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Some useful ''conventions'' as $0!=1$ or $a^0=1$ are particular cases of an empty product, i.e. a product between elements of the empty set. I know that such product is defined as a convention by: $$ \prod _{x_i\in \emptyset} x_i=1 $$ This convention is usually motivated (see here) by the fact that the definition of the product of $n$ numbers $\{x_1,\cdots x_i,\cdots x_n\}$: $$ P_n=\prod _{i=1}^n x_i $$ can be given recursively as $P_n=x_n P_{n-1}$ and the recursive definition become simpler and ''universal'' if we assume $P_0=1$ i.e.: the product with no factors is $1$.

But i don't understand how this convention can be reconciled with the axiomatic definition of the product operation in a field ( or ring) $R$. As far as I know the product is defined as a binary operation $\cdot :R\times R \rightarrow R$ and the empty set is not an element of $R\times R$. So, my question is: the definition of the empty product is only a notation convention that has not an exact mathematical meaning ( since it can not be derived by the axioms) or it can be given in some way as an axiom when we define a ring? And, in this case what kind of axiom we must introduce so that $0!=1$ and $a^0=1$ became theorems and not simply notational conventions?

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    $\begingroup$ A definition is always a convention (and never(?) an axiom) $\endgroup$ – Hagen von Eitzen Jun 8 '15 at 15:18
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    $\begingroup$ @HagenvonEitzen: But we define an operation by means of axioms (or not?) $\endgroup$ – Emilio Novati Jun 8 '15 at 21:00
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It can be instructive to look at the set-theoretic product here. Indeed, when the natural numbers are formalised as sets, it's true that the product of natural numbers is (the isomorphic to) the product of the corresponding sets; and multiplication in a field, commutative ring, algebra or whatever, can be considered to generalise that of natural numbers. So the set-theoretic empty product tells us what the empty product in a ring 'should be'.

Given a set $S$ of sets, the product $\prod S$ is the set of choice functions for $S$, i.e. the set of functions $f : S \to \bigcup S$ such that $f(X) \in X$ for all $X \in S$. In more familiar terms, i.e. if $S = \{ X_i \mid i \in I \}$ for some index set $I$, the product $\prod S = \prod_{i \in I} X_i$ is the set of functions $f : I \to \bigcup_{i \in I} X_i$ such that $f(i) \in X_i$ for all $i \in I$... but for talking about the empty set, the $\prod S$ notation will be more useful.

Now, consider the case when $S = \varnothing$. Then $\bigcup S = \varnothing$ too, so $\prod \varnothing$ is a set of functions $\varnothing \to \varnothing$. There is only one such function, namely $\varnothing$ itself (when functions are formalised as sets of ordered pairs) and it vacuously satisfies the 'choice function' condition; hence $\prod \varnothing = \{ \varnothing \} = 1$.

So considering multiplication in an arbitrary commutative ring (or module or monoid or algebra or ...) as a generalisation of multiplication of natural numbers, in addition to the useful properties mentioned by other users, it makes sense to consider the empty product as being to the multiplicative unit.

This carries over to factorials and exponentials immediately. Indeed:

  • $n!$ is the number of permutations of a set of size $n$. Thus $0!$ is the number of permutations of the empty set, namely $1$ (the empty function).
  • $m^n$ is the number of functions $X \to Y$ where $|X|=n$ and $|Y|=m$. When $n=0$, $m^0$ is the number of functions $\varnothing \to Y$, which is... $1$, again! (This definition also yields $0^0=1$.)
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  • $\begingroup$ I understand that this set-theoretic reasoning justify the definition of the empty product. But I don't see how we can formulate it in a rigorous manner. It can be derived as a theorem from Peano's axioms? I don't know such a proof. More: the statement that a function $f:\emptyset \rightarrow \emptyset$ is $\emptyset$ itself is an example of vacuous truth, in my opinion, and I have always some care using such kind of deductive reasoning.... $\endgroup$ – Emilio Novati Jun 8 '15 at 21:45
  • $\begingroup$ ...As an example: you notice that the same set-theoretic definition yields $0^0=1$, but this is not so symple to reconcile with an axiomatic definition of exponential fields where it seams more ''natural'' $0^0=0$, see my answer to:math.stackexchange.com/questions/11150/… $\endgroup$ – Emilio Novati Jun 8 '15 at 21:49
  • $\begingroup$ @EmilioNovati: I'm not sure what you mean regarding the rigour of the set theoretic argument... the set theory itself is rigorous in the sense that it is extremely close to the axioms. The argument that ties it to algebra is simply a heuristic (as are all answers to such a question about conventions), thus is in the realm of philosophy than mathematics. Of course the identity $\prod \varnothing = 1$ or whatever can't be proved, it must somehow be defined; I just suggested one more argument for defining it to be so! $\endgroup$ – Clive Newstead Jun 8 '15 at 22:34
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    $\begingroup$ ...I should add: from the ZF axioms you can derive that the only function $\varnothing \to \varnothing$ is $\varnothing$, since functions in ZF are identified with their graphs. As for exponential fields, every definition I've seen for an exponential field requires $E(x+y)=E(x)E(y)$ and $E(0)=1$, and the 'trivial exponential function' has $E(x)=1$ for all $x$, not $0$... so I don't believe your argument in that post :P $\endgroup$ – Clive Newstead Jun 8 '15 at 22:43
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While the product operation is defined by the axioms of a field or ring, nothing in those axiom defines what is meant by the operation $\prod$ which acts on an ordered subset $S$ of the field or ring. So the meaning of the product over zero elements (or for that matter over one element) is a matter of how definition of the operation $\prod (S)$ deals with those cases.

One almost-neat definition is that on any ring or field $R$ $$ \forall S \in R^n, r \in r : \prod(S \cup_\times r) = \prod(S) \cdot r $$ where the $R^n$ is saying $S$ is an ordered subset, and the $\cup_\times$ means append $r$ at the end of that ordered subset.

For this definition to give the usual products for $S$ having one or more elements, we must also start with the definition

$$ \prod(\emptyset) = I$$

and that is an underlying reason why this convention is so useful.

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  • $\begingroup$ I agree with this and with its usefulness . But the question is: is this only a notational convention or it it is a theorem that we can deduce from some set of axioms? Or it is an independent axiom that we must add to the axioms of rings? $\endgroup$ – Emilio Novati Jun 8 '15 at 21:58
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    $\begingroup$ It is part of a definition. As such, it cannot be deduced from other axioms. $\endgroup$ – Mark Fischler Jun 9 '15 at 14:52
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Note that the multiplication in a ring is a binary operation and we want to define an n-ary operation where $n$ is allowed to range over ... whatever we want and as far as we can consistantly go. The usual recursive definiion works for all finite $n$ from a starting point of our choice. Maybe the most natural chioce for a start would be $n=2$, where the $n$-ary operation we want to define can simply be taken to be the given binary miultiplication. However, the way we extend this to larger $n$ allows us to certainly take $n=1$ (with the identity map) as starting point - but if you complain about the empty product you should already complain about this! Provided that $R$ has a multiplicative neutral element, we can even go further and take $n=0$ as starting point (and necessarily with the definition that the empty product equals $1$ in order to make our recursive definition work in its simple form).

In other words: This definition allows us to have a simplerecursive definition, which in turn allows us to use this simple recursion furmula in induction proofs. Any other definition for the empty product would result in unnecessariuly complicated arguments with case distinctions in proofs about products.

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  • $\begingroup$ If I well understand in this vision the product $P_n$ of $n$ elements of a ring is a new $n$-ary operation that is defined in such a way that coincide with the underlying product for $n>1$ but require a new definition axioms for $n=0$ (and maybe for $n=1$?). So we have a new kind of algebraic structure where is defined a new operation with new axioms and one of this axioms define the empty product. $\endgroup$ – Emilio Novati Jun 8 '15 at 21:16

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