1
$\begingroup$

In the definition of a field one of the required properties is that

every element other than zero has a multiplicative inverse.

It's vague whether the zero is forced not to have an inverse or not, as in is the property

every element other than zero has a multiplicative inverse and zero does not

or

every element other than zero has a multiplicative inverse and zero may or may not?

Regardless, what differences would occur if we adopted one definition as opposed to the other?

$\endgroup$
1
  • 2
    $\begingroup$ To whomever, the down vote seems unnecessary $\endgroup$
    – user34832
    Commented May 1, 2013 at 20:05

4 Answers 4

10
$\begingroup$

If you kept all of the other usual axioms, then

$$ 0 = 0 \cdot 0^{-1} = 1 $$

$$ x = x \cdot 1 = x \cdot 0 = 0 $$

and so all numbers are zero. This is rather degenerate. This is called the "zero ring"; usually we do not consider this ring to be a field.

$\endgroup$
2
  • $\begingroup$ Why does $aa^{-1} = 0$ hold? $\endgroup$
    – Don Larynx
    Commented Sep 14, 2013 at 17:46
  • $\begingroup$ @Jossie: because everything is zero. $\endgroup$
    – user14972
    Commented Sep 15, 2013 at 1:29
4
$\begingroup$

The fact that $0$ has no inverse in a ring (with identity $1\ne0$) is a result of the distributive property of multiplication and mentioning it in a definition is redundant:

$$\quad ab=(0+a)b=0b+ab \implies 0b=ab-ab=0$$

$\endgroup$
0
1
$\begingroup$

If $0$ had a multiplicative inverse, then $0=0*0^{-1}=1$, so your ring would be the trivial one. By the way, this holds for any ring, not just for fields.

$\endgroup$
1
$\begingroup$

If in a ring $R$ the zero $0$ has an inverse, then $R=0$ (in fact, $1 = 0 \cdot 0^{-1} = 0$, hence $r = 1r = 0r = 0$ for all $r \in R$). One wants fields to be non-zero for many reasons, for example for the uniquenes of the dimension (remark that $R^n \cong R^m$ for all $n,m$ when $R=0$). Therefore the second definition is correct (i.e. coincides with the usual definition of a field), whereas the first and the third also include the zero ring, which is not a field.

$\endgroup$
1
  • $\begingroup$ Why $R^n\cong R^m$ gives a problem with uniqueness of dimension? Wouldn't we have $dim R^n=dim R^m =0$, $=$ also to $n\times0=m\times 0=0$? $\endgroup$
    – user119256
    Commented Jan 7, 2014 at 15:39

You must log in to answer this question.