# Commutative Ring Theory/Printable version

Commutative Ring Theory

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# Divisibility and principal ideals

Definition (principal ideal):

Let ${\displaystyle R}$  be a commutative ring. A principal ideal is a left principal ideal of ${\displaystyle R}$ . Equivalently, it is a right principal ideal or a two-sided principal ideal of ${\displaystyle R}$ .

Proposition (characterisation of divisibility by principal ideals):

Let ${\displaystyle R}$  be a commutative ring, and let ${\displaystyle a,b\in R}$ . Then ${\displaystyle a|b\Leftrightarrow \langle a\rangle \geq \langle b\rangle }$ .

Proof: Both assertions are equivalent to the existence of a ${\displaystyle c\in R}$  such that ${\displaystyle b=ac}$ . ${\displaystyle \Box }$

Definition (similarity):

Let ${\displaystyle R}$  be a commutative ring. Two elements ${\displaystyle a,b\in R}$  are called similar if and only if there exists a unit ${\displaystyle u\in R}$  such that ${\displaystyle a=ub}$ .

Proposition (similarity is an equivalence relation):

Given a ring ${\displaystyle R}$ , the relation of similarity defines an equivalence relation on the elements of ${\displaystyle R}$ .

Proof: For reflexivity, use the identity, and for symmetry, use the inverse. Suppose that ${\displaystyle a=ub}$  and ${\displaystyle b=vc}$ , where ${\displaystyle u,v\in R^{\times }}$ . Then ${\displaystyle a=uvc}$ , where of course ${\displaystyle uv\in R^{\times }}$ . ${\displaystyle \Box }$

Proposition (in an integral domain, the generating element of a principal ideal is unique up to similarity):

Let ${\displaystyle R}$  be an integral domain, and let ${\displaystyle \langle a\rangle \leq R}$  be a principal ideal of ${\displaystyle R}$ . Then if ${\displaystyle \langle a\rangle =\langle b\rangle }$  for some element ${\displaystyle b\in R}$ , we have ${\displaystyle a=ub}$  for some ${\displaystyle u\in R^{\times }}$ .

Proof: The equation ${\displaystyle \langle a\rangle =\langle b\rangle }$  implies that ${\displaystyle a=xb}$  and ${\displaystyle b=ya}$  for certain ${\displaystyle x,y\in R}$ . Hence, ${\displaystyle a=xya}$ . By cancellation (which is applicable because ${\displaystyle R}$  is an integral domain), ${\displaystyle xy=1}$  and hence ${\displaystyle x}$  is a unit, so that ${\displaystyle a}$  and ${\displaystyle b}$  are similar. ${\displaystyle \Box }$

# Greatest common divisors

Definition (divisor):

Let ${\displaystyle R}$  be a ring, and let ${\displaystyle a\in R}$ . A divisor of ${\displaystyle a}$  is an element ${\displaystyle b\in R}$  such that there exists ${\displaystyle c\in R}$  such that ${\displaystyle a=bc}$ . The notation ${\displaystyle b|a}$  indicates that ${\displaystyle b}$  is a divisor of ${\displaystyle a}$

Definition (greatest common divisor):

Let ${\displaystyle R}$  be a commutative ring, and let ${\displaystyle a_{1},\ldots ,a_{n}\in R}$ . A greatest common divisor is an element ${\displaystyle d\in R}$  such that ${\displaystyle d|a_{j}}$  for all ${\displaystyle j\in \{1,\ldots ,n\}}$ , and such that for any other element ${\displaystyle c\in R}$  such that ${\displaystyle c|a_{j}}$  for all ${\displaystyle j\in \{1,\ldots ,n\}}$ , we have ${\displaystyle c|d}$ .

Definition (coprime):

Let ${\displaystyle R}$  be a commutative ring, and let ${\displaystyle a_{1},\ldots ,a_{n}\in R}$ . These elements ${\displaystyle a_{1},\ldots ,a_{n}}$  are said to be coprime if and only if whenever ${\displaystyle d\in R}$  is such that ${\displaystyle d|a_{j}}$  for all ${\displaystyle j\in \{1,\ldots ,n\}}$ , then ${\displaystyle d\in R^{\times }}$ .

Proposition (a set of elements of a commutative ring divided by their greatest common divisor is coprime):

Let ${\displaystyle }$

# Bézout domains

Definition (Bézout domain):

A Bézout domain is an integral domain whose every finitely generated ideal is principal, ie. generated by a single element.

Proposition (Every Bézout domain is a GCD domain):

Let ${\displaystyle R}$  be a Bézout domain. Then ${\displaystyle R}$  is a GCD domain.

Proof: Given any two elements ${\displaystyle a,b\in R}$ , we may consider the ideal ${\displaystyle I=\langle a,b\rangle }$  generated by ${\displaystyle a}$  and ${\displaystyle b}$ . By the definition of Bézout domains, ${\displaystyle I=\langle c\rangle }$  for at least one ${\displaystyle c\in R}$  (which is moreover unique up to similarity). Then ${\displaystyle \langle c\rangle \leq \langle a\rangle }$  and ${\displaystyle \langle c\rangle \leq \langle b\rangle }$ , so that by the characterisation of divisibility by principal ideals, ${\displaystyle c}$  is a common divisor of ${\displaystyle a}$  and ${\displaystyle b}$ . Moreover, if ${\displaystyle d\in R}$  is another common divisor of ${\displaystyle a}$  and ${\displaystyle b}$ , then ${\displaystyle a,b\in \langle d\rangle }$ , so that ${\displaystyle \langle a,b\rangle =\langle c\rangle \leq \langle d\rangle }$ , so that ${\displaystyle d|c}$ . Hence, ${\displaystyle c}$  is a greatest common divisor of ${\displaystyle a}$  and ${\displaystyle b}$ . ${\displaystyle \Box }$

# Principal ideal domains

Definition (principal ideal domain):

A principal ideal domain is an integral domain ${\displaystyle R}$  whose every ideal is principal.

Proposition (a Bézout domain is principal if and only if it is Noetherian or satisfies the ascending chain condition for principal ideals):

Let ${\displaystyle R}$  be a Bézout domain. Then the following are equivalent:

1. ${\displaystyle R}$  is a principal ideal domain
2. ${\displaystyle R}$  is noetherian
3. The principal ideals of ${\displaystyle R}$  satisfy the ascending chain condition
(On the condition of the axiom of dependent choice.)

Proof: The implication "1. ${\displaystyle \Rightarrow }$  2." is obvious. Suppose that 3. holds, and let ${\displaystyle I\leq R}$  be any ideal. If ${\displaystyle I}$  was non-principal, then whenever ${\displaystyle a\in I}$ , we could find a ${\displaystyle b\in I}$  such that ${\displaystyle b\notin \langle a\rangle }$ . Hence, starting with an arbitrary ${\displaystyle a_{1}\in I}$  and invoking the axiom of dependent choice (applied to a set of finite tuples with an adequate relation) yields a sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$  in ${\displaystyle I}$  such that ${\displaystyle a_{n+1}\notin \langle \gcd(a_{1},\ldots ,a_{n})\rangle }$ ; indeed, ${\displaystyle \gcd(a_{1},\ldots ,a_{n})\in I}$  since ${\displaystyle R}$  is a Bézout domain. If we define

${\displaystyle b_{n}:=\gcd(a_{1},\ldots ,a_{n})}$ , we have ${\displaystyle \langle b_{1}\rangle \supsetneq \langle b_{2}\rangle \supsetneq \cdots }$ ;

thus, we have defined an ascending chain of principal ideals of ${\displaystyle R}$  that does not stabilize. Finally, every principal ideal domain must be noetherian, since being noetherian is equivalent to all ideals being finitely generated. ${\displaystyle \Box }$

# Derivations

Proposition (alternative construction of the universal derivation):

Let ${\displaystyle S}$  be a unital ${\displaystyle R}$ -algebra. Note that ${\displaystyle S\otimes _{R}S}$  becomes an ${\displaystyle S}$ -module via the linear extension of the operation ${\displaystyle t(s\otimes u):=(ts)\otimes u}$ . We then have a morphism of ${\displaystyle S}$ -modules

${\displaystyle \phi :S\otimes _{R}S\to S,s\otimes t\mapsto s\cdot t}$ ,

where the dot indicates the algebra multiplication of ${\displaystyle S}$ . Set ${\displaystyle I:=\ker \phi }$  and ${\displaystyle \Omega '_{S/R}:=I/I^{2}}$ . Then

${\displaystyle d':S\to \Omega '_{S/R},~~t\mapsto t\otimes 1-1\otimes t+I^{2}}$

is a derivation, and we have an isomorphism ${\displaystyle \Theta :\Omega _{S/R}\to \Omega '_{S/R}}$  inducing a commutative diagram

Proof: Note first that ${\displaystyle d'}$  is a derivation. This takes some explaining. First, note that for arbitrary ${\displaystyle \alpha ,\beta }$  the element ${\displaystyle \alpha \otimes \beta -\beta \otimes \alpha }$  is in ${\displaystyle I}$ . Moreover, from this follows that the element

${\displaystyle }$

is in ${\displaystyle I^{2}}$  for ${\displaystyle \alpha ,\beta ,\gamma ,\delta \in S}$  arbitrary.

Hence, from the universal property of ${\displaystyle d:S\to \Omega _{S/R}}$ , we obtain a unique morphism of ${\displaystyle S}$ -modules ${\displaystyle \Theta :\Omega _{S/R}\to \Omega '_{S/R}}$  that makes the diagram

commutative. We construct an inverse map to ${\displaystyle \Theta }$ . Namely, on ${\displaystyle B\times B}$  we can define the map

${\displaystyle (\alpha ,\beta )\mapsto \alpha d\beta }$  ${\displaystyle \Box }$