Zero of Inverse Completion of Integral Domain

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.

Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$ as defined in Inverse Completion of Integral Domain Exists.


Let $x \in K: x = \dfrac p q$ such that $p = 0_D$.

Then $x$ is equal to the zero of $K$.


That is, any element of $K$ of the form $\dfrac {0_D} q$ acts as the zero of $K$.


Proof

Let us define $\eqclass {\tuple {a, b} } \ominus$ as in the Inverse Completion of Integral Domain Exists.

That is, $\eqclass {\tuple {a, b} } \ominus$ is an equivalence class of elements of $D \times D^*$ under the congruence relation $\ominus$.

$\ominus$ is the congruence relation defined on $D \times D^*$ by:

$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$


By Inverse Completion of Integral Domain Exists:

$\dfrac p q \equiv \eqclass {\tuple {p, q} } \ominus$

From Equality of Division Products, two elements $\dfrac a b, \dfrac c d$ of $K$ are equal if and only if:

$a \circ d = b \circ c$

This is consistent with the fact that two elements $\eqclass {\tuple {a, b} } \ominus, \eqclass {\tuple {c, d} } \ominus$ of $K$ are equal if and only if $a \circ d = b \circ c$.


Suppose $a = 0_D$ and $a \circ d = b \circ c$.

\(\ds a\) \(=\) \(\ds 0_D\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds 0_D \circ d\) \(=\) \(\ds b \circ c\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds b \circ c\) \(=\) \(\ds 0_D\) Definition of Ring Zero
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds 0_D\) as $b \in D^*$, so $b \ne 0$


Hence:

$\eqclass {\tuple {0_D, b} } \ominus = \eqclass {\tuple {0_D, d} } \ominus$

Thus all elements of $K$ of the form $\eqclass {\tuple {0_D, k} } \ominus$ are equal, for all $k \in D^*$.

To emphasise the irrelevance of the $k$, we will abuse our notation and write:

$\eqclass {\tuple {0_D, k} } \ominus$

as

$\eqclass {0_D} \ominus$


Next, by Product of Division Products, we have that $\ds \frac a b \circ \frac c d = \frac {a \circ b} {c \circ d}$.

Again abusing our notation, we will write:

$\eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {c, d} } \ominus$

to mean:

$\eqclass {\tuple {a \circ c, b \circ d} } \ominus$

So:

\(\ds \eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus\) \(=\) \(\ds \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {a, b} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {0_D \circ a, k \circ b} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {0_D, k \circ b} } \ominus\) Definition of Ring Zero
\(\ds \) \(=\) \(\ds \eqclass {0_D} \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {a \circ 0_D, b \circ k} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus\)

Hence:

$\eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus$

So $\eqclass {0_D} \ominus$ fulfils the role of a zero for $\tuple {K, \circ}$ as required.


Also we have that:

\(\ds \eqclass {0_D} \ominus \circ \eqclass {0_D} \ominus\) \(=\) \(\ds \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {0_D \circ 0_D, k \circ k} } \ominus\)
\(\ds \) \(=\) \(\ds \eqclass {\tuple {0_D, k \circ k} } \ominus\) Definition of Ring Zero
\(\ds \) \(=\) \(\ds \eqclass {0_D} \ominus\)

So $\eqclass {0_D} \ominus$ is idempotent.

It follows that $\eqclass {0_D} \ominus$ can be identified with $0_D$ from the mapping $\psi$ as defined in Construction of Inverse Completion.

$\blacksquare$