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Home , Polynomial, Polynomial ring, Quotient ring, Ring homomorphism, Subring

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. Rings 4 3. Hilbert Theorem 7 4. Hilbert’s Nullstellensatz 8 References 11

1. Rings A is a with two binary structures, say, + and ×, which satisfy: (1)( R, +) is an abelian with 0. (2)( R, ·) is an associative binary structure with identity 1. (3) For all a, b, c ∈ R, (a + b) · c = a · c + b · c, c · (a + b) = c · a + c · b. A subset S of R is a if S is closed under , and , and contains 1. Remark 1.1 : If 1 = 0 then R = {0} : Note first that 0 · a = 0. Hence, if a ∈ R then a = 1 · a = 0 · a = 0.

The set of 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 . The addi- tive identity is the polynomial 0 and the multiplicative identity is the constant polynomial 1.

A complex α is called algebraic if there exists a non-zero p ∈ Z[x] such that p(α) = 0. A number is called transcendental if it is not algebraic. Any is algebraic: If α = m/n for 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 ∈ Z[x] for which p(α√) = 0√. Sometimes, the following trick is helpful. Consider the number α = 3 + −5. Then √ √ √ √ 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 is at most countable.

Let R and R0 denote two rings. A ring or homomorphism φ : R → R0 is a map which preserves addition and multiplication, and sends 1 to 1. An 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 . 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 . 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 ∈ 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 ∈ Z[x] such that φ(p) = φ(q), that is, φ(p − q) = 0 for the non-zero polynomial p − q ∈ Z[x]. That is, α is algebraic.  Next we present a substitution principle. Proposition 1.7. Let φ : R → R0 be a . Given elements 0 0 a1, ··· , an ∈ 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 ∈ R. |α|≤k |α|≤k Clearly, Φ is unique.  Corollary 1.8. There exists a unique isomorphism between the 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 ψ : R[x] → R[x, y]. Apply the Substitution Principle to ψ to get an homo- 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 {x, y}. 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 ker φ of φ is given by {a ∈ R : φ(a) = 0}.

Remark 1.9 : Since a ring homomorphism is also a , ker φ is also an additive group. Note that for a ∈ ker φ and r ∈ R, then φ(a · r) = φ(a) · φ(r) = 0 · φ(r) = 0. Similarly, for a ∈ ker φ and r ∈ R, r · a ∈ ker φ. Note further that ker φ is a subring if and only if 1 ∈ 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 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 disc onto the ring of complex numbers C given by φ(f) = f(0), f ∈ H. The kernel of φ consists precisely of all convergent on the unit disc with vanishing . P A non-empty subset I of a ring R is said to be the left if ri ·ai ∈ I for all finitely many ri ∈ R and ai ∈ I. Similarly, one can define the right ideal. An ideal is the one which is both ideal. In a , left and right ideals coincide. In the remaining part of these notes, all the rings are commutative. Given elements a1, ··· , ak ∈ R, the set

( k ) X < a1, ··· , ak >:= ri · ai : r1, ··· , rk ∈ 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 ∈ 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 = min{deg p : p ∈ I} and let p be in I with degree k. Clearly, < p >⊆ I. Let g ∈ I. By the , g = pq+r, where q, r ∈ F[x] and deg r < k. But then r = g−pq ∈ 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 ∈ R/I then R/I is a ring with multiplication: (a + I) · (b + I) is the (unique) 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 of real numbers with termwise addition and multiplication. The mapping lim : C → R given by lim{cn} = limn→∞ 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 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 ∈ F[x] such that I =< q > . But then p = qr for some r ∈ F[x]. Since p is irreducible and < p >( I, q is a non-zero constant polynomial. Hence I = F[x].  By the previous corollary, the ideal in C[z] generated by z −a is maximal. The following natural question arises: Whether every M in C[z] arises in this way ? The answer is yes. Indeed, by Proposition 1.12, M in C[z] is generated by some f ∈ C[z]. But then f has a root a in C. It follows that M ⊆ < z − a > . Since M is maximal, we must have M =< z − a > . We will later see that the last observation holds also for C[z1, ··· , zn]. This is a version of the celebrated Hilbert’s Nullstellensatz. Proposition 2.3. Let R be a commutative ring. An ideal M of a ring R is maximal if and only if R/M is a field. Proof. Suppose R/M is a field. Let M 0 be an ideal such that M ⊆ M 0. Then there is an a ∈ M 0 \ M. It follows that a + M is a non-zero element of R/M. Thus there exists b + M such that (a + M) · (b + M) = 1 + M, that is, ab − 1 ∈ M ⊆ M 0. Since ab ∈ M 0, we get 1 ∈ M 0, and hence M 0 = R. POLYNOMIAL RINGS IN SEVERAL VARIABLES 5

Conversely, suppose M is a maximal ideal of R. Let a + M be a non-zero element of R/M. Then a∈ / M. Thus the ideal I generated by M and a properly contains M. Since M is maximal, I = R. It follows that there exist r ∈ R and m ∈ M such that 1 = m + r2a. Note that r2 + M is the desired inverse of a + M.  Remark 2.4 : If I is an ideal in F[x] generated by an irreducible polynomial then F[x]/I is a field.

Q[x] 2.5 : Let a ∈ Q. Show that the is isomorphic to Q via the mapping p → p(a).

2 Example 2.6 : Consider the quotient ring Q[x]/I, where I =< x + 1 > . The mapping p → p(i) is an isomorphism between Q[x]/I and C. In partic- ular, Q[x]/I is a field. One can use this identification to find the inverse of a 4 3 given coset in Q[x]/I. For example, the inverse of (x − x + x − 5) + I is the preimage of the inverse of i4 − i3 + i − 5 = −4 + 2i under this isomorphism. Now the inverse of −4 + 2i is −2√−i . Hence the inverse of (x4 − x3 + x − 5) + I 2 5 is the coset −2√−x + I. 2 5 The first isomorphism theorem for rings says that for a surjective ring homomorphism φ : R → R0, the quotient ring R/ ker φ is isomorphic to R0. We skip its routine verification.

Example 2.7 : Consider the ring homomorphism φ : R[x, y] → R[t] given 2 by φ(x) = t , φ(y) = t and φ(a) = a for a ∈ R. We claim that the quotient 2 ring R[x, y]/I is isomorphic to R[t], where I =< x − y > . By the first isomorphism theorem, it suffices to check that ker φ = I. Clearly, x − y2 ∈ ker φ, and hence I ⊆ ker φ. To see that ker φ ⊆ I, let f ∈ ker φ. Consider the quotient ring R[x, y]/I and the coset f +I. Since R[x, y] = R[x][y] (Corollary P2k i 1.8), f(x, y) = i=0 fi(x)y for f1, ··· , f2k ∈ R[x]. It follows that 2k X i f(x, y) = fi(x)y i=0 2 2k−2 = f0(x) + (f1(x) + f3(x)y + ··· + f2k−1(x)y )y 2 4 2k + (f2(x)y + f4(x)y + ··· + f2k(x)y ). Thus f + I is same as the coset containing the polynomial k−1 2 k f0(x)+(f1(x)+f3(x)x+···+f2k−1(x)x )y+(f2(x)x+f4(x)x +···+f2k(x)x ), which is of the form p(x) + yq(x) for some polynomials p, q ∈ R[x]. Thus f(x, y) = p(x) + yq(x) + h(x, y) with h ∈ I. Now since f, h ∈ ker φ, we obtain 0 = φ(f(x, y)) = p(φ(x)) + φ(y)q(φ(x)) + φ(h(x, y)) = p(t2) + tq(t2). Since there are no common powers in the polynomials p(t2) and tq(t2), this is possible provided p = 0 = q. Thus f = h belongs to I, and the claim stands verified. 6 POLYNOMIAL RINGS IN SEVERAL VARIABLES

Example 2.8 : Consider the quotient ring Z[i]/I, where I =< 1 + 3i > . In this quotient ring, 1 + 3i = 0, that is, i = 3 or 10 = 0. Indeed, Z[i]/I is isomorphic to the quotient ring Z/10Z. To see this, define the ring ho- momorphism φ : Z → Z[i]/I by φ(n) = n + I. By the first isomorphism theorem, the of φ is isomorphic to Z/ ker φ. We check that φ is sur- jective and ker φ = 10Z. Since a + bi and a + 3b belong to the same coset in Z[i]/I, φ(a + 3b) = a + ib. Thus φ is surjective. Note further that if n ∈ 10Z then n = 10m for some integer m, and hence φ(n) = 10m + I = I. Also, if n ∈ ker φ then n ∈ I, that is, n = (a + ib)(1 + 3i) = (a − 3b) + (3a + b)i for some integers a, b, and hence 3a + b = 0 forcing n = a − 3b = 10a ∈ 10Z. Exercise 2.9 : Let I (resp. J) denote the ideal < x3 + x + 1, 5 > (resp. 3 < x + x + 1 >) in Z[x]. Verify: 3 (1) The polynomial x + x + 1 is irreducible in Z5[x]. (2) The quotient rings Z[x]/I and Z5[x]/J are isomorphic. (3) The quotient ring Z[x]/I is a field. (4) The ideal I is maximal in Z[x]. A non- R is an if it is without zero divisors: If ab = 0 for a, b ∈ R then either a = 0 or b = 0. An integral domain satisfies the cancellation law: If a·b = a·c with a 6= 0 then a · (b − c) = 0, and hence b = c.

Proposition 2.10. If R is an integral domain, the so is the polynomial ring R[x1, ··· , xn].

Proof. Suppose that f, g ∈ R[x]. Suppose the degree of f is p and the degree of g is q. If R is an integral domain then the degree of fg is p + q. In particular, R[x] is an integral domain. The desired conclusion now follows from Corollary 1.8 by a finite induction. 

For an ideal I in R, consider the quotient ring R/I. Suppose R/I is an integral domain, that is, if (a + I) · (b + I) = I then either a + I = I or b + I = I. This happens if and only if ab ∈ I implies either a ∈ I or b ∈ I. This motivates the definition of : An ideal I is said to be a prime ideal if ab ∈ I then either a ∈ I or b ∈ I.

Proposition 2.11. An ideal M of a ring R is prime if and only if R/M is an integral domain.

2 Example 2.12 : The ideal I =< x − y > in R[x, y] is a prime ideal. This follows from Example 2.7, where we observed that R[x, y]/I being isomorphic to R[t] is an integral domain. POLYNOMIAL RINGS IN SEVERAL VARIABLES 7

3. Hilbert Basis Theorem A ring R is said to be Noetherian if every ideal of R is finitely generated, that is, for any ideal I of R there exist f1, ··· , fk ∈ I such that

I = {g1f1 + ··· + gkfk : g1, ··· , gk ∈ R}. Any field F is Noetherian as the only ideals of F are {0} and F, which are generated by 0 and 1 respectively.

Lemma 3.1. Suppose R is a . Suppose a1, a2, ··· , ∈ R. Then there exists an integer m such that

am ∈ Im :=< a1, ··· , am−1 > . ∞ Proof. Since Il ⊆ Il+1 for each l, the union I := ∪l=1Il is an ideal. Since R is Noetherian, I is finitely generated with generators b1, ··· , bk belonging to k some Il. Thus I = ∪j=1Iij . If m = max{i1, ··· , ik} then am ∈ I = Im.  The following is commonly known as the Hilbert Basis Theorem. Theorem 3.2. If R is Noetherian then so is the polynomial ring R[x]. Proof. (H. Sarges, [4, 1.C.4]) Let J be an ideal in R[x]. Suppose J is not finitely generated. Then one can choose f1, f2, ··· , inductively such that fn is of smallest degree in J \ Jn, where Jn :=< f1, ··· , fn−1 > . Suppose fn is of degree dn and with the leading coefficient an ∈ R. Clearly, d1 ≤ d2 ≤ · · · . Since R is Noetherian, by Lemma 3.1, there exists an integer m such that am = a1b1 + ··· am−1bm−1 for some b1, ··· , bm−1 in R. Note Pm−1 dm−di that g := fm − i=1 bix fi ∈ J \ Jm with degree less than dm. This contradicts the choice of fm.  A finite induction argument combine with Corollary 1.8 immediately yields the following:

Corollary 3.3. If R is Noetherian then so is the polynomial ring R[x1, ··· , xn]. Firstly, the Hilbert’s Basis Theorem (HBT) provides a simple way to generate Noetherian rings. Secondly, its importance lies in its obvious con- nection with the study of common zero sets of polynomials.

Example 3.4 : Let F ⊆ F[x1, ··· , xn] be an arbitrary family of polynomials. Consider the ideal IF generated by the elements in F. By Hilbert Basis Theorem, there exist finitely many polynomials p1, ··· , pk ∈ F[x] such that IF =< p1, ··· , pk > . We claim that the common zero set

Z(IF ) = {x ∈ p(x) = 0 for every p ∈ F} is same as the common zero set Z(p1, ··· , pk) of p1, ··· , pk. Clearly, Z(F) ⊆ Z(p1, ··· , pk). Also, if x ∈ Z(p1, ··· , pk) then p1(x) = 0, ··· , pk(x) = 0, and hence p(x) = 0 for every p in IF . 8 POLYNOMIAL RINGS IN SEVERAL VARIABLES

4. Hilbert’s Nullstellensatz We start with the easier half of Hilbert’s Nullstellensatz.

n Lemma 4.1. For a = (a1, ··· , an) ∈ C , consider the ideal Ia in C[z1, ··· , zn] generated by z1 − z1, ··· , zn − an. Then the ideal Ia is maximal.

Proof. Consider the evaluation ring homomorphism ea : C[z1, ··· , zn] → C given by ea(f) = f(a). Since ea is surjective, by the first isomorphism theorem, C[z1, ··· , zn]/ ker ea is isomorphic to the field C. In particular, ker ea is a maximal ideal in C[z1, ··· , zn]. To prove that Ia is a maximal ideal in C[z1, ··· , zn], it suffices to check that Ia = ker ea. We expand f ∈ C[z1, ··· , zn] about a as follows:

n n X X f(z) = f(a) + βi(zi − ai) + γi,j(zi − ai)(zj − aj) + ··· . i=1 i≤j=1

The right hand side of the last identity consists only finitely many terms as f is a polynomial. It is now clear that f ∈ ker ea if and only if f ∈ Ia. 

The other half of Hilbert’s Nullstellensatz is quite difficult and needs more preparation. The following change of makes life easy.

Lemma 4.2. Suppose that f ∈ C[z1, ··· , zn] is of total degree d. Then d one can find scalars λ1, ··· , λn−1 ∈ C such that the coefficient of zn in f(z1 + λ1zn, ··· , zn−1 + λn−1zn, zn) is non-zero. In particular, the mapping f(z1, ··· , zn) f(z1 + λ1zn, ··· , zn−1 + λn−1zn, zn) is a ring isomorphism from C[z1, ··· , zn] onto itself.

Proof. Let fd denote the homogeneous component of f = g + fd of degree n d. Since fd 6= 0, there exists w ∈ C such that fd(w) 6= 0. By the continuity of fd, we may assume that wn 6= 0. Put λi := wi/wn, i = 1, ··· , n − 1. Note d that the coefficient of zn in f(z1 + λ1zn, ··· , zn−1 + λn−1zn, zn) is

1 fd(λ1, ··· , λn−1, 1) = fd(w1/wn, ··· , wn−1/wn, 1) = d fd(w), wn which is non-zero by our choice. Since the transformation (z1, ··· , zn) (z1+λ1zn, ··· , zn−1+λn−1zn, zn) is bijective, the remaining part follows. 

Lemma 4.3. Let I be an ideal of C[z1, ··· , zn]. Consider the polynomials d e f = f0 + f1zn + ··· fdzn and g = g0 + g1zn + ··· gezn of degree d and e respectively in C[z1, ··· , zn−1][zn]. Define R(f, g) as the of the POLYNOMIAL RINGS IN SEVERAL VARIABLES 9

(d + e) × (d + e)

 f0 f1 ··· fd 0 0 ··· 0       0 f0 ··· fd−1 fd 0 ··· 0       . .   .. ..           0 ··· 0 f0 f1 ··· fd−1 fd         g0 g1 ··· ge−1 ge 0 ··· 0       . .   .. ..         0 ··· 0 g0 g1 ··· ge−1 ge 

If f, g ∈ I then so does R(f, g). Proof. Since the determinant is unchanged by the elementary column op- erations, we take determinant after performing the following operations i C1 ↔ C1 + znCi, i = 2, ··· , d + e − 1. Note that the entries of the first col- 2 d−1 e−1 umn are f, znf, znf, ··· , zn f, g, zng, ··· , zn g. It now clear that R(f, g) belongs to the ideal generated by f and g.  Theorem 4.4. (Hilbert Nullstellensatz, Weak Form) The maximal ideals of the polynomial ring C[z1, ··· , zm] are in bijective correspondence with the n points of the complex n-dimensional C via the mapping

(a1, ··· , an) < z1 − a1, ··· , zn − an > . Proof. (E. Arrondo) In view of Lemma 4.1, it suffices to prove that every maximal ideal I of C[z1, ··· , zn] is of the form Ia :=< z1 − a1, ··· , zn − an > n for some a ∈ C . We prove this by induction on n. The case n = 1 is already discussed in the discussion following Corollary 2.2. Suppose n > 1 and assume that the conclusion holds for n − 1 variables. Let I be an ideal in C[z1, ··· , zn]. By Lemma 4.2, we may assume that I contains a g in zn: 0 0 0 0 e−1 l 0 g(z , zn) = g0(z ) + g1(z )zn + ··· + ge−1(z )zn + zn, z = (z1, ··· , zn−1), where g0, ··· , ge−1 ∈ C[z1, ··· , zn−1]. Consider the ideal 0 I := {f ∈ I : ∂f/∂zn = 0} 0 0 of the subring C[z1, ··· , zn−1]. Since 1 ∈ I if and only if 1 ∈ I ,I is a proper ideal of C[z1, ··· , zn−1]. By the induction hypothesis, there exists 0 n−1 0 a = (a1, ··· , an−1) ∈ C such that I =< z1 − a1, ··· , zn−1 − an−1 > . 10 POLYNOMIAL RINGS IN SEVERAL VARIABLES

Consider now the ideal 00 I := {f(a1, ··· , an−1, zn): f ∈ I} 00 of the ring C[zn]. We claim that I is a proper ideal of C[zn]. Suppose to the 00 contrary that 1 ∈ I . Then there exists f ∈ I such that f(a1, ··· , an−1, zn) = 0 0 0 2 0 d 1. Write f(z , zn) = f0(z ) + f1(z )zn + ··· + fd(z )zn, where f0, ··· , fd ∈ 0 0 C[z1, ··· , zn−1] are such that f1(a ) = 1 and fi(a ) = 0 for i = 2, ··· , d. By Lemma 4.3, R(f, g) ∈ I. Since R(f, g) is independent of zn,R(f, g) ∈ 0 0 I =< z1 − a1, ··· , zn−1 − an−1 > . This is impossible since R(f, g)(a ) = 1. Thus we obtain a contradiction, and hence the claim stands verified. 00 Thus I is a proper ideal of C[zn]. Hence, by the case n = 1, there exists 00 an ∈ C such that I =< zn − an > . It now follows that I is generated by z1 − a1, ··· , zn − an.  Remark 4.5 : Let I be a proper ideal of C[z1, ··· , zn]. Then the common zero set Z(I) of members of I is non-empty. Indeed, since I is contained in some maximal ideal J. By the Hilbert Nullstellensatz, J generated by n z1 − a1, ··· , zn − an for some a ∈ C . It follows that a ∈ Z(I).

Corollary 4.6. Let f1, ··· , fk ∈ C[z1, ··· , zn] be such that they have no common zero. Then there exist g1, ··· , gk ∈ C[z1, ··· , zn] such that

f1g1 + ··· + fkgk = 1.

Proof. Consider the ideal I generated by f1, ··· , fk. If I is a proper ideal then by the last theorem, f1, ··· , fk would have a common zero. In view of the hypothesis, the only possibility left is I = C[z1, ··· , zn]. Thus 1 ∈ I, and the desired conclusion follows.  2 2 Example 4.7 : Consider the polynomials f(z1, z2) = z1 +z2 −1, g(z1, z2) = 2 z1 − z2 + 1, h(z1, z2) = z1z2 − 1. We show that the ideal generated by f, g, h is C[z1, z2]. In view of the last corollary, it suffices to check that the common zero set of f, g, h is empty. To see that, let us solve the system 2 2 2 z1 + z2 = 1, z1 + 1 = z2, z1z2 = 1. 2 2 By solving first two , we obtain z1(3 + z1) = 0. This forces z1 = 0 2 or z1 = 3i. It is easy to see that (z1, z2) does not satisfy z1z2 = 1 and 2 2 z1 + z2 = 1 simultaneously. Note that the conclusion of Remark 4.5 raises the following interesting question: If I is a proper ideal of C[z1, ··· , zn] and J denotes the ideal of all polynomials vanishing on Z(I), then clearly I ⊆ J. Note also that if m f ∈ C[z1, ··· , zn] such that f ∈ I for some positive integer m then f ∈ J. 2 It is evident that, in general, I ( J (e.g. I =< z > then J =< z > in C[z]). How to obtain J from I ? Corollary 4.8. (Hilbert Nullstellensatz) Suppose that I is a proper ideal of C[z1, ··· , zn]. Let J denote the ideal of all polynomials vanishing on Z(I). m Then J = {f ∈ C[z1, ··· , zn]: f ∈ I for some positive integer m}. POLYNOMIAL RINGS IN SEVERAL VARIABLES 11

Proof. (J. Rabinowitsch, [5]) In view of the preceding discussion, it suffices to check that if f ∈ J then f m ∈ I for some positive integer m. Fix f ∈ J and introduce a new variable z0. Consider the ideal K generated by I and 1 − z0f ∈ C[z0, z1, ··· , zn]. Since Z(I) ∩ Z(1 − z0f) = ∅, by Theorem 4.4, K = C[z0, z], where z = (z1, ··· , zn). Thus there exist p ∈ I and q, r ∈ C[z0, z] such that

(4.1) pq + r(1 − z0f) = 1.

Consider the ring homomorphism e : C[z0, z] → C[z] given by φ : g(z0, z) g(−1/f, z) for g ∈ C[z0, z]. Since φ(1 − z0f) = 0 and φ(p) = p, by applying φ on both sides of (4.1), we q˜ obtain p(z)q(−1/f, z) = φ(1) = 1. It follows that p f m = 1 for someq ˜ ∈ C[z] m and positive integer m. Thus f ∈ I as desired.  References [1] E. Arrondo, Another Elementary Proof of the Nullstellensatz, Amer. Math. Monthly, February, 2006, 169-171. [2] M. Artin, , Eastern Economy Edition, 1996. [3] S. Kumaresan, How to work with quotient rings: Expository Articles (Level 2), pri- vate communication. [4] D. Patil and U. Storch, Introduction to algebraic and , IISc Lecture Notes Series, 2010. [5] K. Pommerening, Hilbert’s Nullstellensatz over the complex numbers, available on- line, 1982.