Commutative Ring Theory/Printable version

Commutative Ring Theory

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Divisibility and principal ideals

Definition (principal ideal):

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

Proposition (characterisation of divisibility by principal ideals):

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

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

Definition (similarity):

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

Proposition (similarity is an equivalence relation):

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

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

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

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

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

Greatest common divisors

Definition (divisor):

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

Definition (greatest common divisor):

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

Definition (coprime):

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

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 $R$ be a Bézout domain. Then $R$ is a GCD domain.

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

Principal ideal domains

Definition (principal ideal domain):

A principal ideal domain is an integral domain $R$ 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 $R$ be a Bézout domain. Then the following are equivalent:

1. $R$ is a principal ideal domain
2. $R$ is noetherian
3. The principal ideals of $R$ satisfy the ascending chain condition

(On the condition of the axiom of dependent choice.)

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

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

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

Derivations

Proposition (alternative construction of the universal derivation):

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

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

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

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

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

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

${\displaystyle }$

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

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

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

$(\alpha ,\beta )\mapsto \alpha d\beta$ $\Box$