + 1) = 2; so the degree of sums may drop even with field coefficients. Abstract Algebra, Lecture 12 Jan Snellman Definition n Let f (x) = a0 + a1x + ··· + anx 2 L[x]. Let u 2 L. We define the evaluation of f at u by Polynomial rings n Zero-divisors, f (u) = a0 + a1u + ··· + anu 2 L nilpotents, units Degree Evaluation Coefficients in a Theorem domain For u 2 L, the map Coefficients in a field evu : L[x] L f (x) 7 f (u) ! is a ring homomorphism. ! Abstract Algebra, Lecture 12 Example Jan Snellman A fixed polynomial f (x) 2 L[x] defines a polynomial function f : L L u 7 f (u) Polynomial rings Zero-divisors, ! nilpotents, units That is an importan topic that we will only touch upon in this course. We Degree ! Evaluation note the following oddity: for polynomials in Q[x], different polynomials Coefficients in a give rise to different functions Q Q. However, already for polynomials domain 2 with field coefficients this need not hold. In Z2[x], if f (x) = (x + x)g(x) Coefficients in a for any g(x) 2 [x], then field Z2 ! 2 f ([0]2) = ([0]2 + [0]2)(g([0]2) = [0]2 2 f ([1]2) = ([1]2 + [1]2)(g([1]2) = [0]2 so infinitely many polynomials represent the constantly zero polynomial function. Abstract Algebra, Lecture 12 Jan Snellman Corollary Polynomial rings The set Zero-divisors, nilpotents, units Iu = f f (x) 2 L[x] f (u) = 0 g Degree Evaluation is an ideal in L[x], and Coefficients in a L[x] domain ' L Iu Coefficients in a field The ideals containing Iu are in bijection with the ideals of L. Abstract Algebra, Lecture 12 Jan Snellman Recall: Polynomial rings Theorem Coefficients in a domain n Suppos that L is a domain, and that f = a0 + a1x + ··· + anx 2 L[x]. Divisibility Polynomial rings in 1 L[x] is a domain several variables Coefficients in a 2 f is invertible iff a0 is a unit and deg(f ) = 0.