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Polynomial Rings in Several Variables 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.
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