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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:

  1. zero is a member of the reals
  2. 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!

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    $\begingroup$ Read the field axioms again. $\endgroup$ Feb 15, 2013 at 19:09
  • $\begingroup$ @Paul Read where it says Existence of additive inverses and multiplicative inverses and check rschwieb's answer. $\endgroup$
    – Git Gud
    Feb 15, 2013 at 19:14
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    $\begingroup$ You guys are quite correct... my failing is in taking inadequate notes and then referring back to those rather than the original. Regaining my mental acuity after joining the corporate world is one of my main motivations in reading this book... looks like it's coming along well thanks to the people on this board! $\endgroup$
    – Paul
    Feb 15, 2013 at 19:46
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    $\begingroup$ It's great that you are taking an active approach to learning the material and asking questions when things don't make sense to you. I wish more of the students I teach took such an approach! $\endgroup$ Feb 15, 2013 at 20:43

3 Answers 3

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Zero is always excluded from having a reciprocal.

(The axioms should say that every nonzero element of the field has a reciprocal.)

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    $\begingroup$ It does say non-zero. Did my notes say that? Of course not! Thank you very much! $\endgroup$
    – Paul
    Feb 15, 2013 at 19:41
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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 $1\ne 0$.

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  • $\begingroup$ Name, if possible, three things that would be nicer... I'm interested! $\endgroup$
    – rschwieb
    Feb 15, 2013 at 22:03
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    $\begingroup$ @rschwieb: Um... Matrices with entries in a field would always have inverses. You could cancel factors in equations and fractions without worrying that the factors might be 0. ("Such-and-such must be nonzero" assumptions disappear from theorems all over). Every polynomial function would have a scalar multiple that was monic of any higher degree, so the quotient in long polynomial division could always be a scalar. $\endgroup$ Feb 16, 2013 at 15:12
  • $\begingroup$ @rschwieb: (1) Meromorphic functions are actually holomorphic. (2) You can make spills vanish by dividing them by zero. (3) Pigs can fly, and so can we! $\endgroup$
    – user21820
    Dec 5, 2019 at 17:04
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Here it is from Google Books:

shilov image

That "$\ne 0$" in there is the thing you are missing.

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