A very basic question about the axioms for multiplication in Rudin's "Principles of mathematical analysis, 3rd Ed": On page 5, we have the axioms (M1-M5) for multiplication for a field $F$:
(M4) $F$ contains an element $1\ne 0$ such that $1x=x$ for every $x\in F.$
(M5) If $x\in F$ and $x\ne0$ then there exists an element $1/x\in F$ such that $x(1/x)=1$.
Why do we want to emphasize $1\ne 0$ in (M4) and $x\ne 0$ in (M5)? Is it primarily to make the axioms and the fields so created more useful? E.g. we do not run into situations like the following:
Suppose $1=0$ in (M4). Then $x=1x=0x=(0+0)x=0x+0x$, implying $0x=0$ and hence $x=0$ for every $x\in F$. (which would make the field so created not very useful and fail to model most real world scenarios/applications?)
Or suppose $x=0$ in (M5). Then $1=0(1/0)=(0+0)(1/0)=0(1/0)+0(1/0)$, again implying $0/(1/0)=0$, contradicting $1\ne0$ in (M4).
Are these the primary reasons for requiring $1\ne0$ and not defining $1/0$?