The kernelof a map (homomorphism!) f is the subset of A written as kerf defined by {a∈ A|f(a) = 0}. It is easy to see that the kernel is closed under the addition and multiplication. Here, the closeness means that the result of the operation using two elements in a subset lies in the subset.
(It is pointless to speak of the closedness of an operation without reference to a subset.) There is one more important property as we discuss below.
must be zero, any element (x− 12)·f(x) must be zero in Q[x]/I, too; it must be an element of the ideal we are defining. We require no more elements are identified as zeros. That is, we define I ={f(x)(x−12)| f(x)∈Q[x]} as our ideal, and carry out any computation in the quotient ring Q[x]/I. We write I as (x− 12) by putting the generator inside the parenthesis. Computing in Q[x]/I is the same as computing inQ[x] and evaluate the polynomial atx= 12.
In general, if we write and idealI ofR as (a, b, c), then we mean I={ax+by+cz |x, y, z∈R}.
I is said to be generated by a, b, c. A quick exercise: What is Q[x]/(x− 12, x−1) ? It is a zero ring. Since x−12 and x−1 are “zeros,” their difference 12 is zero. Zero times 2 is zero, so 1 is zero. Therefore, everything is zero. We proved an ideal equality (x− 12, x−1) = (1). The whole ringR viewed as an ideal is called theunit idealdenoted by (1). In fact, when there is any invertible element in an ideal, it is the unit ideal. For this reason, an invertible element in a ring is called aunit. Note that computing the minimal set of generators is in general a hard problem.
For instance, I do not know any algorithmic answer to questions like “can this ideal be generated by three elements?” However, the Gr¨obner basis gives a canonical set of generators for an ideal of a polynomial ring, and we can algorithmically compare two ideals.
Prime, maximal
There is a very important class of ideals that generalizes prime numbers in Z. A prime numberp has a property that ifpdivides a product of two integersabthenpdivides eitheraor b. Aprime1 idealpofR is an ideal not equal to (1) such that
ab∈pimpliesa∈p orb∈p for anya, b∈R. (A.2) The condition is rephrased as a /∈ p and b /∈ p imply ab /∈ p. Is the ideal (0) = {0} prime? It depends. InZthe zero ideal is prime because the product of two nonzero integers is nonzero. In the polynomial ringQ[x], (0) is prime because the degree of nonzero polynomial does not decrease under any nonzero multiplication. If any nonzero elements of a nonzero ringRhas a multiplicative inverse, in which caseRis called afield, then (0) is prime becauseab= 0 meansa= 0 orb= 0. However, in the ring of diagonal 2×2 matrices, (0) is not prime because nonzero matrices
1 0 0 0
and
0 0 0 1
multiply to zero. The ring in which (0) is prime has a special name,(integral) domain. It is easy to verify thatR/pis an integral domain if and only ifpis prime, applying the picture thatpdefines zeros inR/p.
1Do not use “primary” in place of “prime.” The adjective “primary” has a slightly different technical meaning.
There is one more important property of prime ideals. Let f :A →B be a map between two rings. If p⊆B is a prime ideal, then I=f−1(p)⊆A is prime. Proof: aa0 ∈I⇒f(a)f(a0)∈p⇒ f(a)∈p∨f(a0)∈p⇒a∈I∨a0∈I.
A subclass of prime ideals consists of maximal ideals. Amaximal ideal m6= (1) is defined by the maximal property:
m(m0 impliesm0= (1) for any idealm0. (A.3) It is a priori not vivid why maximal ideals are prime. But it is easy to see. Consider R/m. If a∈R/m is nonzero, that isa /∈m, thenI = (m, a))m and therefore I= (1), which means there is an element b ∈ R such that ba+m = 1 for some m ∈ m. By the canonical map, b maps to a multiplicative inverse ofainR/m. (The converse is also true. m6= (1)is a maximal ideal if and only if R/m is a field.) In other words, R/m is a field, and therefore an integral domain. In particular, mis prime.
Modules
Ideals admit another viewpoint. Let us forget the multiplication within an idealI, and treatIas a separate set from the mother ringR. It is still an abelian group, and satisfies (A.1). The condition (∗) looks very similar to the scalar multiplication on vector spaces. Indeed, consider a direct sum R⊕R of abelian groups, the set of all tuples (r, r0) wherer, r0 ∈R are any elements. There is an operation on R⊕R similar to (A.1); we can define r·(s, s0) to be (rs, rs0), similar to the scalar multiplication for a vector space. Indeed, a two-dimensional vector space is precisely obtained in this way by settingR=Q,R,C, etc.
Let us define an abstract notion. Let M be an abelian group with a bilinear operation · such that
∀r∈R, ∀m∈M : r·m∈M. (A.4)
The “multiplication”·here is not the same thing as the multiplication within the ring. We assume expected formulas (rs)·m=r·(s·m), and simply write (rs)·m=rsmwherer, s∈Randm∈M. We call M an R-module or a module over R. It is general than the notion of vector spaces.
A vector space is a module over a field, a commutative ring in which multiplication by a nonzero element has an inverse. An ideal is a subset of R such that it is a module over R.2 Note that R itself is anR-module via the multiplication within R. The above example R⊕R is anR-module.
For anR-moduleM if there exist finitely many elements m1, . . . , mn∈M such that
M ={r1m1+· · ·+rnmn |r1, . . . , rn ∈R}, (A.5)
2 Although an ideal is a valid module, often it is treated differently than a module. One should sometimes be careful to apply things defined for modules, especially when one reads the dimension theory of Eisenbud [76].
then we sayM is finitely generated. All of our modules in the thesis are finitely generated. An ideal is finitely generated if it is finitely generated as a module.
There is a confusingly similar terminology that one must distinguish. Suppose A is a subring of B. That is, A is a ring by itself and contained in a bigger ring B. Or slightly more generally, suppose we are given a ring mapA→B. The subring case is precisely when the map is an inclusion.
We sayB is afinitely generatedA-algebra if there exists finitely many elementsb1, . . . , bn ∈B such that any element ofbcan be written as a polynomial inb1, . . . , bn with coefficients in the image ofA. In this case, B is sometimes written asB =A[b1, . . . , bn]. A typical situation is when Ais a field such asQ,CandB is a polynomial ring over A. For example,B=Q[x, y]. The reason it is a confusing terminology is becauseB is not necessarily finitely generatedA-module. Q[x], a finitely generated Q-algebra, has infinitely many generators {1, x, x2, . . .} as a Q-module. Is Q a finitely generatedZ-algebra? No, because multiplying by integers cannot produce large denominators.
As the rings are understood in relation to others, the modules should be understood via maps.
We required the ring homomorphisms to preserve the defining operations of the rings. The same is true for the module maps. Given two modules M and N over R, we define anR-linear map or R-module homomorphismf :M →N to satisfy
f(m+m0) =f(m) +f(m0) and f(r·m) =r·f(m) for anyr∈R, m, m0∈M. (A.6) Note that the · on the left-hand side is the operation (A.4) for M whereas that on the right-hand side is the operation for N.3 The notions of kernel, image, endomorphism, automorphism, and isomorphism apply to module maps, too. A simple exercise: LetA, B be twoR-algebras; there are ring maps R →A and R →B. The two algebras are naturally R-modules, when a map A → B becomesR-linear? Answer: It becomesR-linear when the following diagram commutes.
A //B
R
__???? ??
Free
An R-module isomorphic to L
αR is called a free module. When there are finitely many sum- mands, it is called finitely generated free module. It is the most convenient type of modules.
In particular, a module map between finitely generated freeR-modules is simply given by a matrix with entries inR, just as linear map between finite dimensional vector space can be described by a matrix with entries in a field. Note that elements of a finitely generated freeR-moduleM =R⊕ncan
3 In group representation theory, the R-linear maps are called equivariant maps, whereRis the group algebra which may not be commutative. The representation space is a module, the subrepresentation space is a submodule, and the irreducible representation space is a simple submodule.
be expressed by column matrices. Letei(i= 1, . . . , n) denote the canonical basis column matrices.
(We could say “column vectors” instead of column matrices. However, an element of a module is not a vector in general.) A mapf fromM to any module is specified if we specify the imagef(ei) because the image of other elementsa1e1+· · ·anenis must bea1f(e1) +· · ·+anf(en) byR-linearity.
Writingf(ei) in columns, we have a matrix representation off. An un-redundant set of generators of a free module is called abasis. The cardinality of a basis is calledrank. (One can show that rank is independent of the choice of a basis.) Only for free modules can we speak of bases. The crucial difference between general modules and vector spaces is that a module is in general not free, while a vector space, a module over a field, is always free. That nonzero elements are invertible makes such a huge difference.
Note that any module can be described by free modules. Take a generating set {mα} of a module M; any element of M is a finite R-linear combination P
irαimαi. A trivial and useless choice would be to take wholeM as a generating set. LetF be a freeR-module whose rank is the same as the cardinality of the generating set, i.e., there is a surjective module mapF →M. The kernelN is a submodule ofF, not necessarily free, andM ∼=F/N. One can carry the same process for the module N. That is, one finds a free module F0 such thatφ:F0 →N is a surjection. Now M is expressed as M ∼= F/imφ. This is conceptually important observation, but not too useful because we do not have any control over the ranks ofF andF0. We need some finiteness conditions.
AnR-moduleM is said to be finitely presented if there is a map φ :F0 → F between finitely generated free modules such that M ∼=F/imφ. The latter expression F/imφis often abbreviated as cokerφ. The mapφis called afinite presentationofM. As we have seen above,φis a matrix with entries inR, andM is expressed by a single matrix φ. In case of a finite dimensional vector space,φcan always be brought to a diagonal matrix with entries of 0 or 1, after basis change ofF andF0. We will discuss more on a finite presentation in SectionA.5.
Noetherian
A technically very important and convenient adjective is Noetherian. It is as important as vector spaces having finite dimensions. A module M is Noetherian if every increasing sequence of submodules is stationary, i.e., if M1 ≤M2 ≤ · · · ≤ Mn ≤ · · · is a sequence of submodules of M, then for all sufficiently large n one has Mn = Mn+1. Often this condition is referred to as the ascending chain conditionora.c.c. A moduleM is Noetherian if and only if any submodule is finitely generated. If a submodule cannot be generated by finitely many elements, one can construct a strictly increasing infinite sequence of submodules using an infinite subset of generators. Conversely, if any submodule of M is finitely generated, then the union ∪∞i=1Mi of any increasing sequence of submodules Mi of M is also finitely generated, say, by m1, . . . , mr. Sincemj is contained in some Mj0, there must be somek such thatm1, . . . , mr∈Mk. It follows thatMk =Mk0 for allk0 ≥k. A
Noetherian module over a field is just a finite dimensional vector space. A Noetherian ringR is a ring that is Noetherian as anR-module, i.e., the a.c.c is satisfied with respect to the ideals ofR.
In a Noetherian ring, any ideal is finitely generated.
Being Noetherian is preserved in many cases. Loosely speaking, it says that a module carries a finite amount of data. A module with a finite amount data manipulated finitely many times would still have a finite amount of data. The following are facts:
• A homomorphic image of Noetherian ring (module) is a Noetherian ring (module).
• A submodule of Noetherian module is Noetherian.
• A finitely generated algebra over a Noetherian ring is a Noetherian ring.
• In particular,a polynomial ring over a field with finitely many variables is a Noetherian ring.
• A finitely generated module over a Noetherian ring is Noetherian.
There are more to mention about being Noetherian using tensor products and localization. See SectionA.4.
Remark that a finitely generated module M over a Noetherian ring R always admits a finite presentation φ : F0 → F. Since M is finitely generated, the module F is of finite rank. F is Noetherian, and therefore, imφ is finitely generated, which means F0 can be taken to be of finite rank. The “matrix”φis of finite size.
Is the ring of all differentiable functions R→RNoetherian? No. Is the ring of all trigonometric functionsR→Rof period 1/n, wherenare positive integers, Noetherian? Yes.