J. K. VERMA
1. Affine varieties of affine space
The main objects of study in algebraic geometry are the sets of solutions of a system of polynomial equations called affine varieties. Let K be a field and R = K[x1, x2, . . . , xn]
be the polynomial ring over K in n indeterminatesx1, x2, . . . , xn.Let f1, f2, . . . , ft ∈R. The
set An(K) = Kn is called the n-dimensional affine space over the field K. The zero set of f1, f2, . . . , ftis the set
Z(f1, f2, . . . , ft) ={a∈An(K)|fi(a) = 0 for i= 1,2, . . . t}.
Note that if g ∈ (f1, f2, . . . , ft) = I then any common zero of f1, f2, . . . , ft is a zero of g.
ThereforeZ(I) ={a∈An(K)|g(a) = 0 for allg∈I}=Z(f1, f2, . . . , ft).
Definition 1.1. Anaffine varietyinAn(K)is the setZ(I)for some idealI of the polynomial ringK[x1, x2, . . . , xn].An affine varietyV is calledirreducibleifV 6=V1∪V2whereV1, V2 < V are affine varieties.
Example 1.2. Let f ∈K[x1, x2, . . . , xn] be a non-constant polynomial. Then Z(f) is called
ahypersurface. If n= 3 thenZ(f) is called a surface. A hypersurface ofA2(K) is called a plane algebraic curve.
Example 1.3. The solutions of finitely many linear equations is called a linear variety. These are the objects of study in linear algebra.
Example 1.4. Theconicsare the plane curves defined by a quadratic polynomials inK[x1, x2].
Example 1.5. An affine varietyV defined by a homogeneous polynomial is called acone. If 06=a∈V then the line{ta|t∈K}is a subset of V.
Example 1.6. Let K be a field and G =GL(n, K) denote the group of all invertible n×n matrices with entries in K. For any A ∈ G, we consider the point (A,detA−1) as a point in An2+1(K).Then GL(n, K) is identified with the hypersurface H:
det(Xij)T−1 = 0
where Xij, i, j ∈ [n] and T are indeterminates. We replaceXij by the matrix entries aij and
T by detA−1.
Example 1.7. Let On be the orthogonal group of order n. The matrices in On have real
entries and they satisfy AtA = I. This condition represents n2 polynomial equations in n2 indeterminates. HenceOn is an affine variety inRn
2
.
Example 1.8. Let SL(n, K) be the group of n×n matrices with determinant 1. Let xij
for i, j = 1,2, . . . , n be n2 indeterminates. Then A(a
ij) satisfies the polynomial equation
det(xij) = 1.HenceSL(n, K) is a hypersurface inAn 2
(K). Proposition 1.9. Let K be an infinite field and n≥1.
(1)Then for any hypersurface H in An(K), An(K)\H is an infinite set.
(2) If V is an affine variety in An(K) andV 6=An(K) thenAn(K)\V is infinite.
(3) If K is algebraically closed andn≥2 thenH is infinite.
Proof. (1) Let the hypersurface H be the zero set of the nonconstantF ∈K[x1, x2, . . . , xn].If
n= 1 then F has at mostnroots inK.HenceK\Z(F) is infinite asK is infinite. Letn≥2. Ifxndoes not occur inF then we are done by induction. So assume thatxnoccurs inF.Write
F =F0+F1xn+· · ·+Ftxtn
where F0, F1, . . . , Ft ∈ K[x1, x2, . . . , xn−1] and t ≥ 1 and Ft 6= 0. By induction hypothesis,
we have a = (a1, a2, . . . , an−1) ∈ An−1(K) so that Ft(a) 6= 0. Then F(a, xn) is a nonzero
polynomial. Hence there are infinitely many b∈K such that F(a, b)6= 0. (2) Clear.
(3) LetKbe algebraically closed andn≥2.There are infinitely (a1, a2, . . . , an−1)∈An(K) for
which Ft(a1, a2, . . . , an−1) 6= 0.As K is algebraically closed, for each such (a1, a2, . . . , an−1),
there is anan∈K such that F(a1, a2, . . . , an) = 0.
Hilbert Basis Theorem and its consequences
Now we show that any affine variety is intersection of finitely many hypersurfaces. This is a consequence of the Hilbert basis theorem for Noetherian rings.
Definition 1.10. If every ideal of a ringRis finitely generated thenRis called aNoetherian ring.
Proposition 1.11. A commutative ring with identity is Noetherian if and only if given any ascending chain of ideals I1 ⊆I2 ⊆ · · · ⊆In⊆ · · ·, there exists an m such that Im =Im+i for alli≥0.In this case we say that the chain {In} is stationary.
Proof. LetRbe Noetherian. Since{In}∞n=1is an ascending chain, I =∪∞n=1Inis an ideal ofR.
Hence we can finda1, a2, . . . , ag ∈I such that I = (a1, a2, . . . , ag). It is easy to see that there
is an m such that ai ∈Im for all i= 1,2, . . . , g. Hence I ⊆Im which implies that Im =Im+i
Conversely let every ascending chain of ideals be stationary. Let I be an ideal of R which is not finitely generated. ThenI is nonzero andI < R. Inductively, we can find a1, a2, . . .∈I
such that In = (a1, a2, . . . , an) and the chain In, n = 1,2, . . . is not stationary. This is a
contradiction. HenceI is finitely generated.
Theorem 1.12(Hilbert basis theorem). LetR be a Noetherian ring. Then the polynomial ringR[x]is Noetherian.
Proof. Let I be an ideal of R[x] which is not finitely generated. Then R 6= I 6= 0. Let f1
be a nonzero polynomial of least degree among polynomials in I. Since I 6= (f1), we can
choose a polynomialf2 ∈I\(f1) which has smallest degree among all polynomials inI\(f1).
Inductively we selectf1, f2, . . . , fn∈I such thatfnhas smallest degree among polynomials in
I\(f1, f2, . . . , fn−1).Let degfi=di fori= 1,2, . . . .Then d1 ≤d2≤d3 ≤. . . and
(f1)<(f1, f2)< . . . <(f1, f2, f3, . . . , fn)< . . .
Writefn=anxdn+· · · forn= 1,2,3, . . . .We claim that the chain of ideals
(a1)<(a1, a2)<(a1, a2, a3)< . . .
does not terminate. Let (a1, a2, . . . , an) = (a1, a2, . . . , an+1) for some n.Then
an+1 =anbn+an−1bn−1+· · ·+a1b1 for someb1, b2, . . . , bn∈R.
Consider the polynomial
g(x) =fn+1(x)− n X
i=1
bifi(x)xdn+1−di.
Then g(x) ∈ I \ (f1, f2, . . . , fn) and degg(x) < dn+1. This is a contradiction. Therefore
the chain {(a1, a2, . . . , an)}∞n=1 is strictly ascending. But R is Noetherian. Hence we get a
contradiction. ThusR[x] is Noetherian.
Corollary 1.13. Let R be a Noetherian and S be a finitely generated R-algebra. Then S is a Noetherian ring.
Proof. SinceS is a finitely generatedR-algebra, there are indeterminatesx1, x2, . . . , xnand an
idealI ofA=R[x1, x2, . . . , xn] such thatS =A/I. Since A is Noetherian, so isS=A/I.
Corollary 1.14. (1) Let g1, g2, . . . , gt∈R=K[x1, x2, . . . , xn]. Then
V(g1, g2, . . . , gt) =V(g1)∩V(g2)∩. . .∩V(gt).
(2)Any affine variety in An(K) is a finite intersection of hypersurfaces. Proof. (1) It is enough to prove this fort= 2.Let f, g∈R. Then
(2) LetV =Z(I) for an idealI inR=K[x1, x2, . . . , xn].AsKis Noetherian,Ris Noetherian
by Hilbert basis theorem. Hence I = (f1, f2, . . . , fn) for some f1, f2, . . . , ft ∈ I. Therefore
V =V(f1, f2, . . . , fn) =V(f1)∩V(f2)∩ · · · ∩V(fn).
Ideal of an affine variety
Definition 1.15. Let V ⊂An(K) be an affine variety. The ideal of V, is the ideal
I(V) ={f ∈K[x1, x2, . . . , xn]|f(a) = 0 for all a∈V}.
Definition 1.16. Let I be an ideal of a commutative ring R. Then the radial of I is defined to be the ideal
√
I ={a∈I |an∈I for some nonnegative integern}.
If I =√I then I is said to be a radical ideal.
Example 1.17. Let P be a prime ideal of a ring R. Then P is a radical ideal. Indeed, if a ∈ √P then there exists n ∈ N such that an ∈ P. As P is a prime ideal, a ∈ P. Hence P is a radical ideal. It is easy to show that √I∩J = √I ∩√J . Hence finite intersection of radical ideals is again a radical ideal. In particular intersection of finitely many prime ideals is a radical ideal. If n is a positive integer then p(n) = (p1p2. . . pn) where p1, p2, . . . , pn are
the distinct prime numbers dividingn.
Corollary 1.18. Every decreasing chain of affine varieties in An is stationary.
Proof. Let V1 ⊃ V2 ⊃ . . . be a decreasing chain of affine varieties. Since R is Noetherian,
the chain of ideals I(V1) ⊂ I(V2) ⊂ . . . . is stationary. Hence there exists an m so that
I(Vn) =I(Vn+i) for alli≥1.Hence V(I(Vn)) =Vn=I(Vn+i) =Vn+i for all i≥1.
Proposition 1.19. Let R=K[x1, x2, . . . , xn].Let V, W ⊆An be affine varieties. Then
(1) I(V) is a radical ideal of R. (2) I(V ∪W) =I(V)∩I(W). (3) Z(I(V)) =V.
(4) V is irreducible if and only if I(V) is a prime ideal.
Proof. (1) Letf ∈Randfm∈I(V) for somem.Thenfm(a) = 0 for alla∈V.Hencef(a) = 0
for all a∈V. This meansf ∈I(V).Thus pI(V) =I(V).
(2) f ∈I(V ∪W) ⇔f(a) = 0 for all a∈ V ∪W ⇔ f(a) = 0 for all a ∈V and a ∈W ⇔
f ∈I(V)∩I(W).
(3) This is clear sinceV is an affine variety.
(4) SupposeV is irreducible. Letf g∈I(V) for somef, g∈R.ThenZ(f g) =Z(f)∪Z(g)⊃
Conversely letI(V) be a prime ideal. Suppose thatV =V1∪V2 where V1 and V2 are affine
varieties properly contained inV.Then I(V) =I(V1)∩I(V2).Since I(V1)I(V2)⊂I(V),either
I(V1) or I(V2) ⊂ I(V). Without loss of generality let I(V1) ⊂ I(V). Then V1 = Z(I(V1)) ⊃
V =Z(I(V)).Thus V =V1 and thereforeV is irreducible.
2. Zariski topology on An(K) and decomposition of an affine variety
We can put a topology onAn(K) =An by declaring the affine varieties inAn as the closed sets. Set R=K[x1, x2, . . . , xn].Note that V(R) =∅ and V(0) =An.If I and J are ideals of
R then V(I ∩J) = V(I)∪V(J). If Vα = V(Iα) are affine varieties then ∩αVα = V(PαIα).
Therefore affine varieties in An are closed sets. This topology is called theZariski topology on An.
Example 2.1. The closed sets in the Zariski topology on A1(K) whereK is an algebraically closed field are the empty set, the whole spaceA1(K) and the zero sets of proper ideals ofK[x]. Any nonzero proper ideal ofK[x] is a principal ideal generated by a nonzero monic polynomial of positive degreensayf(x).SinceKis algebraically closed,f(x) = (x−a1)(x−a2). . .(x−an)
for some a1, a2, . . . , an ∈ K. Hence Z(f(x)) = {a1, a2, . . . , an}. Hence any nonempty Zariski
open set is infinite. Hence any two nonempty open sets intersect. This means that any nonempty open set is dense in Zariski topology.
Theorem 2.2. Every affine variety X ⊂An(K) has a decomposition
X =X1∪X2∪. . . Xr
where X1, X2, . . . , Xr are irreducible affine varieties and there is no containment relation
among them. The irreducible affine varieties X1, . . . , Xr are uniquely determined by X.
Proof. First we show that any affine variety has a decomposition as a finite union of irreducible affine varieties. Let Σ be the set of all affine varieties which cannot be written in this way. Suppose that Σ 6= ∅. As Σ has a minimal element Y. Then Y is reducible. Let Y1, Y2 < Y
be affine varieties so that Y = Y1 ∪Y2. Then Y1, Y2 ∈/ Σ. Hence Y1, Y2 can be written as
finite unions of irreducible affine varieties. This yields such a decomposition for Y. This is a contradiction.
Now we show that such a decomposition is unique. LetX =Y1∪Y2∪· · ·∪Ytbe another such
decomposition. Then Xi =∪tm=1Xi∩Ym. Since Xi is irreducible, Xi∩Ym =Xi for somem.
HenceXi ⊂Ym.By symmetry there is ajsuch thatYm ⊂Xj.HenceXi⊂Ym ⊂Xj.This forces
i=j. ThereforeXi=Ym.This shows that r=tand {X1, X2, . . . , Xr}={Y1, Y2, . . . , Yr}.
3. Hilbert’s Nullstellensatz
LetR=C[x1, x2, . . . , xn] be the polynomial ring overCin the indeterminatesx1, x2, . . . , xn.
Hilbert’s Nullstellensatz establishes a one-to-one correspondence between points in Cn and maximal ideals inR.
Leta= (a1, a2, . . . , an)∈Cn.Then ma = (x1−a1, . . . , xn−an) is a maximal ideal of R as
seen before. Hilbert’s Nullstellensatz establishes that each maximal ideal of R is of the form
ma for somea∈Cn.
Lemma 3.1. Let V be a vector space andW be a subspace which is linear span of a countable set of vectors{v1, v2, . . .}.Then any subset of W of linearly independent vectors is either finite or countable.
Proof. Let Wn = L(v1, v2, . . . , vn), the linear span of v1, v2, . . . vn. Let S be a subset of W.
Then S = ∪∞
n=1S∩Wn. Since S∩Wn has atmostn linearly independent vectors, S is either
finite or countable.
Lemma 3.2. The set { 1
x−a |a∈C} ⊂C(x) is linearly independent over C. Proof. Let b1, b2, . . . , bn, a1, a2, . . . , an∈Cso thata1, a2, . . . , an are distinct. Let
b1
x−a1
+ b2 x−a2
+· · ·+ bn x−an
= 0. Then
n X
i=1
(x−a1)(x−a2). . .(x−ai−1)bi(x−ai+1). . .(x−an) = 0.
Putx=aj to getbj = 0 for j= 1,2, . . . , n.Hence S is linearly independent.
Theorem 3.3 (Hilbert’s Nullstellensatz). There is a one-to-one correspondence between
Cn and the set of maximal ideals in the polynomial ring C[x1, x2, . . . , xn]given by
a= (a1, a2, . . . , an)∈Cn7−→ma= (x1−a1, x2−a2, . . . , xn−an).
Proof. PutR=C[x1, x2, . . . , xn].LetMbe a maximal ideal ofR.We claim thatN =M∩C[x1]
is a maximal ideal ofC[x1].Since N = kerπ whereπ :C[x1]→R/M is the map π(f(x1)) =
f(x1) +M, N is a prime ideal ofC[x1].IfN 6= (0) then it is generated by a linear polynomial
say x −a1 ∈ C[x1]. If N = (0) then π : C(x1) → R/M is an injective map since R/M
is a field, π has a unique extension to an embedding µ : C(x1) → R/M. The monomials in
x1, x2, . . . , xnconstitute a basis ofRas aC-vector space. Hence their residues inR/Mgenerate
it as a C−vector space. Hence R/M is of countable dimension. Hence C(x1) has countable
dimension. But{x1
1−a |a∈C}is an uncountable linearly independent set in C(x1).This is a
contradiction. HenceN 6= (0).ThereforeN = (x1−a1).Similarly there are complex numbers
a2, . . . , an such that xi−ai ∈M for i = 2,3, . . . , n. Hence (x1 −a1, . . . , xn−an) ⊂ M. But
Corollary 3.4. Let f1, f2, . . . , ft ∈ R = C[x1, x2, . . . , xn]. Let Z be the set of common zeros of f1, f2, . . . , ft in Cn and let I = (f1, f2, . . . , ft). Then there is a one-to-one correspondence between Z and the set of maximal ideals of R/I. In particular the system of polynomial
equa-tions f1 = f2 = . . . = fn = 0 has a solution x1 = a1, x2 = a2, . . . , xn = an if and only if
(f1, f2, . . . , fn)⊆(x1−a1, x2−a2, . . . , xn−an).
Proof. Let f1(a) = f2(a) =· · ·=ft(a) = 0 for some a∈Cn.Using Taylor series, we see that
fi ∈ ma for i= 1,2, . . . , t. Hence I ⊂ma which gives the maximal ma/I in R/I. Conversely
any maximal ideal ofR/I has the formM/I whereM is a maximal ideal ofRand I ⊂M.By Nullstellensatz M =ma for somea.Thus fi(a) = 0 for alli= 1,2, . . . , t.
Theorem 3.5 (Strong Nullstellensatz). Let J be an ideal of R=C[x1, x2, . . . , xn].Then
(1) Z(J) =∅ if and only if J =R. (2) I(Z(J)) =√J .
(3) The maps J 7−→Z(J) and V 7−→I(V) are inverses of each other: {affine varieties ⊂Cn} ←→ {radical ideals of R}
(4) The one-to-one correspondence V 7−→I(V) maps irreducible affine varieties in Cn to prime ideals of R.
Proof. (1) Suppose thatZ(J) =∅.IfJ 6=Rthen there is a maximal idealmofRcontainingJ. By Hilbert’s Nullstellensatzm= (x1−a1, x2−a2, . . . , xn−an) fora= (a1, a2, . . . , an) ∈Cn. Hence f(a) = 0 for all f ∈J. Thusa∈Z(J) which is a contradiction. Conversely let J =R. Then 1∈J. Ifa∈Z(J) then 1(a) = 1 = 0,a contradiction.
(2) The containment J ⊂ I(Z(J)) is clear. Let 0 6= g ∈ I(Z(J)). Let t be a new vari-able. Consider the ideal I = (f1, f2, . . . , fm, tg − 1) ⊂ R[t] where J = (f1, f2, . . . , fm).
Let b = (b1, b2, . . . , bn, bn+1) ∈ Cn+1 be in Z(I). Then f1(b) = · · · = fm(b) = 0. Since
g ∈ I(Z(J)), g(b1, . . . , bn) = 0. Since b ∈ Z(I),(tg−1)(b) = −1 = 0. This contradiction
shows thatZ(I) =∅.Hence by (1),I =R[t].Hence
1 =g1f1+g2f2+· · ·+gmfm+gm+1(tg−1))
for someg1, g2, . . . , gm+1 ∈R[t].Putt= 1/g to get
1 =g1(x1, . . . , xn,1/g)f1+g2(x1, . . . , xn,1/g) +. . .+gm(x1, . . . , xn,1/g)fm
After clearing the denominators we see that there is an r such that gr ∈(f
1, f2, . . . , fm) =J.
Thusg∈√J .
(3) LetJ be a radical ideal of R.ThenI(Z(J)) =J.LetV ⊂Cn be an affine variety. Then Z(I(V)) =V. Hence the mapsJ 7−→Z(J) andV 7−→I(V) are inverses of each other.
(4) Let V ⊂ Cn be an affine variety. Then V is irreducible if and only if I(V) is a prime ideal. Hence in the correspondence in (3), prime ideals of R correspond to irreducible affine
Affine varieties of the affine plane A2(C)
Theorem 3.6. Let V be a nonempty irreducible affine variety in the complex planeC2. Then
V is either a point or a plane curve.
Proof. Let P =I(V).As V is irreducible, P is a prime ideal of C[x, y] =R. The ring R is a Noetherian UFD. Let P = (f1, f2, . . . , ft) where f1, . . . , ft are irreducible polynomials. Then
V =Z(P) = V(f)∩. . .∩V(ft). If t= 1, we are done. Let t≥2.We show Z(f, g), where f
and gare irreducible coprime polynomials inR, is a finite set. If f, g∈C[x] orC[y] then it is clear. So letf, g ∈C[x, y] be nonconstant polynomials involving both x and y. Since f, g are irreducible inC[x, y],they are so inC(x)[y].Hence 1 =rf+sg for somer, s∈C(x)[y].Clear the denominators ofrand sto get a polynomialh(x) such thath(x) =r1f+s1gwherer1 and
s1 ∈C[x, y].Let (a, b)∈Z(f, g).Then h(a) = 0. Hence ahas finitely many values. Similarly
b has finitely many values. ThereforeZ(f, g) is a finite set. Therefore V is a point since V is
irreducible.
Corollary 3.7. The nonzero prime ideals ofC[x, y]are the principal prime ideals and maximal ideals.
Proof. Let P be a nonzero prime. Then Z(P) =V is an irreducible affine variety. If V ={a}
is a point then I(V) = P =ma, the maximal ideal ma of R =C[x, y].If V is an irreducible curve then V =Z(f) where f is an irreducible polynomial in R. Hence I(V) = (f) = P by Strong Nullstellensatz.
4. Solutions of polynomial equations via eigenvalues
In this section we discuss criteria for a system of polynomial equations to have finitely many solutions. We also find an upper bound for the number of solutions. We will then discuss how eigenvalues of commuting matrices provide solutions to systems of polynomial equations which have finitely many solutions.
Theorem 4.1(Finiteness Theorem). LetIbe a proper ideal of the ringR=C[x1, x2, . . . , xn]. Then the following are equivalent:
(1) V(I) is a finite set.
(2) R/I is a finite dimensional complex vector space.
(3) I contains polynomials f1(x1), f2(x2), . . . , fn(xn).
Proof. (1) ⇒ (2): SupposeV(I) is finite. Leta1, a2, . . . , ak be theith coordinates of all the
points in V(I). Then f(xi) = (xi−a1)(xi−a2)· · ·(xi−ak) vanishes at every point ofV(I).
By Hilbert’s Nullstellensatz, there is adi so thatf(xi)di ∈I. Letd= max{kd1, kd2, . . . , kdn}.
Then [xdi] ∈ R/I can be expressed in terms of residue classes of lower powers of xi. Hence
(2)⇒ (3): Now letR/I be a finite dimensional complex vector space. Consider the residue classes
[1],[x1],[x21], . . . .
Since R/I is finite dimensional, these must be linearly dependent. Hence there are complex numbersα0, α1, α2, . . . αt,not all zero, such that
α0[1] +α1[x1] +α2[x21] +· · ·+αt[xt1] = 0.
Hencef1(x1) =α0+α1x1+α2x21+· · ·+αtxt1 ∈I.Similar argument givesf2(x2). . . , fn(xn)∈I.
(3) ⇒ (1) : Let degfi(xi) =di for i = 1,2, . . . , n. Let a = (a1, a2, . . . , an) ∈ V(I). Then
fi(ai) = 0 for all i= 1,2, . . . , n.Butfi has at mostdi roots. Hence V(I) is finite.
Number of solutions
Proposition 4.2. Let p1, p2, . . . , pm∈Cn be distinct points. Then there exist m polynomials
g1, g2, . . . , gm so thatgi(pj) = 0 for j6=i andgi(pi) = 1 for all i= 1,2, . . . , n.
Definition 4.3. The polynomials g1, g2, . . . , gm are called the interpolation polynomials for the pointsp1, p2, . . . , pm.
Theorem 4.4. Let I be a zero dimensional ideal of R=C[x1, x2, . . . , xn]. Then |V(I)|= dim
CR/
√
I
and |V(I)|= dim
CR/I if and only if I =
√
I.
Proof. Let V(I) ={p1, p2, . . . , pm} ⊆Cn.Define the linear transformation ϕ:R/I →Cm, ϕ([f]) = (f(p1), f(p2), . . . , f(pm)).
Then Kernel ϕ = {[f] |f(pi) = 0 for all i= 1,2, . . . , m} = √
I/I. The map ϕ is surjective. Indeed, letg1, g2, . . . , gm be the interpolation polynomials for the points p1, p2, . . . , pm.Then
ϕ(gi) =ei where ei is the ith standard basis vector of Cm. Hence m=|V(I)|= dimCR/I−
dim C
√
I/I = dim CR/
√
I. Thus|V(I)|= dim
CR/I if and only ifI =
√
I.
The Eigenvector Theorems
Let I be a zero-dimensional ideal in the ring S = C[x1, x2. . . , xn]. Then R = S/I is a
finite-dimensional complex vector space. We have seen that the number of points in V(I) is atmost dim
CR and equality holds if and only if I is a radical ideal.
Now we discuss the role played by eigenvalues of the multiplication maps mg :R →R. We
will show that the eigenvalues ofmg are the valuesg(p) wherep∈V(I).This observation then
leads to a theorem of Stickelberger which identifies all points inV(I) from common eigenvectors of the mapsmx1, mx2, . . . , mxn.
Example 4.5. Let f(x) = a0xn +a1xn−1 +· · · +anxn ∈ S = C[x] be a polynomial of degree n. The matrix of the map mx : S/(f(x)) → S/(f(x)) with respect to the basis {[1],[x],[x2], . . . ,[xn−1]} is given by
Mx=
0 0 ... −an/a0
1 0 ... −an−1/a0
0 1 ... −an−2/a0
..
. ... ... 0 0 ... −a1/a0
The transpose ofMxis called the companion matrix off(x).Let us show that the eigenvalues
of the linear map mx are precisely the roots of f(x). Let r ∈ C be a root of f(x). Then f(x) = (x −r)g(x) for some g(x) ∈ C[x]. Then [f(x)] = 0 = [xg(x)] −r[g(x)]. Clearly [g(x)] 6= 0 and mx([g(x)]) = r[g(x)]. Thus r is an eigenvalue of mx with eigenvector [g(x)].
Coversely, suppose that mx([g(x)]) = r[g(x)] for a nonzero vector [g(x)] and r ∈ C. Then f(x)h(x) = xg(x)−rg(x) for some h(x) ∈ C[x]. If x−r |h(x) then f(x) | g(x) which is a contradiction. Hencex−r dividesf(x).Thus each eigenvalue ofmx is a root of f(x).
Proposition 4.6. Let f ∈R andmf :A→A be the map mf([g]) = [f g].Then
(1) The mapmf :A→A is a linear transformation.
(2) mf =mg ⇐⇒f−g∈I. In particular mf = 0⇐⇒f ∈I.
(3) mf+g =mf+mg and mf g =mfmg.
(4) If h(t)∈C[t] thenh(mf) =mh(f). Proof. Exercise
Since mf is a linear map on A, it has a minimal polynomial which we denote by hf(t).
Consider the following three sets
(1) The set of roots of the minimal polynomialhf(t)
(2) The set of eigenvalues ofmf and
(3) The set{f(p)|p∈V(I)}
The amazing fact is that these three sets are equal ! This is the content of the next result. Theorem 4.7. Let I be a zero-dimensional ideal of R = C[x1, x2, . . . , xn]. Let hf(t) be the minimal polynomial of the operatormf.Then the following are equivalent for λ∈C.
(a) λis a root of hf(t).
(b)λ is an eigenvalue of mf.
(c) There exists a p∈V(I) so thatλ=f(p).
Proof. The equivalence of (a) and (b) is a fact from linear algebra.
(b) =⇒(c): Letλbe an eigenvalue ofmf and [z] be the corresponding eigenvector inA.Then
for all p∈ V(I). Let g =f −λ. Then g(pi) 6= 0 for all points pi, i = 1,2, . . . , mof V(I). Let
g1, g2, . . . , gm be the interpolation polynomials for the m points p1, p2, . . . , pm. Consider the
polynomial
g0= 1 g(p1)
g1+
1 g(p2)
g2+· · ·+
1 g(pm)
gm.
Theng0(pi)g(pi) = 1 for alli= 1,2, . . . , m.Hence there exists anr so that (g0g−1)r ∈I.Using
the binomial theorem, we get a polynomial g00 ∈ R so that 1−gg” ∈I. Hence [g][g00] = [1]. Hence [g] is invertible in A. But [g][z] = (f−λ)[z] = 0. As [g] is invertible, [z] = 0. This is a contradiction. Henceλis an eigenvalue ofmf.
(c) =⇒ (b): Let λ = f(p) for some p ∈ V(I). Let hf(t) = Pdi=0aiti. Since hf(mf) = Pd
i=0aimfi = 0, we get hf(f) ∈ I. Hence Pdi=0aifi(p) = hf(f(p)) = 0. Hence f(p) is an
eigenvalue of mf.
The Eigenvector Theorem
Theorem 4.8 (The Eigenvector Theorem). Let I be a radical ideal of the polynomial ring R = C[x1, x2, . . . , xn] with V(I) = {p1, p2, . . . , pm}. Let g1, g2, . . . gm be interpolation polynomials for V(I). For any g ∈ R, {[g1],[g2], . . . ,[gm]} is a basis of eigenvectors for mg. Moreover the eigenvalues of mg are {g(p1), g(p2), . . . , g(pm)}.
Proof. We show that [gi] is an eigenvector of mg : R/I → R/I with eigenvalue g(pi) for
i= 1,2, . . . , m. Observe that
(ggi−g(pi)gi)(pj) =g(pj)gi(pj)−g(pi)gi(pj) = 0
for allj.Sof =ggi−g(pi)gi vanishes at each point ofV(I).By Strong Nullstellensatz,f ∈I as
I is a radical ideal. Therefore [ggi−g(pi)gi] = 0.Hencemg[gi] =g(pi)[gi] for alli= 1,2, . . . , n.
Since the mapϕ:R/I →R/I given by ϕ([f]) = (f(p1), . . . , f(pm))∈Cm is an isomorphism andϕ([gi]) =ei for alli,it follows that the residue classes of the interpolation polynomials are
linearly independent. Hence they form a basis of eigenvectors ofR/I.
Theorem 4.9 (Stickelberger). Let I be a zero- dimensional radical ideal of the polynomial ringR=C[x1, x2, . . . , xn].Then
V(I) ={(a1, a2, . . . , an)|mxi([g]) =ai[g]for alli and for some [g]6= 0}.
Proof. Suppose that V(I) = {p1, p2, . . . , pm}, where p1 = (a1, a2, . . . , an). Let g1, g2, . . . , gm
be the interpolation polynomials for p1, p2, . . . , pm. Then (xig1 −aig1)(p1) = 0 and (xig1 −
aig1)(pj) = 0 for j = 1.6 Therefore f =xig1 −aig1 ∈ I for all i and hence mxi([g1]) = ai[g1] for alli.Conversely, let b= (b1, b2, . . . , bn) andmxi([g]) =bi[g] for a nonzero [g]∈R/I.Hence (xi−bi)g∈I. Since g /∈I, There is a p∈V(I) such thatg(p)6= 0.Since (xi−bi)(p)g(p) = 0
for all i,we have xi(p) =bi for all i.Hence p= (b1, b2, . . . , bn).
5. Exercises
(1) Show that if K is infinite then any nonempty open set in An(K) is dense in Zariski topology.
(2) Show that for any two nonempty open sets U, V of an irreducible topological space, U ∩V 6=∅.
(3) If K is not algebraically closed then any varietyV ⊂An(K) can be written as a zero set of a single polynomial inR=K[x1, x2, . . . , xn].
(4) Let F, G∈K[x, y] be relatively prime polynomials. Show thatZ(F, G) is finite.
(5) Let K be a finite field.
(a) For any x∈Kn,there is an f ∈K[x1, x2, . . . , xn] with f(x) = 1 and f(y) = 0 for
all y∈Kn\ {x}.
(b) For any function g : Kn −→ K there is an f ∈ K[x
1, x2, . . . , xn] such that
f(x) =g(x) for all x∈Kn.
(c) Any subset of Kn is the zero set of a suitable polynomialf ∈K[x1, x2, . . . , xn].
(6) Let V ⊂An(C) be an algebraic variety. Show that ifZn⊂V then V =An(C).
(7) Let K be an algebraically closed field. Let p, qbe positive integers. Consider the set
C={(tp, tq)|t∈K}.
Show that C is an affine variety. Find I(C).
(8) Let K be an infinite field andV ⊂An(K) is a finite set of points. Show that I(V) is generated by npolynomials.
(9) Show that a surjective endomorphism f :R → R of a Noetherian ring R is injective. (Hint: Consider Ker (fn).)
(10) Find maximal ideals ofZ[x], F[[x]] where F is a field.
(12) Consider the ideal I = (y2+x3−17) of R =C[x, y].Find generators of all maximal ideals in the quotient ringR/I.
(13) Let M be an ideal of a ring R. Suppose that all the nonzero elements of R\M are units. Show thatM is the only maximal ideal of R.
(14) LetR be a commutative ring with identity in whichan=afor somen∈N.Show that every prime ideal of R is a maximal ideal.
(15) Suppose thatf, g∈C[x, y] are quadratic polynomials such that no polynomial of pos-itive degree is a common factor of f and g. Find the number of maximal ideals in C[x, y]/(f, g).
(16) Let Cbe a subring of a ring R and let R be a finite dimensional vector space over C. Show that R has finitely many maximal ideals.
(17) Determine the points of intersection of the algebraic curvey2 =x3−x2 with the line
y=λx.
(18) Show that the locus ofy= sinx does not lie on any algebraic curve.
(19) Letp1, p2, . . . , prbe points inCn.Show that there existf1, f2, . . . , fr ∈C[x1, x2, . . . , xn]
such that fi(pj) =δij.for all i, i= 1,2, . . . , r.
(20) IsC ={(x, ex)|x∈R}an algebraic curve in A2(R)?
References
[1] David A. Cox, John Little and Donald O’Shea,Using Algebraic Geometry, Springer-Verlag, 2005.
[2] David A. Cox,Introduction to Gr¨obner bases, Applications of computational algebraic geometry (San Diego, CA, 1997), Proc. Sympos. Appl. Math., Vol. 53, 1-24.
[3] Ernst. Kunz,Introduction to Commutative Algebra and Algebraic Geometry, Birkha`‘user, Boston, 1985.
[4] Bernd Sturmfels,Solving Systems of Polynomial Equations, American Mathematical Society, 2002.
[5] Bernd Sturmfels, Polynomial equations and convex polytopes, American Mathematical Monthly 105,
907-922, 1998.
Department of Mathematics, IIT Bombay, Mumbai 400 076
