Advanced Algebra II
Mar. 2, 2007 (Fri.)
1. commutative ring theory
In this chapter, rings are assume to be commutative with identity.
1.1. basic definitions.
We recall some basic definitions in the section.
Definition 1.1.1. An element a 6= 0 ∈ R is said to be a zero divisor if there is an element b 6= 0 ∈ R such that ab = 0.
A ring R 6= 0 and has no zero divisor is called an integral domain.
Proposition 1.1.2. A finite integral domain is a field.
Definition 1.1.3. A subring I ⊂ R is an ideal if rx ∈ R for all r ∈ R, x ∈ I. It will be denoted I C R.
Proposition 1.1.4. A subset I is an ideal if and only if for all a, b ∈ I, r ∈ R, a + b ∈ I and ra ∈ I.
Example 1.1.5.
For a ∈ R, we have (a) := aR, the principal ideal generated by a.
Let N ⊂ R be the subset of all nilpotent elements, i.e. N := {a ∈ R|an = 0, for some n > 0}. Then N is an ideal, called the nilradical.
¤ Definition 1.1.6. An element a ∈ R is said to be a unit if ab = 1 for some b ∈ R.
Note that a is a unit if and only if (a) = R.
Definition 1.1.7. An ideal p in R is prime if p 6= R and if ab ∈ R then either a ∈ R or b ∈ R.
An ideal m in R is maximal if m 6= R and if there is no ideal I ( R such that m ( I.
It’s easy to check that
Proposition 1.1.8. p C R is prime if and only if R/p is an integral domain.
m C R is maximal if and only if R/m is a field.
Also
Proposition 1.1.9. PCR if and only if for any ideal I, J CR, IJ ⊂ p implies that either I ⊂ p or J ⊂ p.
1
Proof. Suppose that P is prime, IJ ⊂ p and I 6⊂ p. We need to show that J ⊂ p. To see this, pick any a ∈ I − p, then ab ∈ p for all b ∈ J.
since p is prime, we must have b ∈ p for all b ∈ J. Hence J ⊂ p.
Conversely, if p satisfies the above property, we need to show that it is prime. For any a, b ∈ R such that ab ∈ p, (ab) = (a)(b) ⊂ p. Thus either (a) ⊂ p or (b) ⊂ p. It follows that a ∈ p or b ∈ p. ¤ By direct application of Zorn’s Lemma, it’s easy to that in a nonzero ring, there exists a maximal ideal. We leave the detail to the readers.
A ring with exactly one maximal ideal is called a local ring. We have the following equivalent conditions:
1. (R, m) is local.
2. The subset of non-units is an ideal.
3. If m C R is a maximal ideal and every element of 1 + m is a unit.
Proposition 1.1.10. The nilradical is the intersection of all prime ideals.
Proof. Let N0 be the intersection of all prime ideals. If x ∈ N, i.e.
xn = 0, then xn = 0 ∈ p for all prime p. Hence x ∈ p for all prime p, x ∈ N0.
On the other hand, if x 3 N. Let Σ := {I C R|I ∩ {x, x2, ...} = ∅}. Σ is non-empty for 0 ∈ Σ. By Zorn’e Lemma, there is a maximal element p in Σ. We claim that p is prime. Then x 3 p, hence x 3 N0.
To see the claim, suppose that ab ∈ p but a, b 6∈ p. We have p ( p + (a), p + (b). By maximality, we have xn∈ p + (a), and xm ∈ p + (b).
So xm+n∈ (p + (a))(p + (b)) = p + (ab) = p, a contradiction. ¤ Exercise 1.1.11. Let I C R be an ideal. Let √
I := {x ∈ R|xn ∈ I}.
Then √
I = ∩p:prime,I⊂pp.
Definition 1.1.12. we define the Jacobson radical of R, denoted J(R), to be the intersection of all maximal ideals.
The Jacobson radical is clearly an ideal. It has the property that:
Proposition 1.1.13. x ∈ J if and only if 1−xy is a unit for all y ∈ R.
To see this, suppose that x ∈ J and 1 − xy is not a unit. Then 1 − xy ∈ m for some maximal ideal m. Since x ∈ J ⊂ m, we have 1 ∈ m, a contradiction.
Conversely, if x 6∈ J, then x 6∈ m for some maximal ideal m. Then m + (x) = R. So we have 1 = xy + u for some u ∈ m, and hence 1 − xy = u is not a unit.
Theorem 1.1.14 (Chinese remainder theorem). Let I1, ..., In be ideals of R such that Ii+ Ij = R for all i 6= j. Given elements x1, ..., xn ∈ R, there exists x ∈ R such that x ≡ xi ( mod Ii) for all i.
The proof is left to the readers as an exercise.
Corollary 1.1.15. Let I1, ..., In be ideals of R such that Ii+ Ij = R for all i 6= j. Let f : R → Qn
i=1R/Ii. Then f is surjective with kernel
∩Ii.
1.2. factorization.
Definition 1.2.1. A non-zero element a ∈ R is said to be divide b ∈ R, denoted a|b if there is c ∈ R such that ac = b.
Elements a, b ∈ R are said to be associate, denoted a ∼ b, if a|b, b|a.
The following properties are immediate.
1. a|b if and only if (a) ⊃ (b).
2. a ∼ b if and only if (a) = (b).
3. a is a unit if and only if (a) = R.
Definition 1.2.2. A non-zero non-unit c ∈ R is said to be irreducible if c = ab then either a or b is unit.
A non-zero non-unit p ∈ R is said to be prime if p|ab then p|a or p|b.
Then we have the following:
Proposition 1.2.3. Let p, c are non-zero non-unit elements in R.
1. p is primes if and only if (p) is prime.
2. c is irreducible if and only if (c) is maximal among all proper prin- cipal ideals.
3. Prime element is irreducible.
4. If R is a PID, then p is prime if and only if p is irreducible.
5. If a = bu with u a unit, then a ∼ b.
6. If R an integral domain, then a ∼ b implies a = bu for some unit u.
Definition 1.2.4. An integral domain is called a unique factorization domain, UFD for short, if every non-zero non-unit element can be factored into products of irreducible elements. And the factorization is unique up to units.
Definition 1.2.5. A ring R is said to be an Euclidean ring if there is a function ϕ : R − {0} → N such that:
1. if a, b 6= 0 ∈ R and ab 6= 0, then ϕ(a) ≤ ϕ(ab).
2. if a, b 6= 0 ∈ R, then there exist q, r ∈ R such that a = qb + r with either r = 0 or r 6= 0 and ϕ(r) < ϕ(b).
Lemma 1.2.6. Let R be a PID, then its ideals satisfies Ascending chain condition, i.e. for an ascending chain of ideals
I1 ⊂ I2. . . there is n such that In = In+1 = . . ..
Hint. Considering I := ∪Ii. It’s an ideal hence I = In for some n. ¤ Theorem 1.2.7. Every Euclidean domain, ED for short, is a PID.
Every PID is a UFD.
1.3. Localization.
Before we going to the study of dimension theory, we need to recall some basic notion of localization.
Definition 1.3.1. A subset S ⊂ R is said to be a multiplicative set if (1) 1 ∈ S,
(2) if a, b ∈ S, then ab ∈ S.
Given a multiplicative set, then one can construct a localized ring S−1R which I suppose the readers have known this. In order to be self-contained, I recall the construction:
In R × S, we define an equivalent relation that (r, s) ∼ (r0, s0) if (rs0 − r0s)t = 0 for some t ∈ S. Let rs denote the equivalent class of (r, s). One can define addition and multiplication naturally. The set of all equivalent classes, denoted S−1R, is thus a ring. There is a natural ring homomorphism ı : R → S−1R by ı(r) = r1.
Remark 1.3.2.
(1) If 0 ∈ S, then S−1R = 0. We thus assume that 0 6∈ S.
(2) If R is a domain, then ı is injective. And in fact, S−1R ,→ F naturally, where F is the quotient field of R.
(3) Let J C S−1R. We will use J ∩ R to denote the ideal ı−1(J).
(If R is a domain, then J ∩ R = ı−1(J) by identifying R as a subring of S−1R).
I would like to recall the most important example and explain their geometrical meaning, which, I think, justify the notion of localization.
Example 1.3.3. Let f 6= 0 ∈ k[x1, ..., xn] and let S = {1, f, f2...}. The localization S−1k[x1, ..., xn] is usually denoted k[x1, ..., xn]f. This ring can be regarded as ”regular functions” on the open set Uf := Ank−V(f ).
One notices that Uf is of course the maximal open subset that the ring k[x1, ..., xn]f gives well-defined functions.
Example 1.3.4. Let x = (a1, ..., an) ∈ Ank and mx = (x1− a1, ..., xn− ak) be its maximal ideal. Take S = k[x1, ..., xn] − mx, then the localiza- tion is denoted k[x1, ..., xn]mx. It is the ring of regular functions ”near x”.
Recall that for a R-module M, one can also define S−1M which is naturally an S−1R-module.
Mar. 9, 2007 (Fri.) We have:
Proposition 1.3.5.
(1) If I C R, then S−1I C S−1R. Moreover, every ideal J C S−1R is of the form S−1I for some I C R.
(2) For J C S−1R, then S−1(J ∩ R) = J.
(3) S−1I = S−1R if and only if I ∩ S 6= ∅.
(4) There is a one-to-one correspondence between {p ∈ Spec(R)|p ∩ S = ∅} and {q ∈ Spec(S−1R)}.
(5) In particular, the prime ideals of the local ring Rp are in one- to-one correspondence with the prime ideals of R contained in p.
Proof.
(1) If xs,yt ∈ S−1I, that is, x, y ∈ I, then xs + yt = xt+ysst ∈ S−1I.
And if rs ∈ S−1R and xt ∈ S−1I, then rsxt = rxst ∈ S−1I. Hence S−1I is an ideal.
Moreover, let J C S−1R. Let I := ı−1(J) C R. We claim that J = S−1I. To see this, for x ∈ I, x1 ∈ J. Hence xs = x11s ∈ J for all s ∈ S. It follows that S−1I ⊂ J. Conversely, if xs ∈ J, then
x
1 = xss1 ∈ J and x1 = ı(x). So xs ∈ S−1I.
(2) If xs ∈ J ∩ R, then xs = y1 for some y ∈ R , i.e. (x − sy)t0 = 0, for some t0 ∈ S. Then look at yt ∈ S−1(J ∩ R). It’s clear that
y
t = xs1t ∈ J.
Conversely, if xs ∈ J, then x1 ∈ J ∩ R. And hence xs ∈ S−1(J ∩ S).
(3) If x ∈ I ∩ S, then 11 = xx ∈ S−1I. Conversely, if 11 = xs ∈ S−1I, then (x − s)t = 0. Therefore, xt = st ∈ S ∩ I.
(4) For q ∈ Spec(S−1R). It’s clear that q ∩ R = ı−1q ∈ Spec(R).
(q ∩ R) ∩ S = ∅, otherwise q = S−1(q ∩ R) = S−1R which is impossible.
Conversely, let p ∈ Spec(R) and p ∩ S = ∅. We would like to show that S−1p is a prime ideal. First of all, if 11 ∈ S−1p, then
1
1 = xs for some x ∈ p. It follows that (x − s)t = 0 for some t ∈ S. Thus st = xt ∈ p which is a contradiction. Therefore, S−1p 6= S−1R.
Moreover, if xsyt ∈ S−1p. Then xyst = xs00 for some x0 ∈ p and s0 ∈ S. Then (xys0 − stx0)t0 = 0 for some t0 ∈ S. Hence xys0t0 ∈ p. It follows that xy ∈ p since s0t0 6∈ p. Thus either x or y in p, and either xs or yt in S−1p.
It remains to show that the correspondence is a one-to-one correspondence. We have seen that
ı−1 : Spec(S−1R) → {p ∈ Spec(R)|p ∩ S = ∅}
is surjective. By (2), it follows that this is injective.
(5) Let S = R − p, then S ∩ q = ∅ if and only q ⊂ p.
¤ But for I C R, then S−1I ∩ R ⊃ I only. Indeed, if x ∈ I C R. Then x = xsx ∈ S−1I ∩ R. Conversely, for x ∈ S−1I ∩ R, then x1 = yt for some y ∈ I. Thus (y − xt)s = 0 for some s, t ∈ S. We can not get y ∈ I in general. However, this is the case if I is prime and S ∩ I = ∅. Thus we have
Proposition 1.3.6. If p C R is a prime ideal and S ∩ p = ∅. Then S−1p ∩ R = p.
Proof. This is an immediate consequence of (1) of the Proposition. ¤ Example 1.3.7.
Let p ∈ Spec(R), then Rp is a local ring with the unique maximal ideal pRp. To see that, if there is a maximal ideal m. By the corre- spondence, m = qRp for some q ⊂ p. Thus m ⊂ pRp and thus must be equal.
A ring with a unique maximal ideal is called a local ring. Thus Rp
is a local ring. ¤
Example 1.3.8.
More explicitly, we can consider the following example. Let R = k[x, y, x] and p = (x, y). Then there is a chain of prime ideals:
0 ⊂ (x) ⊂ (x, y) ⊂ (x, y, z) ⊂ R. (∗) If we look at R/p ∼= k[z], we see a chain of primes ideals 0 ⊂ (z) ⊂ k[z], which corresponds to (x, y) ⊂ (x, y, z) ⊂ R in (∗). This has the following geometric intepreation: by looking at the ring R/p, we understand the ”polynomial functions” on the set defined by x = 0, y = 0, i.e. the z-axis.
On the other hand, if we look at Rp, we see a chain of primes ideals 0 ⊂ (x)Rp ⊂ (x, y)Rp which corresponds to 0 ⊂ (x) ⊂ (x, y) in (∗).
Geometrically, it can be think as ”local functions near a ”generic” point in z-axis.
We will come to more precise description of ”polynomial functions”
and generic point later. ¤
Proposition 1.3.9. The operation S−1 on ideals commutes with for- mation of finite sums, product, intersection and radicals.
Proof. These can be checked directly. ¤
1.4. modules.
It’s essential to study modules in ring theory. One might find that modules not only generalize the notion od ideals but also clarify many things.
Definition 1.4.1. Let R be a ring. An abelian group M is said to be an R-module if there is a map µ : R × M → M such that for all a, b ∈ R, x, y ∈ M, we have:
a(x + y) = ax + ay (a + b)x = ax + bx
a(bx) = (ab)x 1x = x Example 1.4.2.
Let R be a ring and I C R be an ideal. Then I, R/I are R-modules
naturally. ¤
Example 1.4.3.
An abelian group G has a natural Z-module structure by µ(m, g) :=
mg for all m ∈ Z and g ∈ G. ¤
Note that let M, N be R-modules. By a R-module homomorphism, we mean a group homomorphism ϕ : M → N such that ϕ(rx) = rϕ(x).
That is, it’s an R-linear map.
Exercise 1.4.4.
Given an R-module homomorphism f : M → N, then ker(f ), im(f ), coker(f ) are R-modules in a natural way.
Let M be an R-module. Given x ∈ M, then Rx is a submodule of M. Even Ix is a submodule for any ideal I C R. More generally, IN is a submodule of M if N < M and I C R.
On the other hand, given x ∈ M, we may consider the annihilator of x,
Ann(x) := {r ∈ R|rx = 0}.
It’s clear to be an ideal. Also for any submodule N < M . We can define Ann(N) similarly as {r ∈ R|rN = 0}.
A remark is that for a R-module M, the module structure map R × M → M also induces a natural map R/Ann(M) × M → M.
Hence M can also be viewed as R/Ann(M)-module.
Note that for given x ∈ M, the natural map f : R → Rx is a R- linear map. And ker(f ) = Ann(x). So in fact, we have an R-module isomorphism R/ ker(f ) ∼= Rx by the isomorphism theorem.
Definition 1.4.5. A element x ∈ M is said to be torsion if Ann(x) 6=
0. A module is torsion if every non-zero element is torsion. A module is torsion-free if every non-zero element is not torsion.
Exercise 1.4.6.
Given two modules M, N , note that the set of all R-module homo- morphisms, denoted Hom(M, N ), is naturally an R-module.
Another important feature is that
Proposition 1.4.7. The operation S−1 is exact. That is, if M0 −→ Mf −→ Mg 00
is an exact sequence of R-module, then
S−1M0 S−→ S−1f −1M S−→ S−1g −1M00 is exact as S−1R-module.
Proof. By the construction, it follows directly that S−1g ◦ S−1f = 0.
It suffices to check that ker(S−1g) ⊂ im(S−1f ). If xs ∈ ker(S−1g), then g(x)s = 0 ∈ S−1M00. That is tg(x) = 0 for some t ∈ S. We have tg(x) = g(tx) = 0. And then tx = f (y) for some y ∈ M0. Therefore,
x s = xt
st = f (y)
st = S−1f (y st).
¤ Corollary 1.4.8. The operation S−1 commutes with passing to quo- tients by ideals. That is, let I C R be an ideal and ¯S the image of S in R := R/I. Then S¯ −1R/S−1I ∼= ¯S−1R.¯
Proof. By considering 0 → I → R → ¯R → 0 as an exact sequence of R-modules. We have S−1R/S−1I ∼= S−1R as S¯ −1R-modules. We claim that there is a natural bijection from S−1R to ¯¯ S−1R by¯ xs¯ 7→ xs¯¯ which is compatible with all structures.
One can also try to prove this directly. Basically, construct a surjec- tive ring homomorphism S−1R → ¯S−1R and shows that the kernel is¯
S−1I. We leave it as an exercise. ¤
We now introduce a very important and useful Lemma, Nakayama’s Lemma.
Theorem 1.4.9. Let M be a finitely generated R-module. Let a ⊂ J(R) be an ideal contained in the Jacobson radical. Then aM = M implies M = 0.
Perhaps the most useful case is when (R, m) is a local ring. Then J(R) = m is nothing but the unique maximal ideal. The assertion is that if mM = M, then M = 0.
Proof. Suppose M 6= 0. Let x1, ..., xn be a minimal generating set of M. We shall prove by induction on n.
Note that for an ideal I C R, elements in IM can be written as a1x1+ ... + anxn for ai ∈ I.
Since aM = M, we have x1 = Pn
i=1aixi for ai ∈ a. Hence we have (1 − a1)x1 =Pn
i=2aixi. Notice that (1 − a1) is a unit, otherwise it’s in some maximal ideal which leads to a contradiction. So we have either x1 = 0 or x1is generated by x2, ..., xn. Either one is a contradiction. ¤
By applying the Lemma to M/N, and note that a(M/N) = (aM + N)/N, we have the following:
Corollary 1.4.10. Keep the natation as above, if N < M is a sub- module such that M = aM + N, then M = N.
Corollary 1.4.11. Let R be a local ring and M be a finitely gener- ated R-module. If x1, ..., xn generates M/mM as a vector space, then x1, ..., xn generates M.
Proof. Let N be the submodule generated by x1, ..., xn. Then N +
mM = M. ¤
We close this section by considering finitely generated modules over PID. We have two important and interesting examples.
Example 1.4.12.
Let G be a finitely generated abelian group. Then it’s clearly a Z-module while Z is PID.
Moreover, a finite group is clearly a torsion module. ¤ Example 1.4.13.
Let V be a n-dimensional vector space over k. Let A be a n × n matrix over k (or a linear transform from V to V ). Then V can be viewed as a k[t]-module via k[t] × V → V , with f (t)v 7→ f (A)v.
Note that by Caylay-Hamilton Theorem, f (A) = 0 for f (x) being the characteristic polynomial. In fact, Ann(V ) = (p(x)), where p(x) is the minimal polynomial of A. Therefore, V is a torsion module. ¤ Now let M be a finitely generated torsion module over a PID R. One sees that Ann(M) 6= 0. Since R is PID, we have Ann(M) =Q
(paii).
For each pi, we consider M(pi) := {x ∈ M|pnix = 0, for some n}.
One can prove that M = ⊕M(pi).
In fact, for each pi, there exist n1 ≥ n2 ≥ ... ≥ nji such that M(pi) ∼=
⊕jk=1i R/(pnik).
These pnik are called elementary divisors.
Apply this discussion to the example of linear transformation. Then Ann(V ) = (p(x)). We assume that p(x) =Q
(x−λi)ai splits into linear factors.
Then V (λi) is nothing but the generalized eigenspace of λi. And the decomposition V = ⊕V (λi) is the decomposition into general- ized eigenspaces. The further decomposition into cyclic modules corre- sponds to further decomposition of eigensapces into in invariant sub- spaces. The restriction of linear transformation into these invariant subspaces gives a Jordan block.
More explicitly, suppose that we have v ∈ V which corresponds to a generator of R/(x − λi)m. Then we have independent vectors :
v = v0, (A − λi)v =: v1, ..., (A − λi)m−1v = vm−1. Using this set as part of basis, then we see
Av0 = λiv0+ v1, ...
Avm−2 = λivm−2 + vm−1, Avm−1 = λivm−1 This gives the Jordan block.
Mar. 16, 2007 (Fri.) 1.5. tensor product.
In this section, we are going to construct tensor product of modules coming from the multilinear algebra consideration. Then we describe its universal property. Lastly, we regard tensor product as a functor and compare its properties with the functor Hom.
Let R be a ring and M1, M2 be R-modules. We consider a category whose object are (f, N ), where N is R-module and f : M1× M2 → N is a R-bilinear map. A morphism is defined naturally.
Definition 1.5.1. Keep the notation as above, the universal repelling object is called the tensor product of M1, M2, denoted M1⊗RM2.
The existence of tensor product can be constructed as following: Let F be the free R-module generated by the set M1× M2. Let K be the submodule of F generated by R-bilinear relations, that is
(a1+ a2, b) − (a1, b) − (a2, b), (a, b1+ b2) − (a, b1) − (a, b2), (a, rb) − r(a, b),
(ra, b) − r(a, b).
Then we have an induced bilinear map ϕ : M1× M2 → F/K. We claim that (ϕ, F/K) is the universal object.
To see this, note that for any R-bilinear map f : M1 × M2 → N, one easily produce a map h0 : F → N by h0(a, b) 7→ f (a, b). Since f is bilinear, one sees that h0(x) = 0 if x ∈ K. Thus we have an induced map h : F/K → N.
Example 1.5.2.
Z2⊗ZZ3 = 0.
Z2⊗ZZ2 ∼= Z2. ¤
Proposition 1.5.3. Let M1, M2, M3 be R-modules. Then there ex- ists a unique isomorphism (M1⊗M2)⊗M3 → M1⊗(M2⊗M3) such that (x⊗y)⊗z 7→ x⊗(y⊗z).
Proposition 1.5.4. Let M1, M2 be R-modules. There there exists a unique isomorphism M1⊗M2 → M2 → M1 such that x⊗y 7→ y⊗x.
Proposition 1.5.5. Let MR, MR0 be right R-modules and RN,RN0 be left R-modules. And let f : M → M0, g : N → N0 be module homomor- phisms. Then there is a unique group homomorphism f ⊗g : M⊗RN → M0⊗RN0.
Proof. Consider a middle linear map (f, g) : M × N → M0⊗RN by (a, b) 7→ f (a)⊗g(b). By the universal property, we are done. ¤
There are some more properties:
Proposition 1.5.6.
(⊕ni=1Mi)⊗N ∼= ⊕ni=1(M1⊗M).
In fact, this also holds if the index set in infinite.
Also we have
Proposition 1.5.7. M⊗RR ∼= M
Proof. There is a natural map : N → R⊗RN by (x) = x⊗1. It’s clear that this is an R-homomorphism. We then construct f : R × N → N by f (r, x) = rx. It’s clear that this is middle linear and thus induces a group homomorphism ¯f : R⊗RN → N by ¯f (r⊗x) = rx. It’s also easy to see that this is a module homomorphism.
Therefore, it suffices to check that ¯f = 1N (which is clear) and
¯f = 1R⊗RN. This mainly due to Xri⊗xi =X
(1⊗rixi) = 1⊗X rixi.
¤ Combining these two, we have
Proposition 1.5.8. If F is free over R with basis {vi}i∈I. Then every element of M⊗RF can be written as P
i∈Ixi⊗vi, with xi ∈ M and all but finitely many xi = 0.
Moreover,
Proposition 1.5.9. If M, N are free over R with basis {vi}, {wi} re- spectively. Then M⊗RN is free with basis {vi⊗wj}.
We now consider the ”base change”. That is, if f : R → S is a ring homomorphism. Then there are connection between S-modules and R-modules.
First, if N is a S-module, then N can be viewed as an R-module by R × N → N such that (r, x) 7→ f (r)x. This operation is called restriction of scalars.
For example, a vector space V over Q can be viewed as a Z-module.
On the other hand, suppose now that we have M a R-module. S can be viewed as R-module. So we have MS := S⊗RM, which is naturally a S-module. This operation is called base change.
For example, let M = Z[x] be a Z-module and S = Q, then MS = Z[x]⊗ZQ ∼= Q[x].
Exercise 1.5.10.
Let S be a multiplicative set in R. Then we have ı : R → S−1R. Let M be an R-module, then S−1M ∼= S−1R⊗RM. ¤ Exercise 1.5.11.
Show that S−1(M⊗RN) ∼= S−1M⊗s−1RS−1N. In particular, we have (M⊗N)p ∼= Mp⊗RpNp.
Proposition 1.5.12. Let 0 → M1 → M2 → M3 → 0 be an exact sequence of R-modules. And N is an R-module. Then M1⊗N → M2⊗N → M3⊗N → 0 is exact. That is, tensor product ir right exact.
Proof. For y ∈ N3, y = g(z) for some z ∈ N2, thus for x ∈ M, x⊗y = (1⊗g)(x⊗z). Hence im(1⊗g) generate M⊗RN3. It follows that 1⊗g is surjective.
(1⊗g)(1⊗f )(x⊗w) = x⊗gf (w) = x⊗0 = 0.
Therefore, im(1⊗f ) ⊂ ker(1⊗g). There is thus an induced map α : M⊗RN2/im(1⊗f ) → M⊗RN3. It suffices to show that α is an isomor- phism. To this end, we intend to construct the inverse map. Consider x⊗y ∈ M⊗RN3, there is z ∈ N2 such that g(z) = y. We define β0 : M × N3 → M⊗RN2/im(1⊗f ) by β0(x, y) = x⊗z. We first check that this is well-defined. If z, z0 ∈ N2 such that g(z) = g(z0) = y, then z − z0 ∈ ker g = imf . Thus there is w ∈ N1 such that z − z0 = f (w).
One verifies that
x⊗z = x⊗(z0+ f (w)) = x⊗z0+ x⊗f (w)
= x⊗z0+ (1⊗f )(x⊗w) = x⊗z0.
It’s routine to check that β0 is middle linear, hence it induces β : M⊗RN3 → M⊗RN2/im(1⊗f ). One can check that
αβ(x⊗y) = αx⊗z = x⊗g(z) = x⊗y., βα(x⊗z) = β(x⊗g(z)) = x⊗z.
¤ Another way to see it is via the relation with Hom functor.
Lemma 1.5.13. The sequence M1 → M2 → M3 → 0 is exact if and only if 0 → Hom(M3, N ) → Hom(M2, N ) → Hom(M1, N ) is exact for all N.
Lemma 1.5.14. There is a canonical isomorphism Hom(M⊗N, P ) ∼= Hom(M, Hom(N, P )).
Proof of Prop. 1.5.12. Since M1 → M2 → M3 → 0 is exact, we have for all P ,
0 → Hom(M3, Hom(N, P )) → Hom(M2, Hom(N, P )) → Hom(M1, Hom(N, P )), is exact. Thus
0 → Hom(M3⊗N, P ) → Hom(M2⊗N, P ) → Hom(M1⊗N, P ), is exact for all P . And hence M1⊗N → M2⊗N → M3⊗N → 0 is
exact. ¤
This says that the functor ⊗RN is right exact. Similarly, one can see that the functor N⊗R is also right exact.
Definition 1.5.15. A module is said to be flat if the functor ⊗RM is exact.
For example, S−1R is a flat R-module.
We have the following easier criterion for flatness.
Proposition 1.5.16. The following are equivalent:
1. N is flat.
2. If M1 → M2 is injective, then M1⊗N → M2⊗N is injective.
In fact, we can have
Theorem 1.5.17. M is flat if and only if ⊗M is exact with respect to 0 → a → R → R/a → 0 for all ideal a.
We remark that (R/a)⊗M ∼= M/aM. To see this, note that there is a surjective map R⊗M ∼= M → (R/a)⊗M. The kernel of this map if given by the image of ϕ : a⊗M → R⊗M ∼= M. The image consists of
{ϕ(X
rixi)|ri ∈ a, xi ∈ M} = {X
rixi} = aM.
Proof. We introduce the notion of N-flat if ⊗M is exact for any N0 such that 0 → N0 → N. Note that the condition can be rephrased as M is R-flat because every submodule of R is exactly an ideal of R.
Step 1. If M is N-flat, then M is ⊕N-flat.
Step 2. If M is N-flat, then for every submodule S and quotient Q of N, M is S-flat and Q-flat.
Let S0 be a submodule of S. We have that S0⊗M → N⊗M is injective. This map factors through S⊗M, hence S0 → M → S⊗M is also injective and M is S-flat.
Let Q0 be a submodule of Q and N0 be its preimage in N. We have 0 −−−→ S −−−→ N0 −−−→ Q0 −−−→ 0
y
y
y
0 −−−→ S −−−→ N −−−→ Q −−−→ 0 Tensoring with M, we get
0 K
y
y
S⊗M −−−→ N0⊗M −−−→ Q0⊗M −−−→ 0
=
y
y
y 0 −−−→ S⊗M −−−→ N⊗M −−−→ Q⊗M
By Snake Lemma, we have K = 0. Thus M is Q-flat.
Step 3. Since every module is quotient of free modules, i.e there is a surjection ⊕i∈IR → N. So M is R-flat implies that M is ⊕i∈IR-flat by Step 1. And then by Step 2, M is N-flat. Thus M is flat and we
are done. ¤
Mar. 23, 2007 (Fri.) We can consider the following ”local properties”
Proposition 1.5.18. The following are equivalent:
1. M = 0 .
2. Mp = 0 for all prime ideal p 3. .Mm= 0 for all maximal ideal m.
One can think of this as ”a function is zero if its value at each point is zero”.
Proof. It suffices to show that 3 ⇒ 1. Suppose that M 6= 0, let x 6=
0 ∈ M. Then Ann(x) 6= R. Thus Ann(x) ⊂ m for some m. Now
x
1 = 0 ∈ Mm, which means that sx = 0 for some s ∈ (R − m) ∩ Ann(x).
This is a contradiction. ¤
Proposition 1.5.19. Let ϕ : M → N be an R-homomorphism. The following are equivalent:
1. ϕ : M → N is injective.
2. ϕp : Mp → Np is injective for all prime ideal p 3. ϕm: Mm → Nm is injective for all maximal ideal m.
Proof. Since ⊗Rp is exact, we have 1 ⇒ 2. 2 ⇒ 3 is trivial. It suffices to show that 3 ⇒ 1.
Let K := ker(ϕ), then Km= ker(ϕm) = 0 since ⊗Rm is exact. Thus
we have K = 0 by Proposition 1.5.18. ¤
Proposition 1.5.20. The following are equivalent:
1. M is flat.
2. Mp is a flat Rp-module for all prime ideal p 3. .Mm is a flat Rm-module for all maximal ideal m.
Proof. For 1 ⇒ 2, if suffices to check that Mp is Rp-flat. Let b C Rp be an ideal, then b = aRp. Now
aRp⊗Mp = (a⊗M)p = (a⊗M)⊗Rp → M⊗Rp, is injective because M is flat and Rp is flat.
To see 3 ⇒ 1, for any a C R, we consider aM → M. Localize it, we have (a⊗RM)m = aRm⊗Rm⊗Mm → Mm. This is injective by our assumption. By Proposition 1.5.19, we see that aM → M is injective.
¤ 1.6. chain conditions.
In this section, we are going to survey some basic properties of Noe- therian and Artinian modules.
Let (Σ, ≺) be a partial ordered set, then the following two conditions are equivalent
(1) Every increasing sequence of x1 ≺ x2 ≺ ... in Σ is stationary, i.e. there exist n such that xn= xn+1= ....
(2) Every non-empty subset of Σ has a maximal element.
Definition 1.6.1. Let M be an R-module. Let Σ be the set of submod- ules of M. We say that M is a Noetherian (resp. Artinian) R-module if the P.O. set (Σ, ⊆) (resp. (Σ, ⊇)) satisfies the above condition.
Remark 1.6.2. In the case of (Σ, ⊆) we say condition (1) is a.c.c.
(ascending chain condition) and condition (2) is maximal condition.
While in the case of (Σ, ⊆) we say condition (1) is d.c.c. (descending chain condition) and condition (2) is minimal condition.
Proposition 1.6.3. Let 0 → M0 → M → M00 → 0 be an exact sequence of R-modules. Then M is Noetherian (resp. Artinian) if and only if both M0, M00 are Noetherian (resp. Artinian).
Proof. We leave it to the readers as an exercise. ¤ Corollary 1.6.4. If Mi is Noetherian (resp. Artinian), then so is
⊕ni=1Mi.
Proposition 1.6.5. M is a Noetherian R-module if and only if every submodule of M is finitely generated.
Proof. Let N < M be a submodule, let Σ be the set of finitely generated submodules of N. By the maximal condition, there is an maximal element N0 ∈ Σ. We claim that N0 = N then we are done.
To see the claim, let’s suppose on the contrary that N0 N. Pick any x ∈ N − N0, then N0 N0+ Rx < N . And clearly, N0+ Rx is finitely generated. This contradict to the maximality of N0.
Conversely, given an ascending chain M1 < M2 < ... of submodules of M. Let N = ∪Mi. It’s clear that N is a submodule of M. Thus N is finitely generated, say N = Rx1+ ... + Rxr. For each xi, xi ∈ Mji for some ji. Let n = maxi=1,...,r{ji}. Then it’s easy to see that Mn =
Mn+1 = ... and hence we are done. ¤
Definition 1.6.6. A ring R is said to be Noetherian (resp. Artinian) if R is a Noetherian (resp. Artinian) R-module. Or equivalently, the ideals of R satisfies ascending (resp. descending) chain condition.
Example 1.6.7.
1. A field if both Noetherian and Artinian.
2. The ring Z is Noetherian but not Artinian.
3. More generally, a PID is always Noetherian. To see this, suppose that we have a1 ⊂ a2... an ascending chain of ideals. Let a := ∪ai. Then a is an ideal, hence a = (x) for some x. Now x ∈ an for some n,
thus we have a ⊂ an. ¤
Example 1.6.8.
Consider R = k[x1, x2, ....] the polynomial ring of infinitely many inde- terminate. There is an ascending chain of ideal
(x1) ⊂ (x1, x2) ⊂ (x1, x2, x3) ⊂ ...
So R is not Noetherian. Let K be its quotient field, then clearly K is Noetherian. Thus a subring of a Noetherian ring is not necessarily Noetherian.
However, Noetherian and Artinian properties are preserved by taking quotient.
Proposition 1.6.9. If R is Noetherian or Artinian, then so is R/a for any a C R.
Proof. Use the correspondence of ideals. We leave the detail to the
readers. ¤
Indeed, by using the correspondence of ideals one can also show that if R is Noetherian (resp. Artinian) and then so is S−1R.
Proposition 1.6.10. Let R be a Noetherian (resp. Artinian) ring, and M a finitely generated R-module. Then M is Noetherian (resp.
Artinian).
Proof. By Proposition 1.6.3 and 1.6.4. ¤
One important result is the following:
Theorem 1.6.11 (Hilbert’s basis theorem). If R is Noetherian, then so is R[x].
Proof. Let b C R[x] be an ideal. We need to show that it’s finitely generated.
Let a be the set of leading coefficients of b, it’s easy to see that it’s an ideal. Let a1, ..., an be a set of generators. Then there are fi = aixri+ ... ∈ b. Let r = max{ri}. And let b0 = (f1, ..., fn).
For any f = axr+ ... ∈ b of degree r ≥ m, a = P
ciai, for some ci ∈ R. Thus f − cixr−rifi ∈ b has degree < r. Inductively, we get a polynomial g with degree < r and f = g + h with h ∈ b0.
Lastly, consider M = R + Rx + ... + Rxr−1 a finitely generated R-module. Then b ∩ M is a finitely generated R-module. So b = (b ∩ M) + b0 is clearly a finitely generated R[x]-module. ¤
An immediate consequence is
Corollary 1.6.12. If R is Noetherian, then so is R[x1, ..., xn]. Also any finitely generated R-algebra is Noetherian.
By almost the same argument, one can show that R[[x]] is Noetherian if R is Noetherian.
We now turn into the consideration of Artinian rings.
Proposition 1.6.13. Let R be an Artinian ring, then every prime ideal is maximal.
Proof. Let p be a prime ideal, then B := R/p is an Artinian integral domain. For any x 6= 0 ∈ B, we consider (x) ⊃ (x2).... Since B is Artinian, we have (xn) = (xn+1) for some n. In particular, xn= xn+1y.
Hence xy = 1, that is x is a unit. Thus B is a field. ¤ Proposition 1.6.14. Let R be an Artinian ring, then R has only finitely many maximal ideals.
Proof. Let Σ be the set of finite intersection of maximal ideals. There is a minimal element, say µ := m1∩ ... ∩ mn. For any maximal ideal m, one sees that m ∩ µ = µ. Thus
m ⊃ µ ⊃ m1...mn.
It follows that m ⊃ mi, and hence m = mi for some i. Therefore, there
are n maximal ideals. ¤
We now turn into the consideration of Artinian rings.
Proposition 1.6.15. Let R be an Artinian ring, then every prime ideal is maximal.
Proof. Let p be a prime ideal, then B := R/p is an Artinian integral domain. For any x 6= 0 ∈ B, we consider (x) ⊃ (x2).... Since B is Artinian, we have (xn) = (xn+1) for some n. In particular, xn= xn+1y.
Hence xy = 1, that is x is a unit. Thus B is a field. ¤ Proposition 1.6.16. Let R be an Artinian ring, then R has only finitely many maximal ideals.
Proof. Let Σ be the set of finite intersection of maximal ideals. There is a minimal element, say µ := m1∩ ... ∩ mn. For any maximal ideal m, one sees that m ∩ µ = µ. Thus
m ⊃ µ ⊃ m1...mn.
It follows that m ⊃ mi, and hence m = mi for some i. Therefore, there
are n maximal ideals. ¤
We close this section by comparing Artinian rings and Noetherian rings.
Theorem 1.6.17. R is Artinian if and only R is Noetherian and dim R = 0.
Recall that dim R = sup{n|p0 ( p1 ( ... ( pn∈ SpecR}
Proof. Step 1. If R is Artinian, then every prime is maximal. So dim R = 0.
Step 2. Moreover, R has only finitely many maximal ideals say m1, ..., mn. Then N = ∩mi. We claim that Nr = 0 for some r. To see this, notice that we have a descending chain
N ⊃ N2...
We claim that Nr = 0 for some r.
Since R is Artinian, we have Nr = Nr+1 = ..., call it a. We would like to show that a = 0. If not, pick an minimal ideal b such that ab 6= 0.
There is x 6= 0 ∈ b such that xa 6= 0. So by minimality, (x) = b.
Also, (xa)a = xa 6= 0 and (x)a ⊂ (x). So (x) f a = (x) by minimality as well.
Thus there is y ∈ a so that x = xy. Then we have x = xy = xy2 = ....
Since y ∈ N is nilpotent, we have x = 0.
Step 3. We show that R is a Noetherian R-module.
Consider
(Y
mi)r ⊂ Nr = 0, then we have a finite filtration 0 ⊂ ...Q
mi ⊂ ...m1 ⊂ R. And each quotient is Artinian R-module, hence also a Artinian R/mi-module. It follows that each quotient is also Noetherian R/mi-module and hence R-module, too. So R is also Noetherian.
Step 4. If R is Noetherian, we have a primary decomposition of 0 = ∩ni=1qi. Taking radical, we have N = ∩ni=1pi. Since dim R = 0, so every prime is maximal. We see that pi are the only primes ideals.
Step 5. Moreover, since N is finitely generated, we see that Nr = 0 for some r. Then we proceed as above. We are done. ¤
1.7. integral extension.
In this section, we are going to explore more about the notion of integral extension. The goal is to show the going up and going down theorems.
Let A, B be rings. We say B is an extension over A if A ⊂ B.
An element x ∈ B is said to be integral over A if it satisfies a monic polynomial in A[x]. B is integral over A if every element of B is integral over A.
The following Properties are more or less parallel to the theory of algebraic extensions. Proofs are similar. The reader should find no difficulty working them out.
Proposition 1.7.1. Let A ⊂ B be an extension. The followings are equivalent:
(1) x ∈ B is integral over A.
(2) A[x] is a finitely generated A-module.
(3) A[x] is contained is a subring C ⊂ B such that C is a finitely generated A-module.
Corollary 1.7.2. Let A ⊂ B be an extension. If xi ∈ B is integral over A for i = 1, .., n. Then A[x1, ..., xn] is a finitely generated A-module.
Corollary 1.7.3. Let A ⊂ B ⊂ C be extensions. If C is integral over B and B is integral over C, then C is integral over A.
Corollary 1.7.4. Let A ⊂ B be an extension. The integral closure of A in B, which is the set of elements in B integral over A, is a ring (subring of B).
Let A ⊂ B be an extension. A is said to be integrally closed in B if the integral closure of A is A itself.
Corollary 1.7.5. Let A ⊂ B be an extension. And let C be the integral closure of A in B. Then C is integrally closed in B.
Example 1.7.6.
Consider Z ⊂ Q ⊂ Q(√
−1). The integral closure of Z in Q(√
−1) is Z[√
−1]. ¤
Exercise 1.7.7.
Determine the integral closure of Z in Q(√
d) for an integer d. ¤ Proposition 1.7.8. Let A ⊂ B be an integral extension.
(1) If b C B and a := b ∩ A, then B/b is integral over A/a.
(2) Let S be a multiplicative set of A (hence of B), then S−1B is integral over S−1A.
Proof. (1) For ¯b ∈ B/b, one notices that b ∈ B and bn+ an−1bn−1+ ...+a0 = 0 for some ai ∈ A. It’s clear that ¯bn+an−1¯bn−1+... ¯a0 = 0 ∈ ¯B := B/b. And hence ¯b is integral over ¯A := A/a.
(2) For bs ∈ S−1B. One notice that bn+ an−1bn−1+ ... + a0 = 0 for some ai ∈ A. Hence
(b
s)n+an−1 s (b
s)n−1+ ... + a0 sn = 0.
And we are done.
¤ Lemma 1.7.9. Let A ⊂ B be an integral extension and B is an do- main. Then A is a field if and only if B is a field.
Proof. Suppose that A is a field. For any b 6= 0 ∈ B, bn+ an−1bn−1+ ... + a0 = 0 for some ai ∈ A. We may assume that a0 6= 0 ∈ A because B is a domain. Then
b(bn−1+ an−1bn−2+ ... + a1) = −a0. Therefore, b is invertible in B and so B is a field.
Conversely, let B be a field. For a 6= 0 ∈ A, a−1 ∈ B. Thus (a−1)n+ an−1(a−1)n−1+ ... + a0 = 0. In particular, a−1 = −(an−1+ ... +
a0an−1) ∈ A. ¤
Let A ⊂ B be an integral extension, and q ∈ SpecB, p ∈ SpecA. We say that q is lying over p if q ∩ A = p.
We are going to study the relation between prime ideals of integral extension.
Proposition 1.7.10. Keep the notation as above with q is lying over p. Then q is maximal if and only if p is maximal.
Proof. B/q is again integral over A/p. Since B/q is a field if and only
if A/p is a field, we are done. ¤
An consequence is the following corollary which assert the ”unique- ness in a chain of prime ideal”:
Corollary 1.7.11. Keep the notation as above. If q1 ⊂ q2 are prime ideals lying over p. Then q1 = q2.
Proof. Let S = A−p. (Note that qi∩S = ∅ for i = 1, 2.) Then we have Bp integral over Ap. Moreover, q1Bp ⊂ q2Bp. Note that qiBp∩ Ap = (qi ∩ A)Ap = pAp is maximal for i = 1, 2. Hence both q1Bp ⊂ q2Bp are maximal ideal lying over pAp. We then have q1Bp = q2Bp. By the correspondence of prime ideals, we have q1 ⊂ q2. ¤ It’s also desirable to have existence of prime ideal lying over a specific one.
Proposition 1.7.12. Let A ⊂ B be an integral extension. Let p ∈ Spec(A). Then there exist a q ∈ Spec(B) lying over A.
Proof. Let S = A − p. We consider Ap ⊂ Bp. (This is injective since S−1 is exact.) Take a maximal ideal m of Bp. We claim that q := m∩B is a prime ideal lying over p. To see this,
q ∩ A = (m ∩ B) ∩ A = (m ∩ Ap) ∩ A = pAp∩ A = p.
This is because m lying over a maximal ideal of Ap and the only max-
imal ideal of Ap is pAp. ¤
Theorem 1.7.13 (Going-up theorem). Let B be an integral extension over A. Let p1 ⊂ p2 ∈ Spec(A) and q1 ∈ Spec(B) lying over p1. Then there is q2 ∈ Spec(B) containing q1 lying over p2.
Proof. Let ¯A := A/p1and ¯B := B/q1. Then ¯B is integral over ¯A. There is a prime ideal ¯q2 lying over ¯p2. Lift to B, then we are done. ¤ As we have seen, let B be an integral extension over A. Then every chain of distinct prime ideals of B restricts to a chain of distinct prime ideals of A and conversely, every chain of distinct prime ideals of A extends to a chain of distinct prime ideals of B. It follows that
dimA = dimB.
