Theorem: Zero of Inverse Completion of Integral Domain

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 {{iff}}:
:$a \circ d = b \circ c$
This is consistent with the fact that two elements $\eqclass {\tuple {a, b} } \ominus, \eqclass {\tuple {c, d} } \ominus$ of $K$ are equal {{iff}} $a \circ d = b \circ c$.
Suppose $a = 0_D$ and $a \circ d = b \circ c$.
{{begin-eqn}}
{{eqn | l = a
      | r = 0_D
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = 0_D \circ d
      | r = b \circ c
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = b \circ c
      | r = 0_D
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = c
      | r = 0_D
      | c = as $b \in D^*$, so $b \ne 0$
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus
      | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {a, b} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {0_D \circ a, k \circ b} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {0_D, k \circ b} } \ominus
      | c = {{Defof|Ring Zero}}
}}
{{eqn | r = \eqclass {0_D} \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {a \circ 0_D, b \circ k} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {0_D} \ominus
      | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {0_D \circ 0_D, k \circ k} } \ominus
      | c = 
}}
{{eqn | r = \eqclass {\tuple {0_D, k \circ k} } \ominus
      | c = {{Defof|Ring Zero}}
}}
{{eqn | r = \eqclass {0_D} \ominus
      | c = 
}}
{{end-eqn}}
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.
{{qed}}
Category:Integral Domains
Category:Inverse Completions
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