8
$\begingroup$

If I'm not mistaken, in a ring with identity, the additive identity cannot have a multiplicative inverse. I'm trying to prove this.

Here's my attempt so far:


Suppose $0\cdot a=1$ $$0\cdot a=1$$ $$0\cdot (a + n) = 0\cdot a + 0\cdot n$$

It's easy to see here that if $n$ were a positive integer (same works for negative): $$n=1+1+1+\cdots$$ $$0\cdot a + 0\cdot (1 + 1 + 1+\cdots )$$ $$0 \cdot a + 0 + 0 + 0 + \dots = 0\cdot a$$ $$\therefore 0\cdot n = 0$$ If $a$ is some whole number of multiplicative inverses summed together, then this is a contradiction.


This hinges on the assumption that $n$ is some integer multiple of the multiplicative unit, which it may not be. This also got me thinking about whether the distributivity of multiplication over addition is discretized in whole terms (you can only use operators with whole numbers of terms) or whether the concept of distrubitivity actually goes deeper than this.

Mainly I'd like a proof that the additive identity can not have a multiplicative inverse and why $0\neq 1$. But some deep insight into why the properties of a ring lead to this would be a bonus.

$\endgroup$
  • 3
    $\begingroup$ In a ring, one does not necessarily insist that $0\ne 1$. But if $0=1$ then there is only one element. For your calculation, I would look at $(0+0)a$. Expand, and it's almost over. $\endgroup$ – André Nicolas Jun 13 '14 at 17:44
11
$\begingroup$

$$0\cdot a= (0+0)\cdot a=0\cdot a+0\cdot a$$ Now substract $(0\cdot a)$ on both sides (we can do this because $\forall r\ \exists(-r)\mid r-r=0)$:$$0=0\cdot a$$ This means that $0$ can only have a multiplicative inverse if $0\cdot \tilde a=0=1$ for some $\tilde a$. This then implies that we have $x=x\cdot1=x\cdot0=0\ \ \forall x$, hence we live in the zero-ring $\{0\}$.

So for any ring $R$, we have that $0$ has a multiplicative inverse $\iff$$R=\{0\}$

$\endgroup$
6
$\begingroup$

Suppose there exists $a = 0^{-1}$. Then $$ 1 = 0\cdot a = (a-a)\cdot a = a^2 - a^2 = 0 $$

Then, for all $b$ in the ring, $$b = b\cdot 1 = b \cdot 0 = 0 $$ so zero is the only element of the ring.

Consequence: If there is nonzero element of the ring, $0$ does not have a multiplicative inverse.

$\endgroup$
  • $\begingroup$ To be careful it is really $a^2+ (-1)a^2$ and then you must use/show $x+(-1)x = 0$. $\endgroup$ – abnry Jun 13 '14 at 17:57
  • $\begingroup$ @nayrb Additive inverses are defined for everything in a ring so I don't see the need to use $a^2+(-1)a^2$ instead of $a^2-a^2$. I like the fact that additional structure allows more terse notation and embrace it. $\endgroup$ – Myridium Jun 14 '14 at 0:19
  • $\begingroup$ @Myridium: I believe nayrb is noting that $(-a)(a) = -(a^2)$ is also usually another theorem. $\endgroup$ – Eric Towers Jun 14 '14 at 18:17
  • $\begingroup$ @EricTowers Doesn't a ring require associativity? Then $(-a)(a) = (-1)(a)(a) = (-1)a^2 = -(a^2)$ $\endgroup$ – Myridium Jun 15 '14 at 4:58
  • 2
    $\begingroup$ @Myridium: In the ring axioms, "$-$" is an additive property and "$(-1)\cdot$" is a multiplicative thing. Showing they're the same is a lemma. $\endgroup$ – Eric Towers Jun 15 '14 at 19:37
2
$\begingroup$

Distributivity implies absorption into the additive identity, i.e. $$\forall a,b,c \in R\quad a(b+c)=ab+ac \implies ab=a(b+0)=ab+a0 \implies a0=0.$$ So if everything times zero equals zero, then obviously you can't have something times zero equal one, unless of course zero equals one and you are looking at the zero ring.

$\endgroup$

protected by user296602 May 1 '18 at 4:56

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.