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

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

That is, any element of $$K$$ of the form $$\frac {0_D} q$$ 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, $$\frac p q \equiv \left[\!\left[{\left({p, q}\right)}\right]\!\right]_\ominus$$.

From Equality of Division Products, two elements $$\frac a b, \frac c d$$ 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 $$\frac a b \circ \frac c d = \frac {a \circ b} {c \circ d}$$.

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 $$\left[\!\left[{\left({a \circ c, b \circ d}\right)}\right]\!\right]_\ominus$$.

$$ $$ $$ $$ $$ $$ $$

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.