Question: 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 $1/0$?

Answer: 
You are basically right about both axioms (M4 and M5). Let me clarify a bit.

* **M4:** As you have shown, if $0 = 1$ then $F = \{0\}$ (the field has only a single element). So this would be a rather trivial field. You may still wonder, why don't we just let $\{0\}$ be a field? Technically, we could, but it ends up being more useful to disallow it (similar to why $1$ isn't a prime number). You can read more about it in [this question](https://math.stackexchange.com/questions/427078/is-0-a-field).
* **M5:** What you said (as I understand it) is that we leave $(1/0)$ undefined, because if it were defined then the field would have to have $0 = 1$. While you are correct, axiom M5 itself does *not* imply that $(1/0)$ can't be defined; in particular, axiom M5 does not imply $0 \ne 1$ (M4). M5 simply requires that every nonzero number have an inverse; it does not say that $0$ doesn't have an inverse as well.

What is true, however, is that if we were to leave out the condition $x \ne 0$ in M5, then $0$ would have to have an inverse, and the *only* field would be the trivial $\{0\}$. Which we certainly don't want: at the very least, $\mathbb{R}$ and $\mathbb{Q}$ should be fields, but they wouldn't satisfy the axiom if $x \ne 0$ were left out. This seems to be basically what you are getting at.

In fact if M4 were kept an axiom *and* we took out the condition $x \ne 0$ from M5, then there would be no fields at all -- do you see why?
If $F = \{0\}$ then addition and multiplication would be the same operation, as $x+y = xy, \forall x,y \in F$, and therefore $F$ could be described completely with group theory and, therefore, it would make no sense trying to define a whole new structure to describe it.