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I have a fairly straightforward question about defining exponentiation in rings: Given any element $a \in R$ of any ring $R$, what is $a^0$ defined as? Is it the additive identity or the multiplicative identity (if it exists)?

In the context of groups we define $a^0 = e$ for whatever the identity of that one operation is. But in the context of rings, $a^n$ would refer to the ring product, so I would assume $a^0$ to be the multiplicative identity, but this doesn't always exist.

Is there any convention, or is it simply left undefined/context dependent?

Edit: In regards to being a duplicate of the $0^0$ question, I had already read through both the question and some answers, but I thought this was a 'fairly separate' question. In particular, this is about rings in general, and while there is a relationship, most of the discussion in the other question pertains explicitly to $\mathbb{R}$. For example discussion of non-integer powers, and limits as a motivating value for the definition, and the fact $\mathbb{R}$ has unity.

masiewpao
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    I believe it is the multiplicative identity, when it exists. – Pranav Chinmay Nov 28 '20 at 19:45
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    It’s the multiplicative identity, and this sort of thing is why you want rings to have a multiplicative identity. – Qiaochu Yuan Nov 28 '20 at 19:59
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    If the ring as a unity, it is the multiplicative identity; that’s to ensure the general associativity law (the empty sum is the identity of the sum, the empty product is the identity of the multiplication). If the ring does not have a unity, then the expression is undefined. – Arturo Magidin Nov 28 '20 at 20:00
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    Note that you only define $a^0=e$ in groups when the operation is multiplication. For additive groups, you define $0a=0$, as expected. – Arturo Magidin Nov 28 '20 at 20:00
  • Thank you for the explanations, they all make sense! I think I was a bit hasty in asking because I was wondering how to interpret a theorem in the case the exponent was 0. But it turns out it was only defined for primes anyway. – masiewpao Nov 28 '20 at 20:20

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Loosely speaking, for $a\in R$, $a^n$ means the product $a\cdots a$ consisting of $n$ factors. Rigorously, we could establish a right $\mathbb{N}^*$-pseudomodule over $(R,\cdot)$ whose module operation is $^\wedge:R\times \mathbb{N}^*\to R$. Here, a pseudomodule is a generalization of a semimodule that does not require the monoid to be commutative.

The properties of $R_{\mathbb{N}^*}$ are then

  1. $a^{n+m} = a^n\cdot a^m$
  2. $a^1 = a$

When $R$ is unitary, then we can use the right $\mathbb{N}$-pseudomodule which includes the property that $$a^0=1_R$$ for $a\neq 0_R$. Thus in a unitary ring, a zeorth power is indeed the multiplicative identity of the ring.

If $R$ is a division ring, then we upgrade to a right $\mathbb{Z}$-pseudomodule and enable negative exponents. If $R$ is commutative then we can go to a $\mathbb{N}^*$-semimodule. If $R$ is a field, then we have the familiar $\mathbb{Z}$-semimodule of the rational numbers and the real numbers. Rational and then irrational exponents require further properties.

Cam White
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    What about when $1_R$ does not exist? – Shaun Nov 28 '20 at 21:22
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    Most algebraists will call what you have labeled as “loosely speaking” as the standard definition. I’ve never heard of a $\mathbb Z$ pseudo-module, but it would make sense to consider the submonoid of $(R\cdot)$ generated by $a$, which incidentally would amount to an $\mathbb N$ semimodule, but only because it is an abelian monoid. – rschwieb Nov 29 '20 at 00:12
  • Certainly there is no way to conclude $(ab)^n=a^nb^n$. Did you mean commuting powers of a instead? – rschwieb Nov 29 '20 at 00:14
  • I’m hoping you agree on some of this and will make edits. As written now I feel like there are too many misrepresentations in this answer and am on the fence about downvoting it. – rschwieb Nov 29 '20 at 01:32
  • @rschwieb Yes you are correct. I thought of this idea a couple of years ago, so the few properties I was trying to recall were in a different context. Yes I'll make the edits. – Cam White Nov 29 '20 at 09:02