Lovely little lemmas

Or maybe ‘amusing observations’

Cardinality of fraction field

Posted on 24 June 2017 by Remy

This is a ridiculous lemma that I came up with.

Lemma. Let $R$ be a (commutative) ring, and let $K$ be its total ring of fractions. Then $R$ and $K$ have the same cardinality.

Proof. If $R$ is finite, my previous post shows that $R = K$ . If $R$ is infinite, then $K$ is a subquotient of $R^2$ , hence $|K| \leq |R^2| = |R|$ . But $R$ injects into $K$ , so $|R| \leq |K|$ . $\qedsymbol$

Corollary. If $R$ is a domain, then $|\operatorname{Frac}(R)| = |R|$ .

Proof. This is a special case of the lemma. $\qedsymbol$

Finite domains are fields

Posted on 23 June 2017 by Remy

This is one of the classics.

Lemma. Let $R$ be a finite commutative ring. Then every element is either a unit or a zero-divisor.

Proof. If $x \in R$ is not a zero-divisor, then the map $x \colon R \to R$ is injective. Since $R$ is finite, it is also surjective, so there exists $y \in R$ with $xy = 1$ . $\qedsymbol$

Corollary 1. Let $R$ be a finite commutative ring. Then $R$ is its own total ring of fractions.

Proof. The total ring of fractions is the ring $R[S^{-1}]$ , where $S$ is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change $R$ . $\qedsymbol$

Corollary 2. Let $R$ be a finite domain. Then $R$ is a field.

Proof. In this case, the total ring of fractions is the fraction field. Therefore, $R$ is its own fraction field by Corollary 1. $\qedsymbol$