3
$\begingroup$

Are there any typically studied structures (say $\mathfrak D$) where the additive identity, $0_\mathfrak D$ has a multiplicative inverse?

In most usual settings, i.e, complex and real numbers as fields, ring of matrices, etc, this isn't true, what about non-field/ring structures?

E: I'll try to be more clear, I'm looking for structures that may be less known, not fields/rings.

$\endgroup$
3
$\begingroup$

Actually there is one way to save this, even in a "ring" setting. If $R$ is a ring, then extend $R$ by appending the symbol $\infty$ which is to be the multiplicative inverse of $0$. You then have to define how $\infty$ plays with every other element. The obvious thing to do is to require that $a.\infty = \infty$ for any $a\neq 0$ and $a+\infty = \infty$ for any $a$. Note that this forces $\infty$ to not have an additive inverse.

This still does not fix the issue raised by Reveillark. Since you still want a group structure going on underneath (since you're talking about $0$), you still have that $0.x = 0$ for $x\neq\infty$. The real issue is in the second line of his answer:

$$ 1 = \infty.0 = \infty.(0.x) = (\infty.0).x = 1.x = x.$$

There is only one place where we have any possible leeway and that is in the equality $\infty.(0.x) = (\infty.0).x$. The associativity of multiplication with respect to $\infty$ ruins the ring structure (and forces you into conclude exactly as Rev did). The way to save this is not to allow associativity of multiplication when it comes to $0$. We will still have associativity for any other $x,y\neq 0$ since $\infty.(x.y) = \infty = (\infty.x).y$ by definition.

In summary, it can be done, you just have to break associativity a little bit by not allowing $\infty$ to have an additive inverse and not allowing associativity of multiplication with $0$ and $\infty$ to make it work. I'm not sure what you'd call such a space, maybe a rigged ring or something.


Note that if you allowed $\infty$ to have an additive inverse, say $-\infty$, then you'd break associativity of addition since you'd get

$$a+(\infty-\infty) = a = (a+\infty)-\infty = 0.$$

This structure can actually be found in measure theoretic contexts since there they allow values of $\infty$ and $-\infty$ but not simultaneously (as you lose all structure that way).

$\endgroup$
2
$\begingroup$

I don't think this is a definite answer, but it shows that with a few assumptions regarding the structure of $\mathfrak{D}$, then the set becomes quite trivial.

Consider a "structure" $(\mathfrak{D},+,·)$. Let $0$ be the additive identity and $1$ be the multiplicative identity. Assuming right distributivity, we have that, for any $x\in D$:

$$0x=(0+0)x=0x+0x$$

Assuming that $(\mathfrak{D},+)$ is a group, it follows that $0x=0$.

Let $a$ be the multiplicative inverse of $0$. Then, for any $x\in \mathfrak{D}$, if $·$ is associative:

$$1=a0=a(0x)=(a0)x=1x=x$$

Hence $D=\{0\}$, which isn't a very interesting object.

$\endgroup$
  • $\begingroup$ Thanks for your answer, I think I was pretty unclear, I was trying to consider, bring in to the light, some structures outside the usual ones (fields,rings, groups), which may be unknown to me, as, as you said, the only ring that has this property is the trivial ring $0_R$. $\endgroup$ – YoTengoUnLCD Oct 7 '15 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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