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Facultatea de Matematic˘a Anul II, Curbe Algebrice Plane

POLYNOMIAL RINGS

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Basic properties

Let R be a ring. Definition 1.1 The ring of polynomials with coefficients in R, denoted R[X], is the set of formal sums f = a0 + a1 X + . . . + an X n + . . . with ai ∈ R (i ≥ 0), for which there exists n0 ∈ N such that an = 0 for all n > n0 . The degree of the polynomial f is deg f = max{n ∈ N / an 6= 0} . In this case adeg f is called the dominant coefficient. If this is 1, the polynomial is called monic. It is easy to see that R[X] is naturally endowed with a ring structure, and is an integral domain if R is. We can define inductively the ring of polynomials in several variables: R[X1 , X2 , . . . , Xn ] = R[X1 , X2 , . . . , Xn−1 ][Xn ] . Such a polynomial has a unique expression X f= a(i) X (i) where (i) = (i1 , . . . , in ), X (i) = X1i1 · . . . · Xnin are monomials and a(i) ∈ R, only finitely many being not zero. The degree of f is deg f = max{i1 + . . . + in / a(i) 6= 0} . 1

We also have deg(f g) = deg f + deg g. If, for all a(i) 6= 0, i1 + . . . + in = m, we say that the polynomial f is homogenous of degree m. Any polynomial has a unique expression f = f0 + f1 + . . . + fm with fi homogenous of degree i. In other words, if we denote R[X1 , X2 , . . . , Xn ]d = {f ∈ R[X1 , X2 , . . . , Xn ] / deg f = d} ∪ {0} , then R[X1 , X2 , . . . , Xn ] = ⊕d≥0 R[X1 , X2 , . . . , Xn ]d , R[X1 , X2 , . . . , Xn ]0 = R and R[X1 , X2 , . . . , Xn ]d is a R-module for all d > 0. That is, R[X1 , X2 , . . . , Xn ] is a graduated ring. Proposition 1.2 Let k be an infinite field. A polynomial f ∈ k[X1 , X2 , . . . , Xn ] is homogeneous of degree d if and only if f (λx1 , λx2 , . . . , λxn ) = λd f (x1 , x2 , . . . , xn )

(1.1)

for all λ ∈ k and x1 , x2 , . . . , xn ∈ k. Proof. The condition is obviously necessary. Let now f = f0 + f1 + . . . + fd with fi homogenous of degree i. Then f (λx1 , λx2 , . . . , λxn ) = f0 + λf1 (x1 , x2 , . . . , xn ) + . . . + λd fd (x1 , x2 , . . . , xn ) . Then (1.1) implies f0 +λf1 (x1 , x2 , . . . , xn )+. . .+λd−1 fd−1 (x1 , x2 , . . . , xn )+λd (f0 +f1 +. . .+fd−1 ) = 0 for all λ ∈ k and x1 , x2 , . . . , xn ∈ k. As k is an infinite field, it follows that f0 = 0, . . . , fd−1 = 0, f = fd so f is homogeneous. If R is a unique factorization domain, than R[X1 , X2 , . . . , Xn ] is also a unique factorization domain. So every non-zero polynomial can be factored uniquely, up to invertible elements of R and ordering of factors, into a product of irreducible elements. Moreover R[X1 , X2 , . . . , Xn ] is an euclidean ring.

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In particular, this is true for k[X1 , X2 , . . . , Xn ] for any field k. The quotient field of k[X1 , X2 , . . . , Xn ] is called the field of rational functions in n variables over k and is denoted k(X1 , X2 , . . . , Xn ). The set k[X1 , X2 , . . . , Xn ]+ = {f ∈ k[X1 , X2 , . . . , Xn ] / f = 0 or f0 = 0} is an ideal in k[X1 , X2 , . . . , Xn ], that will be called in the sequent the irrelevant ideal. Proposition 1.3 Let k be an algebraically closed field and f ∈ k[X, Y ] homogenous of degree d. Then f can be written as a product of d linear polynomials: d Y f= (αi X + βi Y ) i=1

αi , βi ∈ k (i ∈ {1, . . . , d}). Proof. Let n ∈ N be maximal such that Y n divides the polynomial f . Then f can be written:
f = Y n ad−n X d−n + ad−n−1 X d−n−1 Y + . . . + a0 Y d−n with ad−n 6= 0. Then f = Y d ad−n



X Y

!

d−n + . . . + a0

.

Let t = X . As the field k is algebraically closed, the polynomial ad−n td−n + Y . . . + a0 decomposes in linear terms, and the proposition follows. Example 1.4 X 3 + X 2 Y + XY 2 + Y 3 = (X + Y )(X + iY )(X − iY )  2    X 1 X X 4 2 3 3 2 4 5 6 6 4X Y + 4X Y + 5X Y + 4XY + Y = 4Y + +i −i Y 2 Y Y = Y 2 (2X + Y )2 (X + iY )(Xi Y ) The following theorem is a classic and shall be extensively used: Theorem 1.5 (Hilbert Basis Theorem) If the ring R is noetherian, then the ring R[X] is also noetherian. It follows that if R is a noetherian ring, then R[X1 , . . . , Xn ] is also noetherian. 3

2

Deriving a polynomial

The following proposition is pretty obvious: Proposition 2.1 i ∈ {1, . . . , n}. There exists a unique function ∂Xi : R[X1 , . . . , Xn ] −→ R[X1 , . . . , Xn ] verifying the properties: (i) ∂Xi (a) = 0 for  all a ∈ R; 1 if j = i (ii) ∂Xi (Xj ) = ; 0 if j 6= i (iii) ∂Xi (F + G) = ∂Xi F + ∂Xi G for all F, G ∈ R[X1 , . . . , Xn ]; (iv) ∂Xi (F G) = F ∂Xi G + G∂Xi F for all F, G ∈ R[X1 , . . . , Xn ]. The function ∂Xi is called the derivative by respect to Xi . We shall ∂F . The following properties are easy: also write ∂Xi F = FXi or ∂Xi F = ∂X i 1. (aF + bG)X = aFX + bGX , a, b ∈ R. P 2. F (G1 , . . . , Gn )X = ni=1 FXi (G1 , . . . , Gn )(Gi )X .
3. (FXi )Xj = FXj Xi for all i, j ∈ {1, . . . , n}. Proposition 2.2 (Euler) If F ∈ R[X1 , . . . , Xn ] is homogenous of degree d, then n X dF = Xi FXi . i=1

Proof. Proposition 1.2 states that F (λx1 , λx2 , . . . , λxn ) = λd F (x1 , x2 , . . . , xn ) for all λ ∈ k and x1 , x2 , . . . , xn ∈ R. By deriving this relation by respect to λ we obtain n X

xi FXi (λx1 , λx2 , . . . , λxn ) = dλd−1 F (x1 , x2 , . . . , xn ) .

i=1

But FXi are homogeneous polynomials of degree d − 1 for all i ∈ {1, . . . , n}. By applying once again the relation (1.1) the result follows. 4

3

Resultant and discriminant

Definition 3.1 Let R be a ring (commutative, with unity) and F, G ∈ R[X], F = a0 X m + . . . + am

;

G = b0 X n + . . . + bn .

We call resultant of the polynomials F and G the following determinant with mn lines and mn columns: a0 a1 . . . . . . . . . am a0 a1 . . . . . . . . . am . . . . . . . . . . . . a a . . . . . . . . . a 0 1 m ∈R RF,G = (3.2) b0 b1 . . . . . . b n b0 b1 . . . . . . bn ... ... ... ... b0 b1 . . . . . . bn where the coefficients of F are on the first n lines and the coefficients of G on the remaining m lines. Theorem 3.2 Suppose that the ring R is an unique factorisation domain and that a0 6= 0, b0 6= 0. Then the following conditions are equivalent: 1. The polynomials F and G have a common factor of degree at least 1. 2. There exist two polynomials ϕ, ψ ∈ R[X], deg ϕ < deg F , deg ψ < deg G such that ψF + ϕG = 0 . 3. RF,G = 0. Proof. (1) ⇒ (2) If F and G have the common factor D ∈ R[X], deg D ≥ 1, then F = F1 D and G = G1 D with deg F1 < deg F , deg G1 < deg G. Then we can take ϕ = F1 and ψ = −G1 . (2) ⇔ (3) It suffices to prove the equivalence in the context of K[X] where K = F rR is the field of fractions of R. The statement over R follows after the elimination of the denominators. Let V be the linear space over K whose elements are the polynomials P ∈ K[X] of degree at most m + n − 1. If we look at the following elements of V : X n−1 F, X n−2 F, . . . , XF, F, X m−1 G, X m−2 G, . . . , XG, G 5

then the lines of the determinant giving RF,G contain respectively their components by respect to the basis {X m+n−1 , . . . , X, 1} of V . So RF,G = 0 if and only if these elements of V are linearly dependent, that is, there exist a0 , a1 , . . . , an−1 , b0 , b1 , . . . , bm−1 ∈ K such that a0 X n−1 F +a1 X n−2 F +. . .+an−1 F +b0 X m−1 G+b1 X m−2 G+. . .+bm−1 G = 0 . Then we can take ϕ = b0 X m−1 + b1 X m−2 + . . . + bm−1 ψ = a0 X n−1 + a1 X n−2 + . . . + an−1

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