Consequently, θp is an integral domain. Furthermore, θp has only one prime element p,

and every x ∈ θp is a power of p, up to multiplication by a unit.

Proof. The series representation for x has a nonzero term an pn of lowest degree n, where

an can be taken between 1 and p ’ 1. Factor out pn to obtain x = pn u, where u is a unit

by (7.9.2). ™

7.9.4 De¬nitions and Comments

The quotient ¬eld Qp of θp is called the ¬eld of p-adic numbers. By (7.9.3), each ± ∈ Qp

has the form pm u, where m is an integer (possibly negative) and u is a unit in θp . Thus

± has a “Laurent expansion”

a’r a’1

+ ··· + + a0 + a1 p + · · · .

r

p p

Another representation is ± = x/pr , where x is a p-adic integer and r ≥ 0. This version

is convenient for doing addition and multiplication in Qp .

The rationals Q are a sub¬eld of Qp . To see this, let a/b be a rational number in

lowest terms (a and b relatively prime). If p does not divide b, then by (7.9.2), b is a unit

of θp . Since a ∈ Z ⊆ θp , we have a/b ∈ θp . If b = pt b where p does not divide b , we

can factor out pt and reduce to the previous case. Thus a/b always belongs to Qp , and

a/b ∈ θp i¬ p does not divide b. Rational numbers belonging to θp are sometimes called

p-integers.

We now outline a procedure for constructing the p-adic numbers formally.

7.9.5 De¬nitions and Comments

The p-adic valuation on Qp is de¬ned by

vp (pm u) = m.

In general, a valuation v on a ¬eld F is a real-valued function on F \ {0} satisfying:

(a) v(xy) = v(x) + v(y);

(b) v(x + y) ≥ min(v(x), v(y)).

By convention, we take v(0) = + ∞.

The representation x = pm u shows that vp is indeed a valuation on Qp . If c is any

real number greater than 1, then the valuation v induces an absolute value on F , namely,

|x| = c’v(x) .

When v = vp , the constant c is usually taken to be p, and we obtain the p-adic absolute

value

|x|p = p’vp (x) .

32 CHAPTER 7. INTRODUCING ALGEBRAIC NUMBER THEORY

Thus the p-adic absolute value of pn is p’n , which approaches 0 exponentially as n ap-

proaches in¬nity.

In general, an absolute value on a ¬eld F is a real-valued function on F such that:

(i) |x| ≥ 0, with equality if and only if x = 0;

(ii) |xy| = |x||y|;

(iii) |x + y| ¤ |x| + |y|.

By (b), an absolute value induced by a valuation satis¬es a property that is stronger

than (iii):

(iv) |x + y| ¤ max(|x|, |y|).

An absolute value satisfying (iv) is said to be nonarchimedian.

7.9.6 Proposition

Let F be the quotient ¬eld of an integral domain R. The absolute value | | on F is

nonarchimedian if and only if |n| ¤ 1 for every integer n = 1 ± · · · ± 1 ∈ R.

Proof. Assume a nonarchimedian absolute value. By property (ii) of (7.9.5), | ± 1| = 1. If

|n| ¤ 1, then by property (iv), |n±1| ¤ 1, and the desired conclusion follows by induction.

Conversely, assume that the absolute value of every integer is at most 1. To prove (iv), it

su¬ces to show that |x + 1| ¤ max(|x|, 1) for every x ∈ F . [If y = 0 in (iv), divide by |y|.]

By the binomial theorem,

n n

nr n

|x + 1| = | x |¤ | ||x|r .

n

r r

r=0 r=0

n

has absolute value at most 1. If |x| > 1, then |x|r ¤ |x|n

By hypothesis, the integer

r

for all r = 0, 1, . . . , n. If |x| ¤ 1, then |x|r ¤ 1. Consequently,

|x + 1|n ¤ (n + 1) max(|x|n , 1).

Take nth roots and let n ’ ∞ to get |x + 1| ¤ max(|x|, 1). ™

The next result may seem innocuous, but it leads to a remarkable property of nonar-

chimedian absolute values.

7.9.7 Proposition

If | | is a nonarchimedian absolute value, then

|x| = |y| implies |x + y| = max(|x|, |y|).

7.9. P-ADIC NUMBERS 33

Proof. First note that | ’ y| = |(’1)y| = | ’ 1||y| = |y|. We can assume without loss of

generality that |x| > |y|. Using property (iv) of (7.9.5), we have

|x| = |x + y ’ y| ¤ max(|x + y|, |y|) = |x + y|.

[Otherwise, max(|x + y|, |y|) = |y|, hence |x| ¤ |y| < |x|, a contradiction.] Since |x + y| ¤

max(|x|, |y|) = |x|, the result follows. ™

Any absolute value determines a metric via d(x, y) = |x ’ y|. This distance function

can be used to measure the length of the sides of a triangle.

7.9.8 Corollary

With respect to the metric induced by a nonarchimedian absolute value, all triangles are

isosceles.

Proof. Let the vertices of the triangle be x, y and z. Then

|x ’ y| = |(x ’ z) + (z ’ y)|.

If |x ’ z| = |z ’ y|, then two side lengths are equal. If |x ’ z| = ||z ’ y|, then by (7.9.7),

|x ’ y| = max(|x ’ z|, |z ’ y|), and again two side lengths are equal. ™

We now look at the p-adic numbers from the viewpoint of valuation theory.

7.9.9 De¬nitions and Comments

Let | | be a nonarchimedian absolute value on the ¬eld F . The valuation ring of | | is

θ = {x ∈ F : |x| ¤ 1}.

In the p-adic case, θ = {x ∈ Qp : vp (x) ≥ 0} = θp . By properties (ii) and (iv) of (7.9.5), θ

is a subring of F .

The valuation ideal of | | is

β = {x ∈ F : |x| < 1}.

In the p-adic case, β = {x ∈ Qp : vp (x) ≥ 1} = pθp , those p-adic integers whose series

representation has no constant term. To verify that β is an ideal of θ, note that if x, y ∈ β

and r ∈ θ, then |rx| = |r||x| ¤ |x| < 1 and |x + y| ¤ max(|x|, |y|) < 1.

Now if x ∈ θ \ β, then |x| = 1, hence |x’1 | = 1/|x| = 1, so x’1 ∈ θ and x is a

unit of θ. On the other hand, if x ∈ β, then x cannot be a unit of θ. [If xy = 1, then

1 = |x||y| ¤ |x| < 1, a contradiction.] Thus the ideal β is the set of all nonunits of θ. No

proper ideal I of θ can contain a unit, so I ⊆ β. It follows that β is the unique maximal

ideal of θ. A ring with a unique maximal ideal is called a local ring. We will meet such

rings again when we examine the localization process in Section 8.5.

To construct the p-adic numbers, we start with the p-adic valuation on the integers,

and extend it to the rationals in the natural way: vp (a/b) = vp (a) ’ vp (b). The p-adic

valuation then determines the p-adic absolute value, which induces a metric d on Q.

34 CHAPTER 7. INTRODUCING ALGEBRAIC NUMBER THEORY

[Because d comes from a nonarchimedian absolute value, it will satisfy the ultrametric

inequality d(x, y) ¤ max(d(x, z), d(z, y)), which is stronger than the triangle inequality.]

The process of constructing the real numbers by completing the rationals using equivalence

classes of Cauchy sequences is familiar. The same process can be carried out using the

p-adic absolute value rather than the usual absolute value on Q. The result is a complete

metric space, the ¬eld of p-adic numbers, in which Q is dense.

Ostrowski™s theorem says that the usual absolute value | |∞ on Q, along with the p-adic

absolute values | |p for all primes p, and the trivial absolute value (|0| = 0; |x| = 1 for

x = 0), essentially exhaust all possibilities. To be more precise, two absolute values on a

¬eld F are equivalent if the corresponding metrics on F induce the same topology. Any

nontrivial absolute value on Q is equivalent to | |∞ or to one of the | |p .

Problems For Section 7.9

1. Take p = 3, and compute the standard representation of (2 + p + p2 )(2 + p2 ) in two

ways, using (1) and (2) of (7.9.1). Check the result by computing the product using

ordinary multiplication of two integers, and then expanding in base p = 3.

2. Express the p-adic integer -1 as an in¬nite series of the form (1), using the standard

representation.

3. Show that every absolute value on a ¬nite ¬eld is trivial.

4. Show that an absolute value is archimedian i¬ the set S = {|n| : n ∈ Z} is unbounded.

5. Show that a ¬eld that has an archimedian absolute value must have characteristic 0.

zn of p-adic numbers converges if and only if zn ’ 0 as

6. Show that an in¬nite series

n ’ ∞.

7. Show that the sequence an = n! of p-adic integers converges to 0.

8. Does the sequence an = n converge in Qp ?

Chapter 8

Introducing Algebraic

Geometry

(Commutative Algebra 2)

We will develop enough geometry to allow an appreciation of the Hilbert Nullstellensatz,

and look at some techniques of commutative algebra that have geometric signi¬cance. As

in Chapter 7, unless otherwise speci¬ed, all rings will be assumed commutative.

8.1 Varieties

8.1.1 De¬nitions and Comments

We will be working in k[X1 , . . . , Xn ], the ring of polynomials in n variables over the ¬eld

k. (Any application of the Nullstellensatz requires that k be algebraically closed, but we

will not make this assumption until it becomes necessary.) The set An = An (k) of all

n-tuples with components in k is called a¬ne n-space. If S is a set of polynomials in

k[X1 , . . . , Xn ], then the zero-set of S, that is, the set V = V (S) of all x ∈ An such that

f (x) = 0 for every f ∈ S, is called a variety. (The term “a¬ne variety” is more precise,

but we will use the short form because we will not be discussing projective varieties.)

Thus a variety is the solution set of simultaneous polynomial equations.

If I is the ideal generated by S, then I consists of all ¬nite linear combinations gi fi

with gi ∈ k[X1 , . . . , Xn ] and fi ∈ S. It follows that V (S) = V (I), so every variety is the

variety of some ideal. We now prove that we can make An into a topological space by

taking varieties as the closed sets.

8.1.2 Proposition

(1) If V± = V (I± ) for all ± ∈ T , then V± = V ( I± ). Thus an arbitrary intersection of

varieties is a variety.

1

2 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

r

(2) If Vj = V (Ij ), j = 1, . . . , r, then j=1 Vj = V ({f1 · · · fr : fj ∈ Ij , 1 ¤ j ¤ r}). Thus a

¬nite union of varieties is a variety.

(3) An = V (0) and … = V (1), so the entire space and the empty set are varieties.

Consequently, there is a topology on An , called the Zariski topology, such that the

closed sets and the varieties coincide.

Proof. (1) If x ∈ An , then x ∈ V± i¬ every polynomial in every I± vanishes at x i¬

x ∈ V ( I± ).

r

(2) x ∈ j=1 Vj i¬ for some j, every fj ∈ Ij vanishes at x i¬ x ∈ V ({f1 · · · fr : fj ∈ Ij

for all j}).

(3) The zero polynomial vanishes everywhere and a nonzero constant polynomial van-

ishes nowhere. ™

Note that condition (2) can also be expressed as

«

r

=V Ij = V ©r Ij .

∪r Vj

j=1 j=1

j=1

[See (7.6.1) for the de¬nition of a product of ideals.]

We have seen that every subset of k[X1 , . . . , Xn ], in particular every ideal, determines

a variety. We can reverse this process as follows.

8.1.3 De¬nitions and Comments

If X is an arbitrary subset of An , we de¬ne the ideal of X as I(X) = {f ∈ k[X1 , . . . , Xn ] :

f vanishes on X}. By de¬nition we have:

(4) If X ⊆ Y then I(X) ⊇ I(Y ); if S ⊆ T then V (S) ⊇ V (T ).

Now if S is any set of polynomials, de¬ne IV (S) as I(V (S)), the ideal of the zero-set

of S; we are simply omitting parentheses for convenience. Similarly, if X is any subset

of An , we can de¬ne V I(X), IV I(X), V IV (S), and so on. From the de¬nitions we

have:

(5) IV (S) ⊇ S; V I(X) ⊇ X.

[If f ∈ S then f vanishes on V (S), hence f ∈ IV (S). If x ∈ X then every polynomial in

I(X) vanishes at x, so x belongs to the zero-set of I(X).]

If we keep applying V ™s and I™s alternately, the sequence stabilizes very quickly:

(6) V IV (S) = V (S); IV I(X) = I(X).

[In each case, apply (4) and (5) to show that the left side is a subset of the right side. If

x ∈ V (S) and f ∈ IV (S) then f (x) = 0, so x ∈ V IV (S). If f ∈ I(X) and x ∈ V I(X)

then x belongs to the zero-set of I(X), so f (x) = 0. Thus f vanishes on V I(X), so

f ∈ IV I(X).]

Since every polynomial vanishes on the empty set (vacuously), we have:

8.1. VARIETIES 3

(7) I(…) = k[X1 , . . . , Xn ].

The next two properties require a bit more e¬ort.

(8) If k is an in¬nite ¬eld, then I(An ) = {0};

(9) If x = (a1 , . . . , an ) ∈ An , then I({x}) = (X1 ’ a1 , . . . , Xn ’ an ).

Property (8) holds for n = 1 since a nonconstant polynomial in one variable has only

¬nitely many zeros. Thus f = 0 implies that f ∈ I(A1 ). If n > 1, let f = ar X1 + · · · +

r

/

a1 X1 + a0 where the ai are polynomials in X2 , . . . , Xn and ar = 0. By the induction

hypothesis, there is a point (x2 , . . . , xn ) at which ar does not vanish. Fixing this point,

we can regard f as a polynomial in X1 , which cannot possibly vanish at all x1 ∈ k. Thus

f ∈ I(An ).

/

To prove (9), note that the right side is contained in the left side because Xi ’ ai is 0

when Xi = ai . Also, the result holds for n = 1 by the remainder theorem (2.5.2). Thus

assume n > 1 and let f = br X1 + · · · + b1 X1 + b0 ∈ I({x}), where the bi are polynomials

r

in X2 , . . . , Xn and br = 0. By the division algorithm (2.5.1), we have

f = (X1 ’ a1 )g(X1 , . . . , Xn ) + h(X2 , . . . , Xn )

and h must vanish at (a2 , . . . , an ). By the induction hypothesis, h ∈ (X2 ’a2 , . . . , Xn ’an ),

hence f ∈ (X1 ’ a1 , X2 ’ a2 , . . . , Xn ’ an ).

Problems For Section 8.1

A variety is said to be reducible if it can be expressed as the union of two proper subva-

rieties; otherwise the variety is irreducible. In Problems 1“4, we are going to show that a

variety V is irreducible if and only if I(V ) is a prime ideal.

1. Assume that I(V ) is not prime, and let f1 f2 ∈ I(V ) with f1 , f2 ∈ I(V ). If x ∈ V ,

/

show that x ∈ V (f1 ) implies x ∈ V (f2 ) (and similarly, x ∈ V (f2 ) implies x ∈ V (f1 )).

/ /

2. Show that V is reducible.

3. Show that if V and W are varieties with V ‚ W , then I(V ) ⊃ I(W ).

4. Now assume that V = V1 V2 , with V1 , V2 ‚ V . By Problem 3, we can choose

fi ∈ I(Vi ) with fi ∈ I(V ). Show that f1 f2 ∈ I(V ), so I(V ) is not a prime ideal.

/

5. Show that any variety is the union of ¬nitely many irreducible subvarieties.

6. Show that the decomposition of Problem 5 is unique, assuming that we discard any

subvariety that is contained in another one.

7. Assume that k is algebraically closed. Suppose that An is covered by open sets An \

V (Ii ) in the Zariski topology. Let I is the ideal generated by the Ii , so that I = Ii ,

the set of all ¬nite sums xi1 + · · · xir with xij ∈ Iij . Show that 1 ∈ I. (You may appeal

to the weak Nullstellensatz, to be proved in Section 8.4.)

8. Show that An is compact in the Zariski topology.

4 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.2 The Hilbert Basis Theorem

If S is a set of polynomials in k[X1 , . . . , Xn ], we have de¬ned the variety V (S) as the

zero-set of S, and we know that V (S) = V (I), where I is the ideal generated by S. Thus

any set of simultaneous polynomial equations de¬nes a variety. In general, in¬nitely many

equations may be involved, but as Hilbert proved, an in¬nite collection of equations can

always be replaced by a ¬nite collection. The reason is that every ideal of k[X1 , . . . , Xn ]

has a ¬nite set of generators, in other words, k[X1 , . . . , Xn ] is a Noetherian ring. The

¬eld k is, in particular, a PID, so k is Noetherian. The key step is to show that if R is a

Noetherian ring, then the polynomial ring in n variables over R is also Noetherian.

8.2.1 Hilbert Basis Theorem

If R is a Noetherian ring, then R[X1 , . . . , Xn ] is also Noetherian.

Proof. By induction, we can assume n = 1. Let I be an ideal of R[X], and let J be the ideal

of all leading coe¬cients of polynomials in I. (The leading coe¬cient of 5X 2 ’ 3X + 17

is 5; the leading coe¬cient of the zero polynomial is 0.) By hypothesis, J is ¬nitely

generated, say by a1 , . . . , an . Let fi be a polynomial in I whose leading coe¬cient is ai ,

and let di be the degree of fi . Let I — consist of all polynomials in I of degree at most

d = max{di : 1 ¤ i ¤ n}. Then I — is an R-submodule of the free R-module M of all

polynomials b0 + b1 X + · · · + bd X d , bi ∈ R. Now a ¬nitely generated free R-module is a

¬nite direct sum of copies of R, hence M , and therefore I — , is Noetherian. Thus I — can

be generated by ¬nitely many polynomials g1 , . . . , gm . Take I0 to be the ideal of R[X]

generated by f1 , . . . , fn , g1 , . . . , gm . We will show that I0 = I, proving that I is ¬nitely

generated.

First observe that fi ∈ I and gj ∈ I — ⊆ I, so I0 ⊆ I. Thus we must show that each

h ∈ I belongs to I0 .

Case 1 : deg h ¤ d

Then h ∈ I — , so h is a linear combination of the gj (with coe¬cients in R ⊆ R[X]), so

h ∈ I0 .

Case 2 : deg h = r > d

n

Let a be the leading coe¬cient of h. Since a ∈ J, we have a = i=1 ci ai with the

ci ∈ R. Let

n

q =h’ ci X r’di fi ∈ I.

i=1

The coe¬cient of X r in q is

n

a’ ci ai = 0

i=1

so that deg q < r. We can iterate this degree-reduction process until the resulting poly-

nomial has degree d or less, and therefore belongs to I0 . But then h is a ¬nite linear

combination of the fi and gj . ™

8.2. THE HILBERT BASIS THEOREM 5

8.2.2 Corollary

Every variety is the intersection of ¬nitely many hypersurfaces (zero-sets of single poly-

nomials).

Proof. Let V = V (I) be a variety. By (8.2.1), I has ¬nitely many generators f1 , . . . , fr .

r

But then V = i=1 V (fi ). ™

8.2.3 Formal Power Series

The argument used to prove the Hilbert basis theorem can be adapted to show that if R is

Noetherian, then the ring R[[X]] of formal power series is Noetherian. We cannot simply

reproduce the proof because an in¬nite series has no term of highest degree, but we can

look at the lowest degree term. If f = ar X r + ar+1 X r+1 + · · · , where r is a nonnegative

integer and ar = 0, let us say that f has degree r and leading coe¬cient ar . (If f = 0,

take the degree to be in¬nite and the leading coe¬cient to be 0.)

If I is an ideal of R[[X]], we must show that I is ¬nitely generated. We will inductively

construct a sequence of elements fi ∈ R[[X]] as follows. Let f1 have minimal degree among

elements of I. Suppose that we have chosen f1 , . . . , fi , where fi has degree di and leading

coe¬cient ai . We then select fi+1 satisfying the following three requirements:

1. fi+1 belongs to I;

2. ai+1 does not belong to (a1 , . . . , ai ), the ideal of R generated by the aj , j = 1, . . . , i;

3. Among all elements satisfying the ¬rst two conditions, fi+1 has minimal degree.

The second condition forces the procedure to terminate in a ¬nite number of steps;

otherwise there would be an in¬nite ascending chain (a1 ) ‚ (a1 , a2 ) ‚ (a1 , a2 , a3 ) ‚ · · · .

If stabilization occurs at step k, we will show that I is generated by f1 , . . . , fk .

Let g = aX d + · · · be an element of I of degree d and leading coe¬cient a. Then

a ∈ (a1 , . . . , ak ) (Problem 1).

Case 1 : d ≥ dk . Since di ¤ di+1 for all i (Problem 2), we have d ≥ di for i = 1, . . . , k.

k

Now a = i=1 ci0 ai with the ci0 ∈ R. De¬ne

k

ci0 X d’di fi

g0 =

i=1

so that g0 has degree d and leading coe¬cient a, and consequently g ’ g0 has degree

r

greater than d. Having de¬ned g0 , . . . , gr ∈ (f1 , . . . , fk ) such that g ’ i=0 gi has degree

greater than d + r, say

r

g’ gi = bX d+r+1 + . . . .

i=0

(The argument is the same if the degree is greater than d + r + 1.) Now b ∈ (a1 , . . . , ak )

(Problem 1 again), so

k

b= ci,r+1 ai

i=1

6 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

with ci,r+1 ∈ R. We de¬ne

k

ci,r+1 X d+r+1’di fi

gr+1 =

i=1

r+1

so that g ’ i=0 gi has degree greater than d + r + 1. Thus

∞ ∞ k

cir X d+r’di fi

g= gr =

r=0 r=0 i=1

and it follows upon reversing the order of summation that g ∈ (f1 , . . . , fk ). (The reversal

is legal because the inner summation is ¬nite. For a given nonnegative integer j, there

are only ¬nitely many terms of the form bX j .)

Case 2 : d < dk . As above, a ∈ (a1 , . . . , ak ), so there is a smallest m between 1 and

k such that a ∈ (a1 , . . . , am ). It follows that d ≥ dm (Problem 3). As in case 1 we have

m

a = i=1 ci ai with ci ∈ R. De¬ne

m

ci X d’di fi ∈ (f1 , . . . , fk ) ⊆ I.

h=

i=1

The leading coe¬cient of h is a, so the degree of g ’ h is greater than d. We replace g by

g ’ h and repeat the procedure. After at most dk ’ d iterations, we produce an element

g ’ hi in I of degree at least dk , with all hi ∈ (f1 , . . . , fk ). By the analysis of case 1,

g ∈ (f1 , . . . , fk ).

Problems For Section 8.2

1. Justify the step a ∈ (a1 , . . . , ak ) in (8.2.3).

2. Justify the step di ¤ di+1 in (8.2.3).

3. Justify the step d ≥ dm in (8.2.3).

4. Let R be a subring of the ring S, and assume that S is ¬nitely generated as an algebra

over R. In other words, there are ¬nitely many elements x1 , . . . , xn ∈ S such that the

smallest subring of S containing the xi and all elements of R is S itself. Show that S

is a homomorhic image of the polynomial ring R[X1 , . . . , Xn ].

5. Continuing Problem 4, show that if R is Noetherian, then S is also Noetherian.

8.3 The Nullstellensatz: Preliminaries

We have observed that every variety V de¬nes an ideal I(V ) and every ideal I de¬nes a

variety V (I). Moreover, if I(V1 ) = I(V2 ), then V1 = V2 by (6) of (8.1.3). But it is entirely

possible for many ideals to de¬ne the same variety. For example, the ideals (f ) and (f m )

need not coincide, but their zero-sets are identical. Appearances to the contrary, the two

statements in part (6) of (8.1.3) are not symmetrical. A variety V is, by de¬nition, always

expressible as V (S) for some collection S of polynomials, but an ideal I need not be of the

8.3. THE NULLSTELLENSATZ: PRELIMINARIES 7

special form I(X). Hilbert™s Nullstellensatz says that if two ideals de¬ne the same variety,

then, informally, the ideals are the same “up to powers”. More precisely, if g belongs to

one of the ideals, then g r belongs to the other ideal for some positive integer r. Thus the

only factor preventing a one-to-one correspondence between ideals and varieties is that

a polynomial can be raised to a power without a¬ecting its zero-set. In this section we

collect some results needed for the proof of the Nullstellensatz. We begin by showing that

each point of An determines a maximal ideal.

8.3.1 Lemma

If a = (a1 , . . . , an ) ∈ An , then I = (X1 ’ a1 , . . . , Xn ’ an ) is a maximal ideal.

Proof. Suppose that I is properly contained in the ideal J, with f ∈ J \ I. Apply the

division algorithm n times to get

f = A1 (X1 ’ a1 ) + A2 (X2 ’ a2 ) + · · · + An (Xn ’ an ) + b

where A1 ∈ k[X1 , . . . , Xn ], A2 ∈ k[X2 , . . . , Xn ], . . . , An ∈ k[Xn ], b ∈ k. Note that b

cannot be 0 since f ∈ I. But f ∈ J, so by solving the above equation for b we have b ∈ J,

/

hence 1 = (1/b)b ∈ J. Consequently, J = k[X1 , . . . , Xn ]. ™

The following de¬nition will allow a precise statement of the Nullstellensatz.

8.3.2 De¬nition

The radical of an ideal I (in any commutative ring R) is the set of all elements f ∈ R

such that f r ∈ I for some positive integer r. √

A popular notation for the radical of I is I. If f r √ g s belong to I, then by the

and

∈ I, and it follows that I is an ideal.

r+s’1

binomial theorem, (f + g)

8.3.3 Lemma

√

I ⊆ IV (I).

If I is any ideal of k[X1 , . . . , Xn ], then

√

Proof. If f ∈ I, then f r ∈ I for some positive integer r. But then f r vanishes on V (I),

hence so does f . Therefore f ∈ IV (I). ™

√

The Nullstellensatz states that IV (I) = I, and the hard part is to prove that

√

IV (I) ⊆ I. The technique is known as the “Rabinowitsch trick”, and it is indeed

very clever. Assume that f ∈ IV (I). We introduce a new variable Y and work in

k[X1 , . . . , Xn , Y ]. If I is an ideal of k[X1 , . . . , Xn ], then by the Hilbert basis theorem,

I is ¬nitely generated, say by f1 , . . . , fm . Let I — be the ideal of k[X1 , . . . , Xn , Y ] gener-

ated by f1 , . . . , fm , 1 ’ Y f . [There is a slight ambiguity: by fi (X1 , . . . , Xn , Y ) we mean

fi (X1 , . . . , Xn ), and similarly for f .] At an appropriate moment we will essentially set Y

equal to 1/f and come back to the original problem.

8 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.3.4 Lemma

If (a1 , . . . , an , an+1 ) is any point in An+1 and (a1 , . . . , an ) ∈ V (I) (in other words, the fi ,

i = 1, . . . , m, vanish at (a1 , . . . , an )), then (a1 , . . . , an , an+1 ) ∈ V (I — ).

/

Proof. We are assuming that f ∈ IV (I), so that f vanishes on the zero-set of {f1 , . . . , fm }.

In particular, f (a1 , . . . , an ) = 0. The value of 1 ’ Y f at (a1 , . . . , an , an+1 ) is therefore

1 ’ an+1 f (a1 , . . . , an ) = 1 ’ an+1 (0) = 1 = 0. But 1 ’ Y f ∈ I — , so (a1 , . . . , an , an+1 ) does

not belong to the zero-set of I — . ™

8.3.5 Lemma

If (a1 , . . . , an , an+1 ) is any point in An+1 and (a1 , . . . , an ) ∈ V (I), then (a1 , . . . , an , an+1 ) ∈

/ /

— —

V (I ). Consequently, by (8.3.4), V (I ) = ….

Proof. By hypothesis, fi (a1 , . . . , an , an+1 ) = 0 for some i, and since fi ∈ I — , (a1 , . . . , an+1 )

cannot belong to the zero-set of I — . ™

At this point we are going to assume what is called the weak Nullstellensatz, namely

that if I is a proper ideal of k[X1 , . . . , Xn ], then V (I) is not empty.

8.3.6 Lemma

There are polynomials g1 , . . . , gm , h ∈ k[X1 , . . . , Xn , Y ] such that

m

gi fi + h(1 ’ Y f ).

1= (1)

i=1

This equation also holds in the rational function ¬eld k(X1 , . . . , Xn , Y ) consisting of quo-

tients of polynomials in k[X1 , . . . , Xn , Y ].

Proof. By (8.3.4) and (8.3.5), V (I — ) = …, so by the weak Nullstellensatz, I — = k[X1 , . . . ,

Xn , Y ]. In particular, 1 ∈ I — , and since I — is generated by f1 , . . . , fm , 1 ’ Y f , there is an

equation of the speci¬ed form. The equation holds in the rational function ¬eld because

a polynomial is a rational function. ™

8.3.7 The Rabinowitsch Trick

The idea is to set Y = 1/f , so that (1) becomes

m

1= gi (X1 , . . . , Xn , 1/f (X1 , . . . , Xn ))fi (X1 , . . . , Xn ). (2)

i=1

√

Is this legal? First of all, if f is the zero polynomial, then certainly f ∈ I, so we

can assume f = 0. To justify replacing Y by 1/f , consider the ring homomorphism

from k[X1 , . . . , Xn , Y ] to k(X1 , . . . , Xn ) determined by Xi ’ Xi , i = 1, . . . , n, Y ’

1/f (X1 , . . . , Xn ). Applying this mapping to (1), we get (2). Now the right side of (2) is a

sum of rational functions whose denominators are various powers of f . If f r is the highest

8.3. THE NULLSTELLENSATZ: PRELIMINARIES 9

power that appears, we can absorb all denominators by multiplying (2) by f r . The result

is an equation of the form

m

r

f= hi (X1 , . . . , Xn )fi (X1 , . . . , Xn )

i=1

where the hi are polynomials in k[X1 , . . . , Xn ]. Consequently, f r ∈ I. ™

The ¬nal ingredient is a major result in its own right.

8.3.8 Noether Normalization Lemma

Let A be a ¬nitely generated k-algebra, where k is a ¬eld. In other words, there are

¬nitely many elements x1 , . . . , xn in A that generate A over k in the sense that every

element of A is a polynomial in the xi . Equivalently, A is a homomorphic image of the

polynomial ring k[X1 , . . . , Xn ] via the map determined by Xi ’ xi , i = 1, . . . , n.

There exists a subset {y1 , . . . , yr } of A such that the yi are algebraically independent

over k and A is integral over k[y1 , . . . , yr ].

Proof. Let {x1 , . . . , xr } be a maximal algebraically independent subset of {x1 , . . . , xn }.

If n = r we are ¬nished, since we can take yi = xi for all i. Thus assume n > r, in which

case x1 , . . . , xn are algebraically dependent over k. Thus there is a nonzero polynomial

f ∈ k[X1 , . . . , Xn ] such that f (x1 , . . . , xn ) = 0. We can assume n > 1, for if n = 1 and

r = 0, then A = k[x1 ] and we can take {y1 , . . . , yr } to be the empty set.

We ¬rst assume that k is in¬nite and give a proof by induction on n. (It is possible to

go directly to the general case, but the argument is not as intricate for an in¬nite ¬eld.)

Decompose f into its homogeneous components (sums of monomials of the same degree).

Say that g is the homogeneous component of maximum degree d. Then, regarding g as a

polynomial in Xn whose coe¬cients are polynomials in the other Xi , we have, relabeling

variables if necessary, g(X1 , . . . , Xn’1 , 1) = 0. Since k is in¬nite, it follows from (8.1.3)

part (8) that there are elements a1 , . . . , an’1 ∈ k such that g(a1 , . . . , an’1 , 1) = 0. Set

zi = xi ’ ai xn , i = 1, . . . , n ’ 1, and plug into f (x1 , . . . , xn ) = 0 to get an equation of the

form

g(a1 , . . . , an’1 , 1)xd + terms of degree less than d in xn = 0.

n

A concrete example may clarify the idea. If f (x1 , x2 ) = g(x1 , x2 ) = x2 x3 and x1 =

12

z1 + a1 x2 , then the substitution yields

(z1 + 2a1 z1 x2 + a2 x2 )x3

2

12 2

which indeed is g(a1 , 1)x5 plus terms of degree less than 5 in x2 . Divide by

2

g(a1 , . . . , an’1 , 1) = 0 to conclude that xn is integral over B = k[z1 , . . . , zn’1 ]. By

the induction hypothesis, there are elements y1 , . . . , yr algebraically independent over

k such that B is integral over k[y1 , . . . , yr ]. But the xi , i < n, are integral over B since

xi = zi + ai xn . By transitivity (see (7.1.4)), x1 , . . . , xn are integral over k[y1 , . . . , yr ].

Thus (see (7.1.5)) A is integral over k[y1 , . . . , yr ].

Now we consider arbitrary k. As before, we produce a nonzero polynomial f such

that f (x1 , . . . , xn ) = 0. We assign a weight wi = sn’i to the variable Xi , where s is a

10 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

large positive integer. (It su¬ces to take s greater than the total degree of f , that is,

the sum of the degrees of all monomials in f .) If h = »X1 1 · · · Xnn is a monomial of f ,

a a

n

we de¬ne the weight of h as w(h) = i=1 ai wi . The point is that if h = µX1 1 · · · Xnn ,

b b

then w(h) > w(h ) i¬ h > h in the lexicographic ordering, that is, for some m we have

ai = bi for i ¤ m, and am+1 > bm+1 . We take h to be the monomial of maximum weight.

(If two monomials di¬er in the lexicographic ordering, they must have di¬erent weights.)

Set zi = xi ’ xwi , 1 ¤ i ¤ n ’ 1, and plug into f (x1 , . . . , xn ) = 0 to get

n

cxw(h) + terms of lower degree in xn = 0.

n

For example, if f (x1 , x2 ) = h(x1 , x2 ) = x3 x2 , then x1 = z1 + xw1 gives

12 2

(z1 + 3z1 xw1 + 3z1 x2w1 + x3w1 )x2

3 2

2

2 2 2

and w(h) = 3w1 + 2w2 = 3w1 + 2 since sn’2 = s0 = 1. Thus xn is integral over

B = k[z1 , . . . , zn’1 ], and an induction argument ¬nishes the proof as in the ¬rst case. ™

8.3.9 Corollary

Let B be a ¬nitely generated k-algebra, where k is a ¬eld. If I is a maximal ideal of B,

then B/I is a ¬nite extension of k.

Proof. The ¬eld k can be embedded in B/I via c ’ c + I, c ∈ k. [If c ∈ I,

c = 0, then c’1 c = 1 ∈ I, a contradiction.] Since A = B/I is also a ¬nitely gener-

ated k-algebra, it follows from (8.3.8) that there is a subset {y1 , . . . , yr } of A with the

yi algebraically independent over k and A integral over k[y1 , . . . , yr ]. Now A is a ¬eld

(because I is a maximal ideal), and therefore so is k[y1 , . . . , yr ] (see the Problems in Sec-

tion 7.1). But this will lead to a contradiction if r ≥ 1, because 1/y1 ∈ k[y1 , . . . , yr ]. /

(If 1/y1 = g(y1 , . . . , yr ) ∈ k[y1 , . . . , yr ], then y1 g(y1 , . . . , yr ) = 1, contradicting algebraic

independence.) Thus r must be 0, so A is integral, hence algebraic, over the ¬eld k.

Therefore A is generated over k by ¬nitely many algebraic elements, so by (3.3.3), A is a

¬nite extension of k. ™

8.3.10 Corollary

Let A be a ¬nitely generated k-algebra, where k is a ¬eld. If A is itself a ¬eld, then A is

a ¬nite extension of k.

Proof. As in (8.3.9), with B/I replaced by A. ™

Problems For Section 8.3

1. Let S be a multiplicative subset of the ring R (see (2.8.1)). If I is an ideal that is

disjoint from S, then by Zorn™s lemma, there is an ideal J that is maximal among

ideals disjoint from S. Show that J must be prime.

8.4. THE NULLSTELLENSATZ: EQUIVALENT VERSIONS AND PROOF 11

2. Show that the radical of the ideal I is the intersection of all prime ideals containing I.

√

[If f r ∈ I ⊆ P , P prime, then f ∈ P . Conversely, assume f ∈ I. With a clever

/

choice of multiplicative set S, show that for some prime ideal P containing I, we have

f ∈ P .]

/

3. An algebraic curve is a variety de¬ned by a nonconstant polynomial in two variables.

Show (using the Nullstellensatz) that the polynomials f and g de¬ne the same algebraic

curve i¬ f divides some power of g and g divides some power of f . Equivalently, f and

g have the same irreducible factors.

4. Show that the variety V de¬ned over the complex numbers by the two polynomials

Y 2 ’ XZ and Z 2 ’ X 2 Y is the union of the line L given by Y = Z = 0, X arbitrary,

and the set W of all (t3 , t4 , t5 ), t ∈ C.

5. The twisted cubic is the variety V de¬ned over the complex numbers by Y ’ X 2 and

Z ’ X 3 . In parametric form, V = {(t, t2 , t3 ) : t ∈ C}. Show that V is irreducible. [The

same argument works for any variety that can be parametrized over an in¬nite ¬eld.]

6. Find parametrizations of the following algebraic curves over the complex numbers. (It

is permissible for your parametrizations to fail to cover ¬nitely many points of the

curve.)

(a) The unit circle x2 + y 2 = 1;

(b) The cuspidal cubic y 2 = x3 ;

(c) The nodal cubic y 2 = x2 + x3 .

7. Let f be an irreducible polynomial, and g an arbitrary polynomial, in k[x, y]. If f does

not divide g, show that the system of simultaneous equations f (x, y) = g(x, y) = 0 has

only ¬nitely many solutions.

8.4 The Nullstellensatz: Equivalent Versions

And Proof

We are now in position to establish the equivalence of several versions of the Nullstellen-

satz.

8.4.1 Theorem

For any ¬eld k and any positive integer n, the following statements are equivalent.

(1) Maximal Ideal Theorem The maximal ideals of k[X1 , . . . , Xn ] are the ideals of

the form (X1 ’ a1 , . . . , Xn ’ an ), a1 , . . . , an ∈ k. Thus maximal ideals correspond to

points.

(2) Weak Nullstellensatz If I is an ideal of k[X1 , . . . , Xn ] and V (I) = …, then I =

k[X1 , . . . , Xn ]. Equivalently, if I is a proper ideal, then V (I) is not empty.

(3) Nullstellensatz If I is an ideal of k[X1 , . . . , Xn ], then

√

IV (I) = I.

12 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

(4) k is algebraically closed.

Proof. (1) implies (2). Let I be a proper ideal, and let J be a maximal ideal containing I.

By (8.1.3), part (4), V (J) ⊆ V (I), so it su¬ces to show that V (J) is not empty. By (1),

J has the form (X1 ’ a1 , . . . , Xn ’ an ). But then a = (a1 , . . . , an ) ∈ V (J). [In fact

V (J) = {a}.]

(2) implies (3). This was done in Section 8.3.

(3) implies (2). We use the fact that the radical of an ideal I is the intersection

of all prime ideals containing I; see Section 8.3, Problem 2. Let I be a proper ideal of

k[X1 , . . .√Xn ]. Then I is contained in a maximal, hence prime, ideal P . By the result just

, √

quoted, I is also contained in P , hence I is a proper ideal. By (3), IV (I) is a proper

ideal. But if V (I) = …, then by (8.1.3) part (7), IV (I) = k[X1 , . . . , Xn ], a contradiction.

(2) implies (1). If I is a maximal ideal, then by (2) there is a point a = (a1 , . . . , an ) ∈

V (I). Thus every f ∈ I vanishes at a, in other words, I ⊆ I({a}). But (X1 ’ a1 , . . . ,

Xn ’ an ) = I({a}); to see this, decompose f ∈ I({a}) as in the proof of (8.3.1). Therefore

the maximal ideal I is contained in the maximal ideal (X1 ’ a1 , . . . , Xn ’ an ), and it

follows that I = (X1 ’ a1 , . . . , Xn ’ an ).

(4) implies (1). Let I be a maximal ideal of k[X1 , . . . , Xn ], and let K = k[X1 , . . . ,

Xn ]/I, a ¬eld containing an isomorphic copy of k via c ’ c + I, c ∈ k. By (8.3.9), K

is a ¬nite extension of k, so by (4), K = k. But then Xi + I = ai + I for some ai ∈ k,

i = 1, . . . , n. Therefore Xi ’ ai is zero in k[X1 , . . . , Xn ]/I, in other words, Xi ’ ai ∈ I.

Consequently, I ⊇ (X1 ’ a1 , . . . , Xn ’ an ), and we must have equality by (8.3.1).

(1) implies (4). Let f be a nonconstant polynomial in k[X1 ] with no root in k. We

can regard f is a polynomial in n variables with no root in An . Let I be a maximal ideal

containing the proper ideal (f ). By (1), I is of the form (X1 ’ a1 , . . . , Xn ’ an ) = I({a})

for some a = (a1 , . . . , an ) ∈ An . Therefore f vanishes at a, a contradiction. ™

8.4.2 Corollary

If the ideals I and J de¬ne the same variety and a polynomial g belongs to one of the

ideals, then some power of g belongs to the other ideal.

√ √ √

J. If g ∈ I ⊆ I, then g r ∈ J

Proof. If V (I) = V (J), then by the Nullstellensatz, I=

for some positive integer r. ™

8.4.3 Corollary

The maps V ’ I(V ) and I ’ V (I) set up a one-to-one correspondence between varieties

√

and radical ideals (de¬ned by I = I).

√

Proof. By (8.1.3) part 6, V I(V ) = V . By the Nullstellensatz, IV (I) = I = I for radical

ideals. It remains to prove that for any variety V , I(V ) is a radical ideal. If f r ∈ I(V ),

then f r , hence f , vanishes on V , so f ∈ I(V ). ™

8.5. LOCALIZATION 13

8.4.4 Corollary

Let f1 , . . . , fr , g ∈ k[X1 , . . . , Xn ], and assume that g vanishes wherever the fi all vanish.

Then there are polynomials h1 , . . . , hr ∈ k[X1 , . . . , Xn ] and a positive integer s such that

g s = h1 f1 + · · · + hr fr .

Proof. Let I be the ideal generated by f1 , . . . , fr . Then V (I) is the set of points at which

all fi vanish, so that IV (I) is the set of polynomials that vanish wherever all fi vanish.

√

Thus g belongs to IV (I), which is I by the Nullstellensatz. Consequently, for some

positive integer s, we have g s ∈ I, and the result follows. ™

Problems For Section 8.4

1. Let f be a polynomial in k[X1 , . . . , Xn ], and assume that the factorization of f into

irreducibles is f = f1 1 · · · fr r . Show that the decomposition of the variety V (f ) into

n n

irreducible subvarieties (Section 8.1, Problems 5 and 6) is given by V (f ) = ∪r V (fi ).

i=1

2. Under the hypothesis of Problem 1, show that IV (f ) = (f1 · · · fr ).

3. Show that there is a one-to-one correspondence between irreducible polynomials in

k[X1 , . . . , Xn ] and irreducible hypersurfaces (see (8.2.2))in An (k), if polynomials that

di¬er by a nonzero multiplicative constant are identi¬ed.

4. For any collection of subsets Xi of An , show that I(∪i Xi ) = ©i I(Xi ).

5. Show that every radical ideal I of k[X1 , . . . , Xn ] is the intersection of ¬nitely many

prime ideals.

6. In Problem 5, show that the decomposition is unique, subject to the condition that

the prime ideals P are minimal, that is, there is no prime ideal Q with I ⊆ Q ‚ P .

7. Suppose that X is a variety in A2 , de¬ned by equations f1 (x, y) = · · · = fm (x, y) = 0,

m ≥ 2. Let g be the greatest common divisor of the fi . If g is constant, show that X

is a ¬nite set (possibly empty).

8. Show that every variety in A2 except for A2 itself is the union of a ¬nite set and an

algebraic curve. 9. Give an example of two distinct irreducible polynomials in k[X, Y ]

with the same zero-set, and explain why this cannot happen if k is algebraically closed.

10. Give an explicit example of the failure of a version of the Nullstellensatz in a

non-algebraically closed ¬eld.

8.5 Localization

8.5.1 Geometric Motivation

Suppose that V is an irreducible variety, so that I(V ) is a prime ideal. A polynomial g

will belong to I(V ) if and only if it vanishes on V . If we are studying rational functions

f /g in the neighborhood of a point x ∈ V , we must have g(x) = 0. It is very convenient

to have every polynomial g ∈ I(V ) available as a legal object, even though g may vanish

/

at some points of V . The technical device that makes this possible is the construction of

the ring of fractions S ’1 R, the localization of R by S, where R = k[X1 , . . . , Xn ] and S is

the multiplicative set R \ I(V ). We will now study the localization process in general.

14 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.5.2 Notation

Recalling the setup of Section 2.8, let S be a multiplicative subset of the ring R, and

S ’1 R the ring of fractions of R by S. Let h be the natural homomorphism of R into

S ’1 R, given by h(a) = a/1. If X is any subset of R, de¬ne S ’1 X = {x/s : x ∈ X, s ∈ S}.

We will be especially interested in such a set when X is an ideal.

If I and J are ideals of R, the product of I and J, denoted by IJ, is de¬ned (as in

(7.6.1)) as the set of all ¬nite sums i xi yi , xi ∈ I, yi ∈ J. It follows from the de¬nition

that IJ is an ideal. The sum of two ideals has already been de¬ned in (2.2.8).

8.5.3 Lemma

If I is an ideal of R, then S ’1 I is an ideal of S ’1 R. If J is another ideal of R, the

(i) S ’1 (I + J) = S ’1 I + S ’1 J;

(ii) S ’1 (IJ) = (S ’1 I)(S ’1 J);

(iii) S ’1 (I © J) = S ’1 I © S ’1 J;

(iv) S ’1 I is a proper ideal i¬ S © I = ….

Proof. The de¬nition of addition and multiplication in S ’1 R implies that S ’1 I is an

ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse

inclusions in (i) and (ii) follow from

ab at + bs ab ab

+= , =.

s t st st st

To prove (iii), let a/s = b/t where a ∈ I, b ∈ J, s, t ∈ S. There exists u ∈ S such that

u(at ’ bs) = 0. Then a/s = uat/ust = ubs/ust ∈ S ’1 (I © J).

Finally, if s ∈ S © I then 1/1 = s/s ∈ S ’1 I, so S ’1 I = S ’1 R. Conversely, if

S ’1 I = S ’1 R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that

t(s ’ a) = 0, so at = st ∈ S © I. ™

Ideals in S ’1 R must be of a special form.

8.5.4 Lemma

If J is an ideal of S ’1 R and I = h’1 (J), then I is an ideal of R and S ’1 I = J.

Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S ’1 I, with

a ∈ I and s ∈ S. Then a/1 ∈ J, so a/s = (a/1)(1/s) ∈ J. Conversely, let a/s ∈ J, with

a ∈ R, s ∈ S. Then h(a) = a/1 = (a/s)(s/1) ∈ J, so a ∈ I and a/s ∈ S ’1 I. ™

Prime ideals yield sharper results.

8.5. LOCALIZATION 15

8.5.5 Lemma

If I is any ideal of R, then I ⊆ h’1 (S ’1 I), with equality if I is prime and disjoint from

S.

Proof. If a ∈ I, then h(a) = a/1 ∈ S ’1 I. Thus assume that I is prime and disjoint from

S, and let a ∈ h’1 (S ’1 I). Then h(a) = a/1 ∈ S ’1 I, so a/1 = b/s for some b ∈ I, s ∈ S.

There exists t ∈ S such that t(as ’ b) = 0. Thus ast = bt ∈ I, with st ∈ I since S © I = ….

/

Since I is prime, we have a ∈ I. ™

8.5.6 Lemma

If I is a prime ideal of R disjoint from S, then S ’1 I is a prime ideal of S ’1 R.

Proof. By (8.5.3), part (iv), S ’1 I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S ’1 I, with

a, b ∈ R, s, t ∈ S. Then ab/st = c/u for some c ∈ I, u ∈ S. There exists v ∈ S such that

v(abu ’ cst) = 0. Thus abuv = cstv ∈ I, and uv ∈ I because S © I = …. Since I is prime,

/

ab ∈ I, hence a ∈ I or b ∈ I. Therefore either a/s or b/t belongs to S ’1 I. ™

The sequence of lemmas can be assembled to give a precise conclusion.

8.5.7 Theorem

There is a one-to-one correspondence between prime ideals P of R that are disjoint from

S and prime ideals Q of S ’1 R, given by

P ’ S ’1 P and Q ’ h’1 (Q).

Proof. By (8.5.4), S ’1 (h’1 (Q)) = Q, and by (8.5.5), h’1 (S ’1 P ) = P . By (8.5.6), S ’1 P

is a prime ideal, and h’1 (Q) is a prime ideal by the basic properties of preimages of sets.

If h’1 (Q) meets S, then by (8.5.3) part (iv), Q = S ’1 (h’1 (Q)) = S ’1 R, a contradiction.

Thus the maps P ’ S ’1 P and Q ’ h’1 (Q) are inverses of each other, and the result

follows. ™

8.5.8 De¬nitions and Comments

If P is a prime ideal of R, then S = R \ P is a multiplicative set. In this case, we write

R(P ) for S ’1 R, and call it the localization of R at P . (The usual notation is RP , but it™s

easier to read without subscripts.) If I is an ideal of R, we write I(P ) for S ’1 I. We are

going to show that R(P ) is a local ring, that is, a ring with a unique maximal ideal. First

we give some conditions equivalent to the de¬nition of a local ring.

8.5.9 Proposition

For a ring R, the following conditions are equivalent.

(i) R is a local ring;

(ii) There is a proper ideal I of R that contains all nonunits of R;

16 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

(iii) The set of nonunits of R is an ideal.

Proof. (i) implies (ii). If a is a nonunit, then (a) is a proper ideal, hence is contained in

the unique maximal ideal I.

(ii) implies (iii). If a and b are nonunits, so are a + b and ra. If not, then I contains

a unit, so I = R, a contradiction.

(iii) implies (i). If I is the ideal of nonunits, then I is maximal, because any larger

ideal J would have to contain a unit, so that J = R. If H is any proper ideal, then H

cannot contain a unit, so H ⊆ I. Therefore I is the unique maximal ideal. ™

8.5.10 Theorem

R(P ) is a local ring.

Proof. Let Q be a maximal ideal of R(P ). Then Q is prime, so by (8.5.7), Q = I(P ) for

some prime ideal I of R that is disjoint from S, in other words, contained in P . Thus

Q = I(P ) ⊆ P (P ). If P (P ) = R(P ), then by (8.5.3) part (iv), P is not disjoint from

S = R\P , which is impossible. Therefore P (P ) is a proper ideal containing every maximal

ideal, so it must be the unique maximal ideal. ™

If R is an integral domain and S is the set of all nonzero elements of R, then S ’1 R is

the quotient ¬eld of R. In this case, S ’1 R is a local ring, because any ¬eld is a local ring.

({0} is the unique maximal ideal.) Alternatively, we can appeal to (8.5.10) with P = {0}.

8.5.11 Localization of Modules

If M is an R-module and S a multiplicative subset of R, we can essentially repeat the

construction of Section 2.8 to form the localization S ’1 M of M by S, and thereby divide

elements of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t)

equivalent if for some u ∈ S, u(tx ’ sy) = 0. The equivalence class of (x, s) is denoted by

x/s, and addition is de¬ned by

xy tx + sy

+= .

s t st

If a/s ∈ S ’1 R and x/t ∈ S ’1 M , we de¬ne

ax ax

= .

st st

In this way, S ’1 M becomes an S ’1 R-module. Exactly as in (8.5.3), if M and N are

submodules of a module L, then

S ’1 (M + N ) = S ’1 M + S ’1 N and S ’1 (M © N ) = S ’1 M © S ’1 N.

Further properties will be given in the exercises.

8.6. PRIMARY DECOMPOSITION 17

Problems For Section 8.5

1. Let M be a maximal ideal of R, and assume that for every x ∈ M , 1 + x is a unit.

Show that R is a local ring (with maximal ideal M ). [Show that if x ∈ M , then x is a

/

unit, and apply (8.5.9).]

2. Show that if p is prime and n is a positive integer, then Zpn is a local ring with maximal

ideal (p).

3. Let R be the ring of all n by n matrices with coe¬cients in a ¬eld F . If A is a nonzero

element of R and 1 is the identity matrix, is {1, A, A2 , . . . } always a multiplicative set?

Let S be a multiplicative subset of the ring R. We are going to construct a mapping

from R-modules to S ’1 R-modules, and another mapping from R-module homomorphisms

to S ’1 R-module homomorphisms, as follows. If M is an R-module, we let M ’ S ’1 M .

If f : M ’ N is an R-module homomorphism, we de¬ne S ’1 f : S ’1 M ’ S ’1 N by

x f (x)

’ .

s s

Since f is a homomorphism, so is S ’1 f .

4. If g : N ’ L and composition of functions is written as a product, show that S ’1 (gf ) =

S ’1 (g)S ’1 (f ), and if 1M is the identity mapping on M , then S ’1 (1M ) = 1S ’1 M . We

say that S ’1 is a functor from the category of R-modules to the category of S ’1 R-

modules. This terminology will be explained in great detail in Chapter 10.

5. If

f g

’ ’

M N L

is an exact sequence, show that

S ’1 f S ’1 g

S ’1 M S ’1 N S ’1 L

’ ’

is exact. We say that S ’1 is an exact functor. Again, we will study this idea in

Chapter 10.

6. Let R be the ring of rational functions f /g with f, g ∈ k[X1 , . . . , Xn ] and g(a) = 0,

where a = (a1 , . . . , an ) is a ¬xed point in An . Show that R is a local ring, and identify

the unique maximal ideal.

7. If M is an R-module and S is a multiplicative subset of R, denote S ’1 M by MS . If

N is a submodule of M , show that (M/N )S ∼ MS /NS .

=

8.6 Primary Decomposition

We have seen that every radical ideal in k[X1 , . . . , Xn ] can be expressed as an intersection

of ¬nitely many prime ideals (Section 8.4, Problem 5). A natural question is whether a

similar result holds for arbitrary ideals. The answer is yes if we generalize from prime to

primary ideals.

18 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.6.1 De¬nitions and Comments

The ideal Q in the ring R is primary if Q is proper and whenever a product ab belongs to

Q, either a ∈ Q or bn ∈ Q for some positive integer n. [The condition on b is equivalent

√

to b ∈ Q.] An equivalent statement is that R/Q = 0 and whenever (a + Q)(b + Q) = 0

in R/Q, either a + Q = 0 or (b + Q)n = 0 for some positive integer n. This says that if

b + Q is a zero-divisor in R/Q, then it is nilpotent, that is, some power of b + Q is 0.

It √

follows from the de¬nition that every prime ideal is primary. Also, if Q is primary,

√

then Q is the smallest prime ideal containing Q. [Since Q is the intersection of all

√

prime ideals containing Q (Section 8.3, Problem 2), it su¬ces to show that Q is prime.

√ if a b ∈ Q, then a ∈ Q or b ∈ Q for some m, so either a or b must belong to

nn n nm

But

Q. Note also that since Q is proper, it is contained in a maximal, hence prime, ideal,

√

so Q is also proper.] √

If Q is primary and Q = P , we say that Q is P -primary.

8.6.2 Examples

1. In Z, the primary ideals are {0} and (pr ), where p is prime. In Z6 , 2 and 3 are

zero-divisors that are not nilpotent, and a similar situation will occur in Zm whenever

more than one prime appears in the factorization of m.

2. Let R = k[X, Y ] where k is any ¬eld, and take Q = (X, Y 3 ), the ideal generated by

X and Y 3 . This is a nice example of analysis in quotient rings. We are essentially setting

X and Y 3 equal to zero, and this collapses the ring R down to polynomials a0 +a1 Y +a2 Y 2 ,

with the ai ∈ k and arithmetic mod Y 3 . Formally, R/Q is isomorphic to k[Y ]/(Y 3 ). The

zero-divisors in R/Q are of the form cY + dY 2 , c ∈ k, and they are nilpotent. Thus Q is

primary. If f ∈ R, then the only way for f not to belong to the radical of Q is for the

√

constant term of f to be nonzero. Thus Q = (X, Y ), a maximal ideal by (8.3.1).

Now we claim that Q cannot be a power of a prime ideal; this will be a consequence

of the next result.

8.6.3 Lemma

√

If P is a prime ideal, then for every positive integer n, P n = P .

√

Proof. Since P is a prime ideal containing P n , P n ⊆ P . If x ∈ P , then xn ∈ P n , so

√

x ∈ P n. ™

Returning to Example 2 of (8.6.2), if Q = (X, Y 3 ) is a prime power P n , then its radical

is P , so P must be (X, Y ). But X ∈ Q and X ∈ P n , n ≥ 2; since Y belongs to P but

/

not Q, we have reached a contradiction.

After a preliminary de¬nition, we will give a convenient su¬cient condition for an

ideal to be primary.

8.6.4 De¬nition

The nilradical N (R) of a ring R is the set of nilpotent elements of R, that is, {x ∈ R :

xn = 0 for some positive integer n}. Thus N (R) is the radical of the zero ideal, which is

8.6. PRIMARY DECOMPOSITION 19

the intersection of all prime ideals of R.

8.6.5 Proposition

If the radical of the ideal Q is maximal, then Q is primary.

√

Proof. Since Q is maximal, it must be the only prime ideal containing Q. By the

correspondence theorem and the fact that the preimage of a prime ideal is a prime ideal

(cf. (8.5.7)), R/Q has exactly one prime ideal, which must coincide with N (R/Q). Any

element of R/Q that is not a unit generates a proper ideal, which is contained in a

maximal ideal, which again must be N (R/Q). Thus every element of R/Q is either a unit

or nilpotent. Since a zero-divisor cannot be a unit, every zero-divisor of R/Q is nilpotent,

so Q is primary. ™

8.6.6 Corollary

If M is a maximal ideal, then M n is M -primary for all n = 1, 2, . . . .

Proof. By (8.6.3), the radical of M n is M , and the result follows from (8.6.5). ™

Here is another useful property.

8.6.7 Proposition

If Q is a ¬nite intersection of P -primary ideals Qi , i = 1, . . . , n, then Q is P -primary.

Proof. First note that the radical of a ¬nite intersection of ideals is the intersection of the

radicals (see Problem 1). It follows that the radical of Q is P , and it remains to show

that Q is primary. If ab ∈ Q but a ∈ Q, then for some i we have a ∈ Qi . Since Qi is

/ /

√

P -primary, b belongs to P = Qi . But then some power of b belongs to Q. ™

We are going to show that in a Noetherian ring, every proper ideal I has a primary

decomposition, that is, I can be expressed as a ¬nite intersection of primary ideals.

8.6.8 Lemma

Call an ideal I irreducible if for any ideals J and K, I = J © K implies that I = J or

I = K. If R is Noetherian, then every ideal of R is a ¬nite intersection of irreducible

ideals.

Proof. Suppose that the collection S of all ideals that cannot be so expressed is nonempty.

Since R is Noetherian, S has a maximal element I, necessarily reducible. Let I = J © K,

where I is properly contained in both J and K. By maximality of I, the ideals J and K are

¬nite intersections of irreducible ideals, and consequently so is I, contradicting I ∈ S. ™

20 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

If we can show that every irreducible proper ideal is primary, we then have the desired

primary decomposition. Let us focus on the chain of reasoning we will follow. If I is

an irreducible proper ideal of R, then by the correspondence theorem, 0 is an irreducible

ideal of the Noetherian ring R/I. If we can show that 0 is primary in R/I, then again by

the correspondence theorem, I is primary in R.

8.6.9 Primary Decomposition Theorem

Every proper ideal in a Noetherian ring R has a primary decomposition. (We can drop

the word “proper” if we regard R as the intersection of the empty collection of primary

ideals.)

Proof. By the above discussion, it su¬ces to show that if 0 is an irreducible ideal of R,

then it is primary. Let ab = 0 with a = 0. Since R is Noetherian, the sequence of

annihilators

ann b ⊆ ann b2 ⊆ ann b3 ⊆ · · ·

stabilizes, so ann bn = ann bn+1 for some n. If we can show that

(a) © (bn ) = 0

we are ¬nished, because a = 0 and the zero ideal is irreducible (by hypothesis). Thus let

x = ca = dbn for some c, d ∈ R. Then bx = cab = dbn+1 = 0 (because ab = 0), so d

annihilates bn+1 , hence d annihilates bn . Thus x = dbn = 0. ™

Problems For Section 8.6

1. If I1 , . . . , In are arbitrary ideals, show that

√

n n

Ii = I i.

i=1 i=1

2. Let I be the ideal (XY ’Z 2 ) in k[X, Y, Z], where k is any ¬eld, and let R = k[X, Y, Z]/I.

If P is the ideal (X + I, Z + I), show that P is prime.

3. Continuing Problem 2, show that P 2 , whose radical is prime by (8.6.3) and which is a

power of a prime, is nevertheless not primary.

4. Let R = k[X, Y ], and let P1 = (X), P2 = (X, Y ), Q = (X 2 , Y ). Show that P1 is prime

2

and P2 and Q are P2 -primary.

5. Continuing Problem 4, let I = (X 2 , XY ). Show that P1 © P2 and P1 © Q are both

2

primary decompositions of I.

Notice that the radicals of the components of the primary decomposition (referred to

√

as the √

primes associated with I) are P1 and P2 in both cases. [P1 is prime, so P 1 = P1 ;

√

P2 ⊆ Q and P2 is maximal, so P2 = Q;] Uniqueness questions involving primary

decompositions are treated in detail in textbooks on commutative algebra.

8.7. TENSOR PRODUCT OF MODULES OVER A COMMUTATIVE RING 21

6. We have seen in Problem 5 of Section 8.4 that every radical ideal in R = k[X1 , . . . , Xn ]

is the intersection of ¬nitely many prime ideals. Show that this result holds in an

arbitrary Noetherian ring R.

7. Let R = k[X, Y ] and let In be the ideal (X 3 , XY, Y n ). Show that for every positive

integer n, In is a primary ideal of R.

8.7 Tensor Product of Modules Over a Commutative

Ring

8.7.1 Motivation

In many areas of algebra and its applications, it is useful to multiply, in a sensible way, an

element x of an R-module M by an element y of an R-module N . In group representation

theory, M and N are free modules, in fact ¬nite-dimensional vector spaces, with bases

{xi } and {yj }. Thus if we specify that multiplication is linear in each variable, then we

need only specify products of xi and yj . We require that the these products, to be denoted

by xi — yj , form a basis for a new R-module T .

If f : R ’ S is a ring homomorphism and M is an S-module, then M becomes an

R-module via rx = f (r)x, r ∈ R, x ∈ M . This is known as restriction of scalars.

In algebraic topology and algebraic number theory, it is often desirable to reverse this

process. If M is an R-module, we want to extend the given multiplication rx, r ∈ R,

x ∈ M , to multiplication of an arbitrary s ∈ S by x ∈ M . This is known as extension of

scalars, and it becomes possible with the aid of the tensor product construction.

The tensor product arises in algebraic geometry in the following way. Let M be the

coordinate ring of a variety V in a¬ne space Am , in other words, M is the set of all

polynomial functions from V to the base ¬eld k. Let N be the coordinate ring of the

variety W in An . Then the cartesian product V — W is a variety in Am+n , and its

coordinate ring turns out to be the tensor product of M and N .

Let™s return to the ¬rst example above, where M and N are free modules with bases

{xi } and {yj }. Suppose that f is a bilinear map from M —N to an R-module P . (In other

words, f is R-linear in each variable.) Information about f can be completely encoded

into a function g of one variable, where g is an R-module homomorphism from T to P .

We take g(xi — yj ) = f (xi , yj ) and extend by linearity. Thus f is the composition of

the bilinear map h from M — N to T speci¬ed by (xi , yj ) ’ xi — yj , followed by g. To

summarize:

Every bilinear mapping on M — N can be factored through T .

The R-module T is called the tensor product of M and N , and we write T = M —R N .

We are going to construct a tensor product of arbitrary modules over a commutative ring,

and sketch the generalization to noncommutative rings.

22 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.7.2 De¬nitions and Comments

Let M and N be arbitrary R-modules, and let F be a free R-module with basis M — N .

Let G be the submodule of F generated by the “relations”

(x + x , y) ’ (x, y) ’ (x , y); (x, y + y ) ’ (x, y) ’ (x, y );

(rx, y) ’ r(x, y); (x, ry) ’ r(x, y)

where x, x ∈ M , y, y ∈ N , r ∈ R. De¬ne the tensor product of M and N (over R) as

T = M —R N = F/G

and denote the element (x, y) + G of T by x — y. Thus the general element of T is a ¬nite

sum of the form

xi — yi

t= (1)

i

with xi ∈ M and yi ∈ N . It is important to note that the representation (1) is not

necessarily unique.

The relations force x — y to be linear in each variable, so that

x — (y + y ) = x — y + x — y , (x + x ) — y = x — y + x — y, (2)

r(x — y) = rx — y = x — ry. (3)

See Problem 1 for details. Now if f is a bilinear mapping from M — N to the R-module

P , then f extends uniquely to a homomorphism from F to P , also called f . Bilinearity

means that the kernel of f contains G, so by the factor theorem, there is a unique R-

homomorphism g : T ’ P such that g(x — y) = f (x, y) for all x ∈ M , y ∈ N . As in

(8.7.1), if we compose the bilinear map h : (x, y) ’ x — y with g, the result is f . Again,

we say that

Every bilinear mapping on M — N can be factored through T .

We have emphasized this sentence, known as a universal mapping property (abbrevi-

ated UMP), because along with equations (1), (2) and (3), it indicates how the tensor

product is applied in practice. The detailed construction we have just gone through can

now be forgotten. In fact any two R-modules that satisfy the universal mapping property

are isomorphic. The precise statement and proof of this result will be developed in the

exercises.

In a similar fashion, using multilinear rather than bilinear maps, we can de¬ne the

tensor product of any ¬nite number of R-modules. [In physics and di¬erential geometry,

a tensor is a multilinear map on a product M1 — · · · — Mr , where each Mi is either a ¬nite-

dimensional vector space V or its dual space V — . This suggests where the terminology

“tensor product” comes from.]

In the discussion to follow, M , N and P are R-modules. The ring R is assumed ¬xed,

and we will usually write — rather than —R .

8.7. TENSOR PRODUCT OF MODULES OVER A COMMUTATIVE RING 23

8.7.3 Proposition

M — N ∼ N — M.

=

Proof. De¬ne a bilinear mapping f : M — N ’ N — M by f (x, y) = y — x. By the UMP,

there is a homomorphism g : M — N ’ N — M such that g(x — y) = y — x. Similarly,

there is a homomorphism g : N — M ’ M — N with g (y — x) = x — y. Thus g is an

isomorphism (with inverse g ). ™

8.7.4 Proposition

M — (N — P ) ∼ (M — N ) — P .

=

Proof. De¬ne f : M — N — P ’ (M — N ) — P by f (x, y, z) = (x — y) — z. The UMP

produces g : M — (N — P ) ’ (M — N ) — P with g((x, (y — z))) = (x — y) — z. [We are

applying the UMP for each ¬xed x ∈ M , and assembling the maps to produce g.] Since g

is bilinear (by Equations (2) and (3)), the UMP yields h : M — (N — P ) ’ (M — N ) — P

with h(x — (y — z)) = (x — y) — z. Exactly as in (8.7.3), we can construct the inverse of

h, so h is the desired isomorphism. ™

8.7.5 Proposition

M — (N • P ) ∼ (M — N ) • (M — P ).

=

Proof. Let f be an arbitrary bilinear mapping from M — (N • P ) to Q. If x ∈ M ,

y ∈ N , z ∈ P , then f (x, y + z) = f (x, y) + f (x, z). The UMP gives homomorphisms

g1 : M —N ’ Q and g2 : M —P ’ Q such that g1 (x—y) = f (x, y) and g2 (x—z) = f (x, z).

The maps g1 and g2 combine to give g : (M — N ) • (M — P ) ’ Q such that

g((x — y) + (x — z)) = g1 (x — y) + g2 (x — z).

In particular, with x = x,

g((x — y) + (x — z)) = f (x, y + z),

so if h : M — (N • P ) ’ M — (N • P ) is de¬ned by

h(x, y + z) = (x — y) + (x — z),

then f = gh. Thus (M — N ) • (M — P ) satis¬es the universal mapping property, hence

must be isomorphic to the tensor product. ™

8.7.6 Proposition

Regarding R as a module over itself, R —R M ∼ M .

=

Proof. The map (r, x) ’ rx of R — M ’ M is bilinear, so there is a homomorphism

g : R — M ’ M such that g(r — x) = rx. De¬ne h : M ’ R — M by h(x) = 1 — x. Then

h(rx) = 1 — rx = r1 — x = r — x. Thus g is an isomorphism (with inverse h). ™

24 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY

8.7.7 Corollary

Let Rm be the direct sum of m copies of R, and M m the direct sum of m copies of M .

Then Rm — M ∼ M m .

=

Proof. By (8.7.5), Rm — M is isomorphic to the direct sum of m copies of R — M , which

is isomorphic to M m by (8.7.6). ™

8.7.8 Proposition

Rm — Rn ∼ Rmn . Moreover, if {x1 , . . . , xm } is a basis for Rm and {y1 , . . . , yn } is a basis

=

for R , then {xi — yj , i = 1, . . . , m, j = 1, . . . , n} is a basis for Rmn .

n

Proof. This follows from the discussion in (8.7.1). [The ¬rst assertion can also be proved

by taking M = Rn in (8.7.7).] ™

8.7.9 Tensor Product of Homomorphisms

Let f1 : M1 ’ N1 and f2 : M2 ’ N2 be R-module homomorphisms. The map (x1 , x2 ) ’

f1 (x1 ) — f2 (x2 ) of M1 — M2 into N1 — N2 is bilinear, and induces a unique f : M1 — M2 ’

N1 — N2 such that

f (x1 — x2 ) = f1 (x1 ) — f2 (x2 ), x1 ∈ M1 , x2 ∈ M2 .

We write f = f1 —f2 , and call it the tensor product of f1 and f2 . Similarly, if g1 : N1 ’ P1

and g2 : N2 ’ P2 , then we can compose g1 — g2 with f1 — f2 , and

(g1 — g2 )(f1 — f2 )(x1 — x2 ) = g1 f1 (x1 ) — g2 f2 (x2 ),

hence

(g1 — g2 ) —¦ (f1 — f2 ) = (g1 —¦ f1 ) — (g2 —¦ f2 ).

When M1 = N1 = V , a free R-module of rank m, and M2 = N2 = W , a free R-

module of rank n, there is a very concrete interpretation of the tensor product of the

endomorphisms f : V ’ V and g : W ’ W . If f is represented by the matrix A and g by

the matrix B, then the action of f and g on basis elements is given by

f (vj ) = aij vi , g(wl ) = bkl wk

i k

where i and j range from 1 to m, and k and l range from 1 to n. Thus

(f — g)(vj — wl ) = f (vj ) — g(wl ) = aij bkl (vi — wk ).

i,k

The mn by mn matrix representing the endomorphism f — g : V — W ’ V — W is denoted

by A — B and called the tensor product or Kronecker product of A and B. It is given by

®

a11 B a12 B · · · a1m B

. . .

A—B =° . .»

. .

···

am1 B am2 B amm B

8.8. GENERAL TENSOR PRODUCTS 25

The ordering of the basis of V — W is

v1 — w1 , . . . , v1 — wn , . . . , vm — w1 , . . . , vm — wn .

To determine the column of A — B corresponding to vj — wl , locate the aij B block

(i = 1, . . . , m; j ¬xed) and proceed to column l of B. As we move down this column, the

indices i and k vary according to the above ordering of basis elements. If this road map

is not clear, perhaps writing out the entire matrix for m = 2 and n = 3 will help.

Problems For Section 8.7

1. Verify Equations (2) and (3) of (8.7.2).

2. If m and n are relatively prime, show that Zm —Z Zn = 0.

3. If A is a ¬nite abelian group and Q is the additive group of rationals, show that

A —Z Q = 0. Generalize to a wider class of abelian groups A.

4. The de¬nition of M —R N via a universal mapping property is as follows. The tensor

product is an R-module T along with a bilinear map h : M — N ’ T such that given

any bilinear map f : M — N ’ P , there is a unique R-homomorphism g : T ’ P such

that f = gh. See the diagram below.

GT

h

M —N

f qqq

g

5

P

Now suppose that another R-module T , along with a bilinear mapping h : M — N ’

T , satis¬es the universal mapping property. Using the above diagram with P = T

and f replaced by h , we get a unique homomorphism g : T ’ T such that h = gh.

Reversing the roles of T and T , we get g : T ’ T such that h = g h .

Show that T and T are isomorphic.

5. Consider the element n—x in Z—Zn , where x is any element of Zn and we are tensoring

over Z, i.e., R = Z. Show that n — x = 0.

6. Continuing Problem 5, take x = 0 and regard n — x as an element of nZ — Zn rather

than Z — Zn . Show that n — x = 0.

7. Let M, N, M , N be arbitrary R-modules, where R is a commutative ring. Show that

the tensor product of homomorphisms induces a linear map from HomR (M, M ) —R

HomR (N, N ) to HomR (M —R N, M —R N ).

8. Let M be a free R-module of rank m, and N a free R-module of rank n. Show that

there is an R-module isomorphism of EndR (M ) —R EndR (N ) and EndR (M — N ).

8.8 General Tensor Products

We now consider tensor products of modules over noncommutative rings. A natural

question is “Why not simply repeat the construction of (8.7.2) for an arbitrary ring R?”.

26 CHAPTER 8. INTRODUCING ALGEBRAIC GEOMETRY