TITLE: How can the real numbers be a field if $0$ has no inverse?
QUESTION [7 upvotes]: I'm reading a linear algebra book (Linear Algebra by Georgi E. Shilov, Dover Books) and the very start of the book discusses fields. 9 field axioms discussing addition and multiplication are given then the author goes on to discuss common sets of numbers.
The integers are identified as being a set of numbers which is not a field because there does not exist a reciprocal element for every integer (axiom # 8 in this book states the existence of a reciprocal element $B$ for a number $A$ such that $AB=1$). The author goes on to call the real numbers a field, and asserts that an axiomatic treatment can be had by supplementing the field axioms with the order axioms and the least upper bound axiom.
My understanding consists of the following statements I believe are facts: 

zero is a member of the reals
there exists no reciprocal element of zero that is a real number

Given those two facts it seems to me that the reals fail the same test for being a field that the author states the integers fail. Yet the author is calling the real numbers a field. To my mind this is a contradiction.
Is there a resolution to this apparent contradiction? I'm a total beginner at this sort of math (I'm an engineer by training, not a mathematician!) and would appreciate any assistance!

REPLY [8 votes]: Many things would be nicer if we could assume that everything in a field, including 0, had a multiplicative inverse. Unfortunately, that is not possible in any useful way.
In any structure that satisfies the field axioms, it must hold for any $a$ that
$$ 0\cdot a = (0+0)\cdot a = 0\cdot a + 0\cdot a $$
and canceling one $0\cdot a$, we see that $0\cdot a = 0$ for all $a$.
If $0$ had an inverse, then we would have $a = 1\cdot a = (0\cdot 0^{-1})\cdot a = 0\cdot(0^{-1}\cdot a) = 0$, and so this is only possible if every element in the field equals $0$, which is to say that $0$ is the only element of the field.
The set $\{0\}$ with $0\cdot 0=0$ and $0+0=0$ is indeed a field according to some formulations of the field axioms. But it is not a very interesting field, and actually it tends to be such a cumbersome special case that by convention it is not considered a "field" at all. Formally this convention is expressed by requiring as an axiom that $1\ne 0$.