meadows and fields, aren't $0^{-1}=0$ can be proven simply from the axioms of fields?

Question

Recall field axioms

In this article http://www-compsci.swan.ac.uk/~csjvt/JVTPublications/RationalsAsADT.pdf

Page 4, we have the SIP \begin{matrix} \left(-x\right)^{-1}=-\left(x^{-1}\right) \\ \left(x\cdot y\right)^{-1}=x^{-1}\cdot y^{-1}\\ \left(x^{-1}\right)^{-1}=x \\ \end{matrix}

And then page 7 we are shown how this and the ring axioms lead to the conclusion $0^{-1}=0$

Q1: But isn't the expressions in SIP also true (can be derived) for fields (at least for $x \neq 0$), and since the proof to show that $0^{-1}=0$ rely only on the SIP and ring axioms (CR in the article), shouldn't we are able to derive the same result from the field axioms, that is concluding that $0^{-1}=0$ even for fields by simply introducing a new element $0^{-1}$ and show via the workings shown in the article that this element must $=0$?

In a follow up paper of the subject where the authors introduce non involutive meadows http://arxiv.org/pdf/1406.2092v1.pdf

Page 6

It is said by putting x=0 in $$(x^{∽1})^{∽1} = x + (1 − x \cdot x^{∽1})$$ and $$x^{∽1} \cdot (x^{∽1})^{∽1} = 1$$

we can conclude

$$0^{∽1} \cdot (0^{∽1})^{∽1} = 1$$ $$(0^{∽1})^{∽1} = 1$$

Hence

$$0^{∽1} = 1$$

Q2 But how does $$(0^{∽1})^{∽1} = 0 + (1 − 0 \cdot 0^{∽1})$$ gives $$(0^{∽1})^{∽1} = 1$$?

Attempted self workings

Start with axiom

$$(x^{∽1})^{∽1} = x + (1 − x \cdot x^{∽1})$$

Put x=0 gives

$$(0^{∽1})^{∽1} = 0 + (1 − 0 \cdot 0^{∽1})$$

Using the additive identity property $x+0=x$

$$(0^{∽1})^{∽1} = 1 − 0 \cdot 0^{∽1}$$

But what implies

$$0 \cdot 0^{∽1}=0$$

Is it because of the derived property of rings that $x \cdot 0 = 0$ for any x in the ring, and that $0^{\sim 1}$ is just one of the elements in the ring thus it has to obey this property also?

Answer 1

But isn't the expressions in SIP also true (can be derived) for fields (at least for $x≠0$)...

Yes, and indeed on page 4 the authors even say so: "Guarded versions of SIP - such as $x \neq 0 \implies (x^{−1})^{−1} = x^{-1}$ can be proved from Gil and Sep." They mention there also that the group CR, Gil and Sep form the ordinary field axioms.

since the proof to show that $0^{−1}=0$ rely only on the SIP and ring axioms...

Yes, this shows that $0^{-1}=0$ in any commutative ring. The problem is that $0^{-1}$ is not a multiplicative inverse for $0$, as the notation suggests. It's simply another symbol for $0$, one which lacks the defining property that multiplicative inverses have.

Is it because of the derived property of rings that $x⋅0=0$ for any $x$ in the ring...?

Yes, all the axiom groups I saw there appeared to include the axiom group CR, and that equation is true for any element $x$ in a commutative ring.