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 ringis 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∈R,
(a+b)·c=a·c+b·c, c·(a+b) =c·a+c·b.
A subset S of R is a subringifS is closed under addition, subtraction and multiplication, and contains 1.
Remark 1.1 : If 1 = 0 then R ={0} : Note first that 0·a= 0. Hence, if a∈R thena= 1·a= 0·a= 0.
The set of integersZ is a ring with usual addition and multiplication.
Example 1.2 : Given a ringR,consider the setR[x1,· · · , xm] of polynomi- als in the variablesx1,· · ·, xm with coefficients fromR.ThenR[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 numberαis calledalgebraicif there exists a non-zerop∈Z[x]
such thatp(α) = 0.A number is calledtranscendental if it is not algebraic.
Any rational number is algebraic: Ifα=m/nfor integerm and non-zero integernthen p(x) =nx−m satisfiesp(α) = 0.
1
Example 1.3 :Note that the imaginary numberiis 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 numberα=√
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 numbers is at most countable.
LetR andR0 denote two rings. Aring homomorphism or homomorphism φ:R→R0 is a map which preserves addition and multiplication, and sends 1 to 1.An isomorphismis a bijective homomorphism.
Remark 1.5 :Ifφis 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 Cthat 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-zerop∈Z[x] such thatφ(p) = 0.Also,φbeing 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. Letφ:R→R0 be a ring homomorphism. Given elements a1,· · · , an ∈R0, there is a unique homomorphism Φ : R[x1,· · ·, xn] → R0, which agrees with φ on constant polynomials, and which sends xi to ai for each i.
Proof. The desired homomorphism is given by Φ(X
|α|≤k
cαxα) = X
|α|≤k
φ(cα)aα forcα∈R.
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 toy.
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],
which agrees on constant polynomials, and which sends x tox 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- morphism Ψ :R[x][y]→R[x, y],which is identity onR[x],and which sends ytoy.Finally, note that Ψ◦Φ is identity onRand{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 kernelkerφ ofφ is given by {a∈R:φ(a) = 0}.
Remark 1.9 : Since a ring homomorphism is also a group homomorphism, kerφ is also an additive group. Note that for a ∈ kerφ and r ∈ R, then φ(a·r) = φ(a)·φ(r) = 0·φ(r) = 0. Similarly, for a ∈ kerφ and r ∈ R, r·a ∈kerφ. Note further that kerφ is a subring if and only if 1 ∈kerφif and only if kerφ=R if and only if φis 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, kerφ 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 numbersCgiven byφ(f) =f(0), f ∈ H.The kernel of φconsists precisely of all power series convergent on the unit disc with vanishing constant term.
A non-empty subsetI of a ringRis said to be theleft idealifP
ri·ai ∈I for all finitely many ri ∈R and ai ∈ I. Similarly, one can define the right ideal. Anidealis 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 elementsa1,· · · , ak∈R, the set
< a1,· · ·, ak>:=
( k X
i=1
ri·ai :r1,· · ·, rk∈R )
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 inR isprincipal if there existsa∈R such that I =< a > . Every ideal inZ is of the formnZ for some integern.
Proposition 1.12. If F is a field then every ideal in F[x]is principal.
Proof. Let I be a non-zero ideal of F[x]. Let k = min{degp : p ∈ I} and let p be in I with degree k. Clearly,< p >⊆I. Let g∈ I. By the Division Algorithm,g=pq+r,whereq, r∈F[x] and degr < k.But thenr =g−pq∈ I with degree less than k. This is possible provided r = 0. This gives the
desired equality I =< p > .
2. Quotient Rings
LetIbe an ideal of a ringR.SinceIis 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) coseta·b+I which contains it. Note that as subsets of R,(a+I)·(b+I) may be strictly contained in the coseta·b+I.
The additive identity ofR/I isI 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 ringCof convergent sequences of real numbers with termwise addition and multiplication. The mapping lim :C →Rgiven by lim{cn}= limn→∞cn defines a ring homomorphism. The kernel of lim is the idealN of null convergent sequences inC.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. Letp be an irreducible polynomial inF[x] and let< p >be the ideal generated byp.Suppose there exists an idealI such that < p >(I ⊆F[x].
By Proposition 1.12, there exists q ∈ F[x] such that I =< q > . But then p=qrfor somer∈F[x].Sincepis irreducible and< p >(I, qis a non-zero
constant polynomial. Hence I =F[x].
By the previous corollary, the ideal inC[z] generated byz−ais maximal.
The following natural question arises: Whether every maximal ideal M in C[z] arises in this way ? The answer is yes. Indeed, by Proposition 1.12,M inC[z] is generated by somef ∈C[z].But thenf has a rootainC.It follows thatM ⊆< z−a > . Since M is maximal, we must haveM =< 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 idealM of a ring R is maximal if and only if R/M is a field.
Proof. Suppose R/M is a field. Let M0 be an ideal such that M ⊆ M0. Then there is ana∈M0\M.It follows thata+M is a non-zero element of R/M. Thus there existsb+M such that (a+M)·(b+M) = 1 +M, that is,ab−1∈M ⊆M0.Since ab∈M0,we get 1∈M0,and henceM0 =R.
Conversely, supposeM is a maximal ideal ofR.Leta+M be a non-zero element of R/M. Then a /∈ M. Thus the ideal I generated by M and a properly containsM.SinceM 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 : IfI is an ideal inF[x] generated by an irreducible polynomial thenF[x]/I is a field.
Exercise 2.5 : Leta∈Q.Show that the quotient ring <x−a>Q[x] is isomorphic toQvia the mappingp→p(a).
Example 2.6 :Consider the quotient ring Q[x]/I, whereI =< x2+ 1> . The mapping p→p(i) is an isomorphism betweenQ[x]/I and C.In partic- ular,Q[x]/I is a field. One can use this identification to find the inverse of a given coset inQ[x]/I.For example, the inverse of (x4−x3+x−5) +I is the preimage of the inverse ofi4−i3+i−5 =−4 + 2iunder this isomorphism.
Now the inverse of−4 + 2iis −2−i
2√
5 .Hence the inverse of (x4−x3+x−5) +I is the coset −2−x
2√ 5 +I.
The first isomorphism theorem for rings says that for a surjective ring homomorphism φ:R →R0,the quotient ring R/kerφ is isomorphic to R0. We skip its routine verification.
Example 2.7 :Consider the ring homomorphism φ:R[x, y]→ R[t] given by φ(x) =t2, φ(y) =t and φ(a) =afora∈R. We claim that the quotient ring R[x, y]/I is isomorphic to R[t], where I =< x−y2 > . By the first isomorphism theorem, it suffices to check that kerφ= I.Clearly, x−y2 ∈ kerφ,and henceI ⊆kerφ.To see that kerφ⊆I,letf ∈kerφ.Consider the quotient ringR[x, y]/I and the cosetf+I.SinceR[x, y] =R[x][y] (Corollary 1.8),f(x, y) =P2k
i=0fi(x)yi forf1,· · ·, f2k∈R[x].It follows that f(x, y) =
2k
X
i=0
fi(x)yi
= f0(x) + (f1(x) +f3(x)y2+· · ·+f2k−1(x)y2k−2)y + (f2(x)y2+f4(x)y4+· · ·+f2k(x)y2k).
Thusf +I is same as the coset containing the polynomial
f0(x)+(f1(x)+f3(x)x+· · ·+f2k−1(x)xk−1)y+(f2(x)x+f4(x)x2+· · ·+f2k(x)xk), 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 ∈ kerφ, 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 polynomialsp(t2) andtq(t2),this is possible provided p = 0 = q. Thus f = h belongs to I, and the claim stands verified.
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 image of φ is isomorphic to Z/kerφ. We check that φ is sur- jective and kerφ= 10Z.Sincea+bianda+ 3bbelong to the same coset in Z[i]/I, φ(a+ 3b) =a+ib.Thusφis surjective. Note further that ifn∈10Z then n= 10m for some integer m,and hence φ(n) = 10m+I =I. Also, if n∈kerφ thenn∈I,that is, n= (a+ib)(1 + 3i) = (a−3b) + (3a+b)ifor some integers a, b, and hence 3a+b= 0 forcingn=a−3b= 10a∈10Z.
Exercise 2.9 :Let I (resp. J) denote the ideal < x3 +x+ 1,5 > (resp.
< x3+x+ 1>) inZ[x].Verify:
(1) The polynomial x3+x+ 1 is irreducible in Z5[x].
(2) The quotient rings Z[x]/I and Z5[x]/J are isomorphic.
(3) The quotient ringZ[x]/I is a field.
(4) The idealI is maximal in Z[x].
A non-zero ring R is an integral domain 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: Ifa·b=a·cwitha6= 0 thena·(b−c) = 0,and henceb=c.
Proposition 2.10. IfR is an integral domain, the so is the polynomial ring R[x1,· · ·, xn].
Proof. Suppose thatf, g ∈R[x].Suppose the degree off ispand the degree of g is q. If R is an integral domain then the degree of f g 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 eithera∈I or b∈I.
This motivates the definition of prime ideal:
An ideal I is said to be a prime idealifab∈I then either a∈I orb∈I.
Proposition 2.11. An ideal M of a ring R is prime if and only ifR/M is an integral domain.
Example 2.12 :The ideal I =< x−y2> inR[x, y] is a prime ideal. This follows from Example 2.7, where we observed thatR[x, y]/Ibeing isomorphic toR[t] is an integral domain.
3. Hilbert Basis Theorem
A ringRis said to be Noetherianif every ideal ofRis 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 Noetherian ring. Suppose a1, a2,· · · ,∈ R.
Then there exists an integer m such that
am ∈Im :=< a1,· · · , am−1 > .
Proof. SinceIl⊆Il+1 for eachl, the union I :=∪∞l=1Il is an ideal. SinceR is Noetherian,I is finitely generated with generatorsb1,· · ·, bk belonging to someIl.ThusI =∪kj=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 that g := fm−Pm−1
i=1 bixdm−difi ∈ 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. IfRis Noetherian then so is the polynomial ringR[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 : LetF ⊆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 everyp∈ F }
is same as the common zero setZ(p1,· · ·, pk) ofp1,· · · , pk.Clearly,Z(F)⊆ Z(p1,· · · , pk).Also, ifx∈Z(p1,· · · , pk) then p1(x) = 0,· · · , pk(x) = 0,and hencep(x) = 0 for every pinIF.
4. Hilbert’s Nullstellensatz We start with the easier half of Hilbert’s Nullstellensatz.
Lemma 4.1. Fora= (a1,· · ·, an)∈Cn,consider the idealIainC[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]/kerea is isomorphic to the field C. In particular, kerea 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 = kerea.We expand f ∈ C[z1,· · · , zn] about aas follows:
f(z) =f(a) +
n
X
i=1
βi(zi−ai) +
n
X
i≤j=1
γi,j(zi−ai)(zj−aj) +· · ·.
The right hand side of the last identity consists only finitely many terms as f is a polynomial. It is now clear that f ∈kerea if and only if f ∈Ia.
The other half of Hilbert’s Nullstellensatz is quite difficult and needs more preparation. The following change of variable makes life easy.
Lemma 4.2. Suppose that f ∈ C[z1,· · ·, zn] is of total degree d. Then one can find scalars λ1,· · · , λn−1 ∈ C such that the coefficient of znd 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 d.Since fd6= 0,there exists w∈Cn such thatfd(w)6= 0.By the continuity offd,we may assume thatwn6= 0.Putλi :=wi/wn, i= 1,· · · , n−1.Note that the coefficient of znd inf(z1+λ1zn,· · · , zn−1+λn−1zn, zn) is
fd(λ1,· · ·, λn−1,1) =fd(w1/wn,· · · , wn−1/wn,1) = 1
wndfd(w), 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 f = f0 +f1zn+· · ·fdznd and g = g0 +g1zn+· · ·gezen of degree d and e respectively in C[z1,· · · , zn−1][zn]. Define R(f, g) as the determinant of the
(d+e)×(d+e) matrix
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 C1↔C1+zniCi, i= 2,· · · , d+e−1.Note that the entries of the first col- umn are f, znf, zn2f,· · · , znd−1f, g, zng,· · ·, zne−1g. It now clear that R(f, g)
belongs to the ideal generated byf andg.
Theorem 4.4. (Hilbert Nullstellensatz, Weak Form) The maximal ideals of the polynomial ring C[z1,· · ·, zm] are in bijective correspondence with the points of the complexn-dimensional space Cn 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 idealI ofC[z1,· · ·, zn] is of the formIa:=< z1−a1,· · ·, zn−an>
for some a ∈ Cn. 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 monic polynomial ginzn:
g(z0, zn) =g0(z0) +g1(z0)zn+· · ·+ge−1(z0)ze−1n +zln, z0 = (z1,· · · , zn−1), whereg0,· · ·, ge−1 ∈C[z1,· · · , zn−1].
Consider the ideal
I0 :={f ∈I :∂f /∂zn= 0}
of the subring C[z1,· · ·, zn−1]. Since 1 ∈ I if and only if 1 ∈ I0, I0 is a proper ideal of C[z1,· · · , zn−1]. By the induction hypothesis, there exists a0= (a1,· · · , an−1)∈Cn−1 such thatI0 =< z1−a1,· · ·, zn−1−an−1 > .
Consider now the ideal
I00:={f(a1,· · · , an−1, zn) :f ∈I}
of the ringC[zn].We claim thatI00is a proper ideal ofC[zn].Suppose to the contrary that 1∈I00.Then there existsf ∈I such thatf(a1,· · ·, an−1, zn) = 1. Write f(z0, zn) = f0(z0) +f1(z0)zn2 +· · ·+fd(z0)znd, where f0,· · ·, fd ∈ C[z1,· · · , zn−1] are such that f1(a0) = 1 and fi(a0) = 0 for i = 2,· · · , d.
By Lemma 4.3, R(f, g) ∈ I. Since R(f, g) is independent of zn, R(f, g) ∈ I0 =< z1 −a1,· · · , zn−1 −an−1 > . This is impossible since R(f, g)(a0) = 1. Thus we obtain a contradiction, and hence the claim stands verified.
Thus I00 is a proper ideal of C[zn]. Hence, by the case n = 1, there exists an ∈C such that I00 =< 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 z1−a1,· · ·, zn−an for somea∈Cn.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.
Example 4.7 : Consider the polynomialsf(z1, z2) =z21+z22−1, g(z1, z2) = z21−z2+ 1, h(z1, z2) =z1z2−1.We show that the ideal generated byf, g, h isC[z1, z2].In view of the last corollary, it suffices to check that the common zero set of f, g, his empty. To see that, let us solve the system
z12+z22 = 1, z12+ 1 =z2, z1z2 = 1.
By solving first two equations, we obtainz12(3 +z21) = 0.This forcesz1 = 0 or z21 = 3i. It is easy to see that (z1, z2) does not satisfy z1z2 = 1 and z21+z22= 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 f ∈C[z1,· · ·, zn] such thatfm ∈I for some positive integerm thenf ∈J.
It is evident that, in general, I ( J (e.g. I =< z2 > then J =< z > in C[z]). How to obtain J fromI ?
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).
Then J ={f ∈C[z1,· · · , zn] :fm ∈I for some positive integer m}.
Proof. (J. Rabinowitsch, [5]) In view of the preceding discussion, it suffices to check that if f ∈J then fm ∈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
pq+r(1−z0f) = 1.
(4.1)
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].
Sinceφ(1−z0f) = 0 andφ(p) =p,by applyingφon both sides of (4.1), we obtainp(z)q(−1/f, z) =φ(1) = 1.It follows thatpfq˜m = 1 for some ˜q∈C[z]
and positive integer m.Thusfm ∈I as desired.
References
[1] E. Arrondo,Another Elementary Proof of the Nullstellensatz, Amer. Math. Monthly, February, 2006, 169-171.
[2] M. Artin,Algebra, 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 geometry and commutative algebra, IISc Lecture Notes Series, 2010.
[5] K. Pommerening,Hilbert’s Nullstellensatz over the complex numbers, available on- line, 1982.