## Commutative Algebra

These notes are based on a lecture on Commutative Algebra at the University of Luxembourg. A variety of sources including Eisenbud, the Princeton Companion to Mathematics and a Classic ‘Algebra’ text by MacLane and Birkhoff have been used to compile these notes. Hopefully it will be of benefit to those studying this course.

1. Lecture Notes:Axioms for Rings

A ring R = (R,+,${\cdotp}$,1) is a set R with two binary operations, addition and multiplication, and a nullary operation, ‘select 1’, such that

• (i) (R,+) is an abelian group under addition;
• (ii) (R,${\cdotp}$,1) is a monoid under multiplication;
• (iii) Multiplication is distributive (on both sides) over addition.

The last requirement means that all triples of elements a,b,c in R satisfy the identities

$\displaystyle a(b+c)=ab+bc,\;\;\;\;\;\;(a+b)c=ac+bc \ \ \ \ \ (1)$

A commutative rign is one in which multiplication is commutative. Familiar systems of numbers are commutative rings under the usual operations of sum and product; examples are

$\displaystyle \begin{array}{rcl} \mathbb{Z},\;\;\;\; the\; ring\; of\; all\; integers, \\ \mathbb{Q}, \;\;\;\; the\; ring\; of\; all \;rational\; numbers, \\ \mathbb{R}, \;\;\;\; \;the\; ring\; of\; all\; real\; numbers, \\ \mathbb{C}, \;\;\;\;\; the\; ring\; of\; all\; complex\; numbers.\; \end{array}$

Lets consider a rather trivial example of a ring, the set containing just the element 0, with addition and multiplicatio given (in the only possible way!) by 0 + 0, 00=0, is a ringl it will be called the ‘trivial ring’. \paragraph*{}The definition of a ring amounts to the statement that a ring is a set R with a selected element ${I\in R}$ and two binary operations (a,b)${\mapsto}$ a + b and (a,b) ${\mapsto}$ ab which are both associative, so that

$\displaystyle a + (b + c) = (a + b) + c,\;\;\; a(bc)=(ab)c, \ \ \ \ \ (2)$

for all ${a, b, c \in R,}$ which have addition commutative, so that

$\displaystyle a + b = b + a \;\;\; \forall a,b \in R, \ \ \ \ \ (3)$

which containts the unit 1 and a zero0 such that

$\displaystyle a + 0 = a, a1=a=1a, \forall a \in R, \ \ \ \ \ (4)$

which cotaints to each element a an additive inverse(-a) with

$\displaystyle a + (-a) = 0, \ \ \ \ \ (5)$

and in which both distributive laws (1) hold. As an example of a non-commutative ring, we can consider the ring of square matrices. Every field – like for instance the real numbers and the complex numbers is an example of a commutative ring. ${\mathbb{C}}$ is algebraically closed every non-constant polynomial with coefficients in ${\mathbb{C}}$ has (at least one root in ${\mathbb{C}}$. This is the famous Gaussian result: the fundamental theorem of algebra.

1.1. Residue Class Rings

: \left(\frac{\mathbb{Z}}{n\mathbb{Z}},+,\cdotp \right)\;\;\;n \in \mathbb{N}, n > 1 p \in \mathbb{P}\;\;prime, then (\frac{\mathbb{Z}}{p\mathbb{Z}},+,\cdotp) = \mathbb{F}_{p}\;\;is\;a\;field\;(finite) Subrings ${(R,+\cdotp)}$ \underline{Subring} ${S\subseteq R}$ subset such that ${+\arrowvert_{S}}$, ${\cdotp\arrowvert_{S}}$

$\displaystyle \begin{array}{rcl} S_{1} + S_{2}\; =\; S_{1}\; + \;S_{2}\; in\; R \\ S_{1},S_{2}\; \in\; S \\ I\; take\; two\; elements\; from\; the\; subsets\; of\; the \;ring.\; \\ s\;\in\;S \rightarrow -s\;\in\;S \\ 0\;\in\;S\;\rightarrow\;\;0\;\in\;S \\ s_{1},s_{2}\;\in\;S\;\rightarrow\;s_{1}\cdotp s_{2}\;\in S \\(1\in S) \\ (\mathbb{Z},+,\cdotp)\; is\; a\; subring\; of\; (\mathbb{Q},+,\cdotp) \end{array}$

As with other algebraic structures, there are several ways of forming new rings from old ones: for instance we can take subrings and direct products of two rings. Slightly less obviously, we can start with a ring R and form the ring of all polynomials with coefficients in R. We can also take QUOTIENTS, but in order to discuss these we must introduce the notion of an ideal.

1.2. Ideals

A typical quotient construction for an algebraic structure A will identify some substructure B and regard two elements of A as ‘equivalent’ if they ‘differ by an element of B.’ If A is a group or a vector space then B will be a subgroup or a subspace. However the situation with rings is slightly different. To see why we need to think of quotients in another way. Lets consider them as images of homomorphisms. The substructures we would like to quotient by are the kernels of these homomorphisms, so we should ask ourselves what the kernel of a ring homomorphism (that is, the set of elements that map to 0) will be like. \paragraph*{} If ${\phi}$:R${\rightarrow}$ R’ is a homomorphism between two rings, and ${\phi(a)}$=${\phi(b)}$ = 0, then ${\phi(a\; +\; b)\;=\;0}$, then ${\phi(a\;+\;b)\;=\;0.}$ Also, if r is any element of R, then \phi(ra)=\phi(a)\phi(b)\;=\;0. Thus, the kernel of a homomorphism is closed under addition, and also under multiplication by any element of the ring. THese two properties define the notion of an ideal. For example, the set of all even integers is an ideal in ${\mathbb{Z}}$. In interesting cases, ideals are not subrings, since if an ideal contains 1 then it must contain r for every r in the ring. (An example which makes the difference very clear is the subset of the ring of all polynomials that consists of all constant polynomials. The constants form a subring, by they certainly do not form an ideal.) It isn’t hard to show that for any ideal I in a ring R there is a homomorphism that has I as its kernel, namely the quotient map from R to the quotient R/I. Here R/I is a construction that as usual we think of as ”R, but with two elements considered the same if they differ by an element of I.” \paragraph*{} Quotients of rings are extremely useful in algebraic number theory because they allow us to rephrase questions about algebraic numbers as questions about polynomials. To get an idea of how this is done, consider the ring ${\mathbb{Z}[X]}$ of all polynomials with integer coefficients, and the ideal that consists of all multiples (by integer polynomials) of the polynomial ${X^{2}+1}$ In the quotient ${\mathbb{Z}[X]}$ by this ideal, we reard two polynomials as the same if they differ by a multiple of ${X^{2}+1}$. In particular, ${X^{2}}$ is the same as -1. In other words, in this quotient ring we have a square root of -1, and in fact the quotient ring is isomorphic to the ring ${\mathbb{Z}[i]}$ that we have met earlier. \paragraph*{} One of the things we like to do to integersis factorize them, and we can try to do the same in rings as well. However, it turns out that, while it is usually possible to factorize an element of a ring into ”irreducible” ones that cannot be factorized further (like for instance the primes in ${\mathbb{Z}}$ ), in many cases the factorization is not unique. At first, this might be rather unexpected, and indeed in the 18th and 19th centuries this was a stumbling block. Here is an example: in the ring ${\mathbb{Z}[\sqrt{-3}]}$, which consists of all complex numbers a + b${\sqrt{-3}]}$, where a and b are integers, the number 4 may be factorized as 2 x 2 and also as (1+${\sqrt{-3}]}$) ${\times}$ (1+${\sqrt{-3}]}$).

1.3. Modules

Modules are to rings as vector spaces are to fields. In other words, they are algebraic structures where the basic operations are addition and scalar multiplication, but now the scalars are allowed to come from a ring rather than a field. For an example of a module over a ring that is not a field, take any Abelian group G. This can be turned into a module over ${\mathbb{Z},}$ with addition given by the group operation and scalar multiplication defined in the obvious way: for instance 3g means g+g+g, and -2g means the inverse of g + g. \paragraph*{} The simplicity of this definition masks the fact that the structure of modules is in general far more subtle than that of vector spaces. For example we can describe a basis of a module to be a linearly independent set of elements that span the module. However, many useful facts about bases in vector spaces do not hold for modules. For instance, in ${\mathbb{Z},}$ which we may consider as a module over itself, the set {2,3} spans the module but does not contain a basis, and similarly the set {2} is linearly independent but cannot be extended to a basis. In fact, modules may be very far from having a basis: for example, if we consider the integers modulo n as a module over ${\mathbb{Z}}$, then even a single element x fails to be linearly independent, since nx = 0. \paragraph*{} The following example of a module is an important one. Let V be a complex vector space and let ${\alpha}$ be a linear map from V to V. This can be made into a module over the ring ${\mathbb{C}[X]:}$ if ${v \in V}$ and P is a complex polynomial, then Pv is defined to be P${(\alpha)}$v. (For instance, if P is the polynomial ${x^{2}+1,}$ then Pv = ${\alpha^{2}v\;+\;v.}$) Applying general structural results about modules to this example one can then go onto prove the Jordan Normal Form Theorem.

2. Varieties

In mathematics, an algebraic variety is the set of solutions of a system of polynomial equations. Algebraic varieties are one of the central objects of study in classical (and to some extent, modern) algebraic geometry. The word “variety” is employed in the sense of a mathematical manifold, for which, in Romance languages, cognates of the word “variety” are used. Historically, the fundamental theorem of algebra established a link between algebra and geometry by saying that a monic polynomial in one variable over the complex numbers is determined by the set of its roots, which can be considered a geometric object. Building on this result, Hilbert’s Nullstellensatz provides a fundamental correspondence between ideals of polynomial rings and subsets of affine space. Using the Nullstellensatz and related results, we are able to capture the geometric notion of a variety in algebraic terms as well as bring geometry to bear on questions of ring theory. Two simple examples of varieties are the circle and the parabola, which can be defined by the polynomial ${x^{2}+y^{2}=1}$ and ${y=x^{2}}$, respectively. With one qualification, a variety is the solution set of a system of polynomial equations. The solution set to a system of polynomial equations is called a algebraic set, and it is called a variety if it cannoy be written as a union of smaller algebraic sets. However the concept is much more general: varieties can live in ${\mathbb{R}^{n}}$, ${\mathbb{C}^{n}}$ for any n. Indeed, the definitions make sense, and are interesting and important in ${\mathbb{F}^{n}}$, where ${\mathbb{F}}$ can be any field.

Affine Varieties

3. Irreducibles and factorization

In this section, we examine the posibility of factorising elements of a ring into ‘irreducible’ elements (which cannot themselves be further factorized), and look at a special class of rings in which the analogue of the Fundamental Theorem of Arithmetic holds. \paragraph{}* First, we will make some simplifying assumptions about the ring R. We always assume that R is commutative, so that we regard ab and ba as essentially the same factorization of a ring element. (So, in a factorization, we do not care about the order of the factors.) Also, we exclude the divisors of zero. For, if ab = 0, then ac=a(b+c) for any element c, and there is little chance of unique factorizations. Accordingly, we assume, in this section and the next two, that R is an integral domain.

Definition 1 Let R be an integral domain

1. An element p ${\in}$ R is irreducibleif p is not zero or a unit, and if, whenever p =ab, either a or b is a unit (and the other is an associate of p).
2. R is a unique factorization domain or UFD if it holds that (a) every element other than zero and units can be factorized into irreducibles; (b) if ${p_{1}...p_{m} = q_{1}...q_{n}}$, where the ${p_{i}}$ and ${q_{j}}$ are irreducibles, then m = n, and (possibly after reordering ${p_{i}}$ and ${q_{j}}$ are associates for i = 1,…,n.

Alternatively we can say that the ‘Fundamental Theorem of Arithmetic’ says that ${\mathbb{Z}}$ is a UFD.

3.1. Zero-Divisors,Nilpotent Elements, Units

A zero-divisor in a ring A is an element x which “divides 0”, i.e., for which there exists y ${\neq}$ 0 in `4 such that xy = 0. A ring with no zero-divisors ${\neq}$ 0 (and in which 1 ${\neq}$0) is called an integral domain. For example, ${\mathbb{Z}}$ and ${\mathbb{K}[x_{1} ..... ,x_{n}]}$ (${\mathbb{K}}$ a field, ${x_{i}}$ indeterminates) are integral domains. An element x ${\in}$,A is nilpotent if ${x^{n} = 0}$ for some n > 0. A nilpotent element is a zero-divisor (unless A = 0), but not conversely (in general). A unit in A is an element x which “divides 1”, i.e., an element x such that xy = 1 for some y ${\in}$ A. The element y is then uniquely determined by x, and is written ${x^{-1}}$. The units in A form a (multiplicative) abelian group.

4. The Nullstellensatz

\cite{Atiyah68,Eisenbud} This is one of the first fundamental theorems of algebraic geometry. It controls the correspondence between affine algebraic sets and ideals; in particular, it enables us to calculate I(V(I)). We have already seen that certain problems arise when k is not algebraically closed. We therefore assume henceforth that:

k is algebraically closed

Definition 2 In commutative ring theory, a branch of mathematics, the radical of an ideal I is an ideal such that an element x is in the radical if some power of x is in I. A radical ideal is an ideal that is its own radical (this can be phrased as being a fixed point of an operation on ideals called ‘radicalization’).

Intuitively the Nullstellensatz allows one to create an algebra-geometry dictionary.

\newpage \appendix

5. Summary of useful results from algebra

5.1. Rings

We denote by (x) or xA the idea generated by x in A, i.e. the set of all elements of the form xa with ${a \in A}$. \subsubsection{Rings}

• a.The isomorphism theorem.Let ${f:A \rightarrow B}$ be a ring homomorphism and set I = Ker f. Let J be an ideal of J contained in I and let ${p:A\rightarrow A/J}$ be the canonical projection. Then:
• 1) There is a unique homomorphism ${\bar{f}:A/J\rightarrow B}$ such that f=${\bar{f}p}$ (we say that f factors through A/J),
• 2)${\bar{f}}$ is injective if and only if J = I,
• 3) ${\bar{f}}$ is surjective if and only if f is. In particular, Im f ${\simeq}$ A/Ker f.
• b. Universal property of rings of polynomials. Let A and B be two rings. Giving a homomorphism f: ${A[X_{1},...,X_{n}] \rightarrow B}$ is equivalent to giving its restriction to A (i.e., a homomorphism from A to B) and the images of the variables ${X_{i}}$ (i.e.,n elements in B)
• c. Euclidean division.Let A be a ring and consider a polynomial P ${\in}$ A[X], ${P \neq 0,}$ whose leading coefficient is a unit. For all ${F\; \in\;A[X]}$ there are Q,R ${\in\;A[X]}$ such that F = PQ + R and ${deg\;R\;<\; deg\;P}$ or R = 0. For example, in k[X,Y] we can divide by ${Y^{2}-X^{3}}$ with respect to Y, but not by XY – 1.
• d.Products of rings. The direct product of two rings A an B is the product set ${A\;\times\;B}$ (i.e. , pairs(a,b) with product laws: (a,b) + (a’,b’) = (a+a’,b+b’),(a,b)(a’,b’)=(aa’,bb’)).

5.2. Ideals

• Operations on ideals. An arbitrary intersection of ideals is an ideal. The sum of a family ${I_{k}}$ of ideals of A is the set of finite sums ${\Sigma x_{k}}$ with ${x_{k}\;\in\;I_{k}}$. This is an ideal which is the upper bound of the ideals ${I_{k}}$ for inclusion. We denote it by ${\Sigma I_{k}}$. In particular, if ${I_{k}}$ = ${(f_{k})}$, then we obtain the ideal generated by the elements ${f_{k}.}$ In ${\mathbb{Z}}$ the sum of the ideals generated by (x) and (y) is the ideal generated by the gcd of x and y.
The product of two ideals I and J is the ideal denoted by IJ generatedby products xy with ${x\;\in\;I}$ ${y\;\in\;J}$. We have IJ ${\subset}$ I ${\cap}$ J, but the converse is false: in ${\mathbb{Z}}$ the product ideal (resp. intersection) of (x) and (y) is the ideal generated by xy (resp. by the lcm of x and y).
• b. Prime ideals. An integral domain is a ring ${A \neq {0}}$ such that ${\forall a, b\;\in\;A,}$ ${ab = 0 \Rightarrow\;a\;=0\;}$ \paragraph{} For example, a field is an integral domain, a subring of an integral domain is an integral domain, and the ring of polynomials over an integral domain is an integral domain. \paragraph{} An ideal p of A is said to be prime if A/p is an integral domain. We note that inverse image of a prime ideal under a homomorphism is a prime ideal. \paragraph{} An ideal m is said to be maximal if it is maximal for inclusion amongst the ideals of A different from A. Equivalently, A/m is a field, called the residue fieldof m. It follows that any maximal ideal is prime. It can be proved using Zorn’s lemma that any ideal is contained in a maximal ideal.
• Ideals of a quotientLet A be a ring, I an ideal and p the canonical projection from A to A/I. The ideals of A/I are in (increasing) bijection with the ideals of A containing I via the maps ${p}$ and ${p^{-1}}$ Moreover, the prime ideals (resp. maximal ideals) correspond under this bijection.
• Nilpotent elements. An element ${a\;\in\;A}$ is said to be nilpotent if there is an integer ${n\;>\;0}$ such that ${a^{n}\;=\;0.}$ The set of nilpotent elements form an ideal called the nilradical of A. This ideal is the intersection of all the prime ideals of A. A ring without non-zero nilpotent elements is said to be reduced

5.3. Kernel: Ring Homomorphisms

Let R and S be unital rings and let f be a ring homomorphism from R to S. If ${0_{S}}$ is the zero element of S, then the kernel of f is the preimage of the singleton set ${{0_{S}}}$; that is, the subset of R consisting of all those elements of R that are mapped by f to the element ${0_{S}}$. ker f:=\; {r\;in\;R:\;f(r)\;=\;0_{S}} Generally the Kernel of R is an ideal of R.

Let us study the affine line ${\mathbb{K}}$. Here ${R_{1}=\mathbb{K}[X].}$ All ideals in ${\mathbb{K}[X]}$ are principal ideals, i.e. generated by just one polynomial. The vanishing set of an ideal consists just of the finitely many zeros of this polynomial (if it is not identically zero). Conversely, for every set of finitely many points there is a polynomial vanishing exactly at these points. Hence, besides the empy set and the whole line the algebraic sets are the sets of finitely many points. At this level there is already a new concept showing up. The polynomial assigned to a certain point set is not unique. For example it is possible to increase the vanishing order of the polynomial at a certain zero without changing the vanishing set. It would be better to talk about point sets with multiplicities to get a closer correspondence to the polynomials. Additionally, if ${\mathbb{K}}$ is not algebraically closed then there are nontrivial polynomials without any zeros at all.

Definition 3 (a) Let V be a closed set. V is called irreducible if for every decomposition V = ${V_{1}\cup V_{2}}$ with ${V_{1}}$,${V_{2}}$ closed we have ${V_{1} = V}$ or ${V_{2} = V}$. (b) An algebraic set which is irreducible is called a variety.