Zero of Inverse Completion of Integral Domain

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

Let $\left({K, \circ}\right)$ be the inverse completion of $\left({D, \circ}\right)$ as defined in Inverse Completion of Integral Domain.

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 $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus$ as in the Inverse Completion of Integral Domain.

That is, $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus$ 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 $\left({x_1, y_1}\right) \ominus \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$.

By the method of its construction, $\dfrac p q \equiv \left[\!\left[{\left({p, q}\right)}\right]\!\right]_\ominus$.

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 correlates with the fact that two elements $\left[\!\left[{\left({a, b}\right)}\right]\!\right], \left[\!\left[{\left({c, d}\right)}\right]\!\right]_\ominus$ of $K$ are equal iff $a \circ d = b \circ c$.

Suppose $a = 0_D$.

Hence:
 * $\left[\!\left[{\left({0_D, b}\right)}\right]\!\right] = \left[\!\left[{\left({0_D, d}\right)}\right]\!\right]_\ominus$

Thus all elements of $K$ of the form $\left[\!\left[{\left({0_D, k}\right)}\right]\!\right]_\ominus$ are equal, for all $k \in D^*$.

To emphasise the irrelevance of the $k$, we will abuse our notation and write:
 * $\left[\!\left[{\left({0_D, k}\right)}\right]\!\right]_\ominus$

as
 * $\left[\!\left[{0_D}\right]\!\right]_\ominus$

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

Again abusing our notation, we will write:
 * $\left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus \circ \left[\!\left[{\left({c, d}\right)}\right]\!\right]_\ominus$

to mean:
 * $\left[\!\left[{\left({a \circ c, b \circ d}\right)}\right]\!\right]_\ominus$

So:

Hence:
 * $\left[\!\left[{0_D}\right]\!\right]_\ominus \circ \left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus = \left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus = \left[\!\left[{\left({a, b}\right)}\right]\!\right]_\ominus \circ \left[\!\left[{0_D}\right]\!\right]_\ominus$

So $\left[\!\left[{0_D}\right]\!\right]_\ominus$ fulfils the role of a zero for $\left({K, \circ}\right)$ as required.

Also we have that:

So $\left[\!\left[{0_D}\right]\!\right]_\ominus$ is idempotent.

It follows that $\left[\!\left[{0_D}\right]\!\right]_\ominus$ can be identified with $0_D$ from the mapping $\psi$ as defined in Construction of Inverse Completion.