7.4: D- Basic Terminology and Results from Algebra

Let us quickly review some basic definitions and a result or two from commutative that we need in chapter 6. See a book such as Zariski–Samuel [ZS] for a full reference.

Definition: Group

A set \(G\) is called a group if it has an operation \(x * y\) defined on it and it satisfies the following axioms:

(G1) If \(x \in G\) and \(y \in G\), then \(x * y \in G\).

(G2) (associativity) \((x*y)*z = x*(y*z)\) for all \(x,y,z \in G\).

(G3) (identity) There exists an element \(1 \in G\) such that \(1*x = x\) for all \(x \in G\).

(G4) (inverse) For every element \(x\in G\) there exists an element \(x^{-1} \in G\) such that \(x * x^{-1} = 0\).
A group \(G\) is called abelian if it also satisfies:

(G5) (commutativity) \(x*y = y*x\) for all \(x,y \in G\).

A subset \(K \subset G\) is called a subgroup if \(K\) is a group with the same operation as the group \(G\). If \(G\) and \(H\) are groups, a function \(f \colon G \to H\) is a group homomorphism if it respects the group law, that is, \(f(a * b) = f(a) * f(b)\). If \(f\) is bijective, then it is a group isomorphism.

An example of a group is a group of automorphisms. For example, let \(U \subset \mathbb{C}\) be open and suppose \(G\) is the set of biholomorphisms \(f \colon U \to U\). Then \(G\) is a group under composition, but \(G\) is not necessarily abelian: If \(U=\mathbb{C}\), then \(f(z)= z+1\) and \(g(z)=-z\) are members of \(G\), but \(f\circ g (z) = -z+1\) and \(g \circ f(z) = -z-1\).

Definition: Commutative Ring

A set \(R\) is called a commutative ring if it has two operations defined on it, addition \(x+y\) and multiplication \(xy\), and if it satisfies the following axioms:

(A1) If \(x \in R\) and \(y \in R\), then \(x+y \in R\).

(A2) (commutativity of addition) \(x+y = y+x\) for all \(x,y \in R\).

(A3) (associativity of addition) \((x+y)+z = x+(y+z)\) for all \(x,y,z \in R\).

(A4) There exists an element \(0 \in R\) such that \(0+x = x\) for all \(x \in R\).

(A5) For every element \(x\in R\) there exists an element \(-x \in R\) such that \(x + (-x) = 0\).

(M1) If \(x \in R\) and \(y \in R\), then \(xy \in R\).

(M2) (commutativity of multiplication) \(xy = yx\) for all \(x,y \in R\).

(M3) (associativity of multiplication) \((xy)z = x(yz)\) for all \(x,y,z \in R\).

(M4) There exists an element \(1 \in R\) (and \(1 \not= 0\)) such that \(1x = x\) for all \(x \in R\).

(D) (distributive law) \(x(y+z) = xy+xz\) for all \(x,y,z \in R\).

The ring \(R\) is called a field if furthermore:

(F) For every \(x\in R\) such that \(x \not= 0\) there exists an element \(\frac{1}{x} \in R\) such that \(x(\frac{1}{x}) = 1\).

In a commutative ring \(R\), the elements \(u \in R\) for which there exists an inverse \(\frac{1}{u}\) as above are called units.

If \(R\) and \(S\) are rings, a function \(f \colon R \to S\) is a ring homomorphism if it respects the ring operations, that is, \(f(a + b) = f(a) + f(b)\) and \(f(ab) = f(a)f(b)\), and such that \(f(1) = 1\). If \(f\) is bijective, then it is called a ring isomorphism.

Namely, a commutative ring is an abelian additive group (by additive group we just mean we use \(+\) for the operation and \(0\) for the respective identity), with multiplication thrown in. If the multiplication also defines a group on the set of nonzero elements, then the ring is a field. A ring that is not commutative is one that does not satisfy commutativity of multiplication. Some authors define ring without asking for the existence of \(1\).

A ring that often comes up in this book is the ring of holomorphic functions. Let \(\mathcal{O}(U)\) be the set of holomorphic functions defined on an open set \(U\). Pointwise addition and multiplication give a ring structure on \(\mathcal{O}(U)\). The set of units is the set of functions that never vanish in \(U\). The set of units is a multiplicative group.

Given a commutative ring \(R\), let \(R[x]\) be the set of polynomials \[P(x) = c_k x^k + c_{k-1} x^{k-1} + \cdots + c_1 x + c_0 ,\] where \(c_0,\ldots,c_k \in R\). The integer \(k\) is the degree of the polynomial and \(c_k\) is the leading coefficient of \(P(x)\). If the leading coefficient is 1, then \(P\) is monic. If \(R\) is a commutative ring, then so is \(R[x]\). Similarly, we define the commutative ring \(R[x_1,\ldots,x_n]\) of polynomials in \(n\) indeterminates.

The most basic result about polynomials, Theorem B.29 the fundamental theorem of algebra, which states that every nonconstant polynomial over \(R=\mathbb{C} \) has a root, is really a theorem in one complex variable.

Definition: Ideal

Let \(R\) be a commutative ring. A subset \(I \subset R\) is an ideal if \(f \in R\) and \(g,h \in I\) implies that \(fg \in I\) and \(g+h \in I\). In short, \(I \subset R\) is an additive subgroup such that \(RI = I\).

Given a set of elements \(S \subset R\), the ideal generated by \(S\) is the intersection \(I\) of all ideals containing \(S\). If \(S = \{ f_1,\ldots,f_k \}\) is a finite set, we say \(I\) is finitely generated, and we write \(I = (f_1,\ldots,f_k)\).

A principal ideal is an ideal generated by a single element. A commutative ring where every ideal is a principal ideal is called a principal ideal domain or a PID.

A commutative ring \(R\) is Noetherian if every ideal in \(R\) is finitely generated.

It is not difficult to prove that “an ideal generated by \(S\)” really is an ideal, that is, the intersection of ideals is an ideal. If an ideal \(I\) is generated by \(f_1,\ldots,f_k\), then every \(g \in I\) can be written as \[g = c_1 f_1 + \cdots + c_k f_k,\] for some \(c_1,\ldots,c_k \in R\). Clearly the set of such elements is the smallest ideal containing \(f_1,\ldots,f_k\).

Theorem \(\PageIndex{1}\)

Hilbert Basis Theorem

If \(R\) is a Noetherian commutative ring, then \(R[x]\) is Noetherian. As the proof is rather short, we include it here.

Proof

Suppose \(R\) is Noetherian, and \(I \subset R[x]\) is an ideal. Starting with the polynomial \(f_1\) of minimal degree in \(I\), construct a (possibly finite) sequence of polynomials \(f_1,f_2,\ldots\) such that \(f_k\) is the polynomial of minimal degree from the set \(I \setminus (f_1,\ldots,f_{k-1})\). The sequence of degrees \(\deg(f_1),\deg(f_2),\ldots\) is by construction nondecreasing. Let \(c_k\) be the leading coefficient of \(f_k\).

As \(R\) is Noetherian, then there exists a finite \(k\) such that \((c_1,c_2,\ldots,c_m) \subset (c_1,c_2,\ldots,c_k)\) for all \(m\). Suppose for contradiction there exists a \(f_{k+1}\), that is, the sequence of polynomials did not end at \(k\). In particular, \((c_1,\ldots,c_{k+1}) \subset (c_1,\ldots,c_k)\) or \[c_{k+1} = a_1 c_1 + \cdots a_k c_k .\] As degree of \(f_{k+1}\) is at least the degree of \(f_1\), through \(f_k\), we can define the polynomial \[g = a_1 x^{\deg(f_{k+1})-\deg(f_1)} f_1 + a_2 x^{\deg(f_{k+1})-\deg(f_2)} f_2 + \cdots + a_k x^{\deg(f_{k+1})-\deg(f_k)} f_k .\] The polynomial \(g\) has the same degree as \(f_{k+1}\), and in fact it also has the same leading term, \(c_{k+1}\). On the other hand \(g \in (f_1,\ldots,f_{k})\) while \(f_{k+1} \notin (f_1,\ldots,f_k)\) by construction. The polynomial \(g-f_{k+1}\) is also not in \((f_1,\ldots,f_k)\), but as the leading terms canceled, \(\deg(g-f_{k+1}) < \deg(f_{k+1})\), but that is a contradiction, so \(f_{k+1}\) does not exist and \(I = (f_1,\ldots,f_k)\).

Definition: Irreducible

TAn element \(f \in R\) is irreducible if whenever \(f = gh\) for two elements \(g,h \in R\), then either \(g\) or \(h\) is a unit. A commutative ring \(R\) is a unique factorization fomain (UFD) if up to multiplication by a unit, every element has a unique factorization into irreducible elements of \(R\).

One version of a result called the Gauss lemma says that just like the property of being Noetherian, the property of being a UFD is retained when we take polynomials.

Theorem \(\PageIndex{2}\)

Gauss Lemma

If \(R\) is a commutative ring that is a UFD, then \(R[x]\) is a UFD.

Proof

The proof is not difficult, but it is perhaps beyond the scope of this book.