Inverse Completion Less Zero of Integral Domain is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({D, +, \circ}\right)$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.

Let $\left({K, \circ}\right)$ be the inverse completion of $\left({D, \circ}\right)$.


Then $\left({K^*, \circ}\right)$ is closed, where $K^* = K \setminus \left\{{0_K}\right\}$.


Proof

Let $\left({K, \circ}\right)$ be the inverse completion of $\left({D, \circ}\right)$.

We define $\left({K, \circ}\right)$ of $\left({D, \circ}\right)$ by Inverse Completion of Integral Domain Exists.

The structure of $\left({K, \circ}\right)$ is such that element of $\left({K, \circ}\right)$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.

From Zero of Inverse Completion of Integral Domain, $0_K$ is all elements of $D \times D^*$ of the form $\dfrac {0_D} x$.

Therefore the elements of $K^*$ are those of the form $\dfrac x \circ y^{-1}$ where $x, y \in D^*$.

By Product of Division Products, $\displaystyle \frac a b \circ \frac c d = \frac {a \circ c} {b \circ d}$.

As $\left({D, +, \circ}\right)$ is an integral domain, none of its non-zero elements are zero divisors.

Therefore $\forall x, y \in D^*: x \circ y \ne 0_D$.

So:

$\displaystyle \forall \frac a b, \frac c d \in K^*: \frac {a \circ c} {b \circ d} \in K^*$

Hence the result.

$\blacksquare$