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I'm trying to understand the "correct" way of raising elements of commutative rings to the power of $a/b,$ where $a$ and $b$ are integers, but not having much luck. Suppose $R$ is a commutative (unital) ring and that $x \in R.$ A first blush attempt at defining $x^{a/b}_R$ would be that it is the subset of $R$ specified as follows.

$$(*) \qquad y \in x^{a/b}_R \iff y^b = x^a$$

But this isn't well-defined. For example, suppose $x \in \mathbb{R}$ is fixed and positive. Then observe that the following are equivalent:

  • $y \in x^{1/1}_\mathbb{R}$
  • $y^1 = x^1$
  • $y=x$

Hence $x^{1/1}_\mathbb{R} = \{x\}$.

But the following are also equivalent:

  • $y \in x^{2/2}_\mathbb{R}$
  • $y^2 = x^2$
  • $y = \{-x,x\}$

So $x^{2/2}_\mathbb{R} = \{-x,x\}$.

Hence since $x^{1/1}_\mathbb{R} = x^{2/2}_\mathbb{R}$, we deduce that $\{x\}=\{-x,x\}$, a contradiction.

I can see at least two possibilities for addressing this issue. One is to weaken $(*)$ by adding the condition that $a$ and $b$ are coprime. Another is to change the number system in which exponents live from $\mathbb{Q}$ to something else, in which $1/1$ doesn't equal $2/2$; in particular, we could try replacing $\mathbb{Q}$ with some kind of a ring equipped with an involution $b \mapsto b^\dagger$ (perhaps defined only for non-zero elements) that does not satisfy $bb^\dagger =1$, and then we could define that $a/b$ equals $ab^\dagger$.

Question. Is there a consensus on the correct way of raising elements of commutative rings to the power of $a/b$?

goblin GONE
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1 Answers1

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I imagine it would be quite hard to reach consensus, since as you pointed out the obvious "definition" does not end up being well-defined. Given the non-standard notation I would think whenever a paper intended to use such objects it would need to explicitly define them.

Another possible workaround the inconsistency problem you found would be to define, for $x\in R$, $p,q\in\mathbb{Z}$, $$f(x;p,q):=\{y\in R\,:\,y^q=x^p\}$$ and then, for $x\in R$, $a\in\mathbb{Q}$, $$x^a:=\bigcap\left\{f(x;p,q)\,:\,a=\frac{p}q\right\}.$$ This would be equivalent to only defining $x^{p/q}$ when $p,q$ are co-prime, but seems a little more satisfying to me.

Jason
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  • Thanks for your answer. Its worth observing that the notation $f(x;p,q)$ could easily be replaced by the notation $x^{(p,q)}_{\mathbb{R}}$ in which case we're thinking of $(p,q)$ as living in its own number system, which is basically $\mathbb{Z} \times \mathbb{Z}$ modified with its own special notion of addition. The only issue I can see is the case $q=0$. – goblin GONE Jan 29 '15 at 06:27
  • Yes, that's true. The case $q\neq0$ doesn't end up being a problem once you define $x^a$ since $a\neq\frac{p}0$ for any $p\in\mathbb{Z}$. You could always exclude $q\neq0$ in the first place if you wanted, as well. – Jason Jan 29 '15 at 06:53