Zero of Inverse Completion of Integral Domain
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.
$\blacksquare$