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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 $1/0$?

darij grinberg
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syeh_106
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2 Answers2

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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.

  • 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?

Community
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Caleb Stanford
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    Thanks a lot for the claification! About your last question, let me try... If M4 is kept as is, and we allow $1/0$ to be defined by M5, then, as we mentioned, we must have $1=0$. Therefore no sets can satisfy both M4 and M5 with $x\ne 0$ removed. Is this what you'd also like to point out? – syeh_106 Nov 11 '16 at 06:21
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    @syeh_106 Yep, that's right, that's what I was getting at. Thank you for the question! – Caleb Stanford Nov 11 '16 at 06:23
  • So to recap our discussions, we have: 1) If we remove $1\ne 0$ from M4, then $F={0}$ and no elements of $F$ will qualify M5 to have its reciprocal defined. (Many sets can be a field, however, since M5 doesn't require every element to have a reciprocal.) 2) If we remove $x\ne0$ from M5, then no sets are qualified as a field. 3) If we remove both, then $F={0}$ and $1/0=1=0$. Many sets can be a field again, albeit not very interesting ones. 4) So the most interesting case is the original M4 and M5. I hope I didn't miss something... – syeh_106 Nov 11 '16 at 07:04
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    @syeh_106 That's almost exactly right. Only issue is your wording of (1); removing $1 \ne 0$ doesn't imply $F = {0}$, but it makes ${0}$ a field. – Caleb Stanford Nov 11 '16 at 16:18
  • You're right. (Same for (3).) Thanks a lot for all your help! – syeh_106 Nov 12 '16 at 01:36
  • @syeh_106 I think concerning (3), you are correct in stating that $F = {0}$, since, if we were to remove the first condition from M4, namely that we let $1 = 0$ be the multiplicative identity, then together with what we removed from M5, we get $1x = 1$, $(1+0)x=x$, $0x=0$ (distributive property), but then this means that $1x = 0$, hence $x =0$, for any $x \in F$. Which is not very interesting, like you said before. – hteica Feb 29 '20 at 04:36
  • @syeh_106 In general, I think, whenever possible, if we modify M5, then we would always get $F = {0 }$, since this would mean that $1 = 0$, and by M4, we get $x=1x=0x=0$ (Since $1x=x$ implies $0x=0$). Tell me if I've got anywhere wrong, I would love to know, thanks! – hteica Feb 29 '20 at 04:48
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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.

Vitor Borges
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