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Let $\langle K, +, * \rangle$ be our field. By definition, we know that every non-zero element, i.e every element except the additive identity, has an multiplicative inverse in the field, and we also do know that every element, including the multiplicative identity, has a additive inverse in the field.

However, giving the fact that additive and multiplicative operations are just binary operations, and they are just represented by different symbols, I would expect both addition and the multiplication to behave in the same way in the sense that if multiplicative identity has an additive inverse, so should the additive inverse to have a multiplicative inverse, i.e a symmetry between them.

So my question is exactly which property of those binary operations or the proper of the field is corresponds to this unsymmetrical behaviour ?

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    $\begingroup$ There is no symmetry because of the distributive law, which holds only in the form $a(b+c) = ab + ac$ and not $a+bc = (a+b)(a+c)$. From this identity and the group axioms for $+$, as well as the fact that $0\neq 1$ you can dedice that $0$ has no multiplicative inverse. Now why do we want this identity to hold in a field (ring more generally) ? Simply because rings are meant to generalize $\mathbb{Z}$ and other common structures ans it turns out that distributivity is essential in those structures to get anything done $\endgroup$ – Max Jul 29 '17 at 8:09
  • $\begingroup$ @Max I advised you to post your comment as an answer. $\endgroup$ – onurcanbektas Jul 29 '17 at 8:26
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    $\begingroup$ In a ring the zero element is absorbing, since $0r= (r-r)r = r^2-r^2=0$. $\endgroup$ – Wuestenfux Jul 29 '17 at 8:36
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    $\begingroup$ @onurcanbektas But multiplication and addition are not just symbols for two abstract binary operations. They are symbols for two abstract binary operations that satisfy certain axioms. And, as it happens, the field axioms do not treat $+$ and $\cdot$ symmetrically; if we swap the two symbols, we do not get an equivalent set of axioms. (On the other hand, in Boolean algebra, the symbols for AND and OR are treated symmetrically, and so we have a duality principle.) $\endgroup$ – bof Jul 29 '17 at 11:32
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    $\begingroup$ This is tangentially related, but this question reminds me of the exercise "which fields have multiplicative group isomorphic to their additive group?" $\endgroup$ – Mark S. Jul 29 '17 at 12:07
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As I said in the comments, the symmetry is broken because of the distributivity, $a(b+c) = ab + ac$, which does not hold symmetrically (indeed $a+bc = (a+b)(a+c)$ is not true in general).

This identity, the group axioms for $+$ and the fact that $0\neq 1$ altogether imply that $0$ has no multiplicative inverse ($0a = (0+0)a = 0a + 0a$ and so $0a=0$, so unless $0=1$, $0$ has no inverse)

These axioms are there to generalize $\mathbb{Z},\mathbb{Q}$ (integral domains, or more generally rings with more than one element), and are therefore "natural", because in particular distributivity is essential in those structures.

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  • $\begingroup$ So, If we create a new algebraic structure similar to field with including the distribution of addition over multiplication, i.e $a + bc = (a+ b)(a+c)$, would zero have an inverse ? $\endgroup$ – onurcanbektas Jul 29 '17 at 11:31
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    $\begingroup$ In mathematics, the phenomena come first, and the definitions come afterwards. It's not the other way around. To put things somewhat differently, mathematics is not a game in which we make definitions up out of nothing, and then see where they lead. $\endgroup$ – Lubin Jul 29 '17 at 11:39
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    $\begingroup$ No, the same proof would still go, as long as every element has an additive inverse. But if you completely remove "classical" distributivity and add your new axiom, then zero could possibly have a multiplicative inverse. $\endgroup$ – Max Jul 29 '17 at 11:40
  • $\begingroup$ @Lubin You sounded like one of my physics professors. $\endgroup$ – onurcanbektas Jul 29 '17 at 11:48
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    $\begingroup$ @onurcanbektas: If you just add that to the ring or field axioms, you would be able to prove that $x=x+0=x+0\cdot(-x)=(x+0)(x-x) = x\cdot 0 = 0$ for every $x$, so your structure would be rather uninteresting. $\endgroup$ – Henning Makholm Jul 29 '17 at 11:48
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It's because $$0r=(r-r)r=r^2-r^2=0,$$ so if $0r=1,$ we would have $0=1$, a contradiction.

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