POLYNOMIAL RINGS IN SEVERAL VARIABLES Abstract. These are the notes prepared for the course MTH 751 to be offered to the PhD students at IIT Kanpur. Contents 1. Rings 1 2. Quotient Rings 4 3. Hilbert Basis Theorem 7 4. Hilbert's Nullstellensatz 8 References 11 1. Rings A ring is a set with two binary structures, say, + and ×; which satisfy: (1)( R; +) is an abelian group with identity 0. (2)( R; ·) is an associative binary structure with identity 1: (3) For all a; b; c 2 R; (a + b) · c = a · c + b · c; c · (a + b) = c · a + c · b: A subset S of R is a subring if S is closed under addition, subtraction and multiplication, and contains 1: Remark 1.1 : If 1 = 0 then R = f0g : Note first that 0 · a = 0. Hence, if a 2 R then a = 1 · a = 0 · a = 0: The set of integers Z is a ring with usual addition and multiplication. Example 1.2 : Given a ring R; consider the set R[x1; ··· ; xm] of polynomi- als in the variables x1; ··· ; xm with coefficients from R: Then R[x1; ··· ; xm] is a ring with usual addition and multiplication of polynomials. The addi- tive identity is the constant polynomial 0 and the multiplicative identity is the constant polynomial 1: A complex number α is called algebraic if there exists a non-zero p 2 Z[x] such that p(α) = 0: A number is called transcendental if it is not algebraic. Any rational number is algebraic: If α = m=n for integer m and non-zero integer n then p(x) = nx − m satisfies p(α) = 0: 1 2 POLYNOMIAL RINGS IN SEVERAL VARIABLES Example 1.3 : Note that the imaginary number i is algebraic: x2 + 1 = 0 at x = i: Given a number α with a priori information that it is algebraic, it may not be easy to find a p 2 Z[x] for which p(αp) = 0p: Sometimes, the following trick is helpful. Consider the number α = 3 + −5: Then p p p p α − 3 = −5 ) α2 − 2 3α + 3 = −5 ) 2 3α = α2 + 8 ) 12α2 = (α2 + 8)2: It follows that the polynomial p(x) = x4 + 4x2 + 64 satisfies p(α) = 0: Exercise 1.4 : Show that the set of algebraic numbers is at most countable. Let R and R0 denote two rings. A ring homomorphism or homomorphism φ : R ! R0 is a map which preserves addition and multiplication, and sends 1 to 1: An isomorphism is a bijective homomorphism. Remark 1.5 : If φ is an isomorphism then the group structures (R; +) and (R0; +) are isomorphic. In particular, φ(0) = 0: Proposition 1.6. Let α be a complex number. Define φ : Z[x] ! Z[α] by φ(p) = p(α); where Z[α] is the smallest subring of C that contains α: Then φ is a surjective homomorphism. Moreover, φ an isomorphism if and only if α is a transcendental number. Proof. The first part is a routine verification. Thus it suffices to check that φ is injective if and only if α is not algebraic. If α is algebraic then there exists a non-zero p 2 Z[x] such that φ(p) = 0: Also, φ being homomorphism, maps the zero polynomial to 0: Hence φ is not injective. Conversely, if φ is not injective then there exist p 6= q 2 Z[x] such that φ(p) = φ(q); that is, φ(p − q) = 0 for the non-zero polynomial p − q 2 Z[x]: That is, α is algebraic. Next we present a substitution principle. Proposition 1.7. Let φ : R ! R0 be a ring homomorphism. Given elements 0 0 a1; ··· ; an 2 R ; there is a unique homomorphism Φ: R[x1; ··· ; xn] ! R ; which agrees with φ on constant polynomials, and which sends xi to ai for each i: Proof. The desired homomorphism is given by X α X α Φ( cαx ) = φ(cα)a for cα 2 R: jα|≤k jα|≤k Clearly, Φ is unique. Corollary 1.8. There exists a unique isomorphism between the polynomial ring R[x; y] and the ring R[x][y] of polynomials in y with coefficients from R[x]; which is identity on R; and which sends x to x and y to y: Proof. Consider the inclusion map φ : R ! R[x][y]: By the Substitution Principle 1.7, there exists a unique homomorphism Φ : R[x; y] ! R[x][y]; POLYNOMIAL RINGS IN SEVERAL VARIABLES 3 which agrees on constant polynomials, and which sends x to x and y to y: We check that Φ is the required isomorphism by displaying the inverse of Φ: Note that R[x] is a subring of R[x; y]: Thus we have an inclusion map : R[x] ! R[x; y]: Apply the Substitution Principle to to get an homo- morphism Ψ : R[x][y] ! R[x; y]; which is identity on R[x]; and which sends y to y: Finally, note that Ψ◦Φ is identity on R and fx; yg: By the uniqueness part of the Substitution Principle, Ψ ◦ Φ is the identity map. This shows that Ψ is surjective. Another application of the Substitution Priciple shows that Φ ◦ Ψ is the identity map. Let φ : R ! R0 be a ring homomorphism. The kernel ker φ of φ is given by fa 2 R : φ(a) = 0g: Remark 1.9 : Since a ring homomorphism is also a group homomorphism, ker φ is also an additive group. Note that for a 2 ker φ and r 2 R; then φ(a · r) = φ(a) · φ(r) = 0 · φ(r) = 0: Similarly, for a 2 ker φ and r 2 R; r · a 2 ker φ. Note further that ker φ is a subring if and only if 1 2 ker φ if and only if ker φ = R if and only if φ is identically zero. Example 1.10 : Consider the ring homomorphism φ : R[x] ! R defined by evaluation at a real number a: By the polynomial factor theorem, ker φ precisely consists of polynomials divisible by x − a: Example 1.11 : Consider the homomorphism φ from the ring H of holo- morphic functions on the unit disc onto the ring of complex numbers C given by φ(f) = f(0); f 2 H: The kernel of φ consists precisely of all power series convergent on the unit disc with vanishing constant term. P A non-empty subset I of a ring R is said to be the left ideal if ri ·ai 2 I for all finitely many ri 2 R and ai 2 I: Similarly, one can define the right ideal. An ideal is the one which is both left and right ideal. In a commutative ring, left and right ideals coincide. In the remaining part of these notes, all the rings are commutative. Given elements a1; ··· ; ak 2 R; the set ( k ) X < a1; ··· ; ak >:= ri · ai : r1; ··· ; rk 2 R i=1 defines a left ideal of R: We will refer to < a1; ··· ; ak > as the left ideal generated by a1; ··· ; ak: For example, if R := R[x1; ··· ; xk] and ai = xi for i = 1; ··· ; k then < a1; ··· ; ak > is the kernel of the homomorphism of the evaluation at 0: The ideal I in R is principal if there exists a 2 R such that I =< a > : Every ideal in Z is of the form nZ for some integer n: Proposition 1.12. If F is a field then every ideal in F[x] is principal. 4 POLYNOMIAL RINGS IN SEVERAL VARIABLES Proof. Let I be a non-zero ideal of F[x]: Let k = minfdeg p : p 2 Ig and let p be in I with degree k: Clearly, < p >⊆ I: Let g 2 I: By the Division Algorithm, g = pq+r; where q; r 2 F[x] and deg r < k: But then r = g−pq 2 I with degree less than k: This is possible provided r = 0: This gives the desired equality I =< p > : 2. Quotient Rings Let I be an ideal of a ring R: Since I is an additive group, so is the quotient R=I: Further, if a + I; b + I 2 R=I then R=I is a ring with multiplication: (a + I) · (b + I) is the (unique) coset a · b + I which contains it. Note that as subsets of R; (a + I) · (b + I) may be strictly contained in the coset a · b + I: The additive identity of R=I is I and the multiplicative identity is 1+I: The quotient map q : R ! R=I given by q(a) = a + I is a ring homomorphism with kernel I: Example 2.1 : Consider the ring C of convergent sequences of real numbers with termwise addition and multiplication. The mapping lim : C! R given by limfcng = limn!1 cn defines a ring homomorphism. The kernel of lim is the ideal N of null convergent sequences in C: It is easy to see that C=N is isomorphic to R: A ideal M of a ring R is said to be maximal if M ( R but M is not contained in any ideals other than M and R: Corollary 2.2. Every ideal in F[x] which is generated by an irreducible polynomial is maximal. Proof. Let p be an irreducible polynomial in F[x] and let < p > be the ideal generated by p: Suppose there exists an ideal I such that < p >( I ⊆ F[x]: By Proposition 1.12, there exists q 2 F[x] such that I =< q > : But then p = qr for some r 2 F[x]: Since p is irreducible and < p >( I; q is a non-zero constant polynomial.