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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: TobyBartels
   Format: MarkdownItexThis came up on [another thread](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1507) (see particularly [#36](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1507&Focus=13043#Comment_13043), [#45](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1507&Focus=13067#Comment_13067), [#47](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1507&Focus=13098#Comment_13098), and [#60](http://www.math.ntnu.no/~stacey/Vanilla/nForum/comments.php?DiscussionID=1507&Focus=13146#Comment_13146)). The usual notation for the group of nonzero elements of a field $K$ is $K^{\times}$. An obvious generalisation is that the group of invertible elements of a monoid $M$ should be $M^{\times}$. (The discussion above is really about whether this should be generalised from a monoid to a category.) I have never liked this. With a field, at least ‘$\times$’ indicates that you're looking at multiplication rather than addition. With a monoid, multiplication is *already* the only operation, so the notation is counterintuitive. I decided once that a better symbol would be ‘$\div$’, so that $M^{\div}$ is the group of invertible elements of the monoid $M$. (The notion of group can be nicely axiomatised using only the operation of division, so in a way, groups are about division while monoids are about multiplication.) As a special case (a case with extra structure, not merely extra property), $R^{\div}$ is the group of invertible elements of a ring $R$. And of course, $K^{\div}$ is the group of invertible elements of a field $K$. However, to accomodate the classical notation $K^{\times}$, I also use $R^{\times}$ for the monoid (*not* necessarily a group) of non-[[zero-divisors]] of $R$. Then $K^{\div} = K^{\times}$ when $K$ is a field, although in general $R^{\div} \subset R^{\times}$. There is no meaning of $M^{\times}$ for an arbitrary monoid $M$, although it does make sense for a monoid with a (necessarily unique) absorbing element (an element $z$ such that $x z = z$ and $z x = z$ for all $x$). This is independent of the question of whether notation for a monoid should be extended to a category.

   This came up on another thread (see particularly #36, #45, #47, and #60).

   The usual notation for the group of nonzero elements of a field $K$ is $K^{\times}$. An obvious generalisation is that the group of invertible elements of a monoid $M$ should be $M^{\times}$. (The discussion above is really about whether this should be generalised from a monoid to a category.) I have never liked this.

   With a field, at least ‘$\times$’ indicates that you’re looking at multiplication rather than addition. With a monoid, multiplication is already the only operation, so the notation is counterintuitive. I decided once that a better symbol would be ‘$\div$’, so that $M^{\div}$ is the group of invertible elements of the monoid $M$. (The notion of group can be nicely axiomatised using only the operation of division, so in a way, groups are about division while monoids are about multiplication.) As a special case (a case with extra structure, not merely extra property), $R^{\div}$ is the group of invertible elements of a ring $R$. And of course, $K^{\div}$ is the group of invertible elements of a field $K$.

   However, to accomodate the classical notation $K^{\times}$, I also use $R^{\times}$ for the monoid (not necessarily a group) of non-zero-divisors of $R$. Then $K^{\div} = K^{\times}$ when $K$ is a field, although in general $R^{\div} \subset R^{\times}$. There is no meaning of $M^{\times}$ for an arbitrary monoid $M$, although it does make sense for a monoid with a (necessarily unique) absorbing element (an element $z$ such that $x z = z$ and $z x = z$ for all $x$).

   This is independent of the question of whether notation for a monoid should be extended to a category.

2. • CommentRowNumber2.
   • CommentAuthorTobyBartels
   • CommentTimeJun 29th 2010
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   Author: TobyBartels
   Format: MarkdownItexI have used this notation in [[zero-divisor]].

   I have used this notation in zero-divisor.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexA note: At least in commutative algebra, the standard notation for the group of units of a ring $R$ is $R^*$. The notation $R^\times$ is meant to denote the entire multiplicative monoid of the ring. These are identical in the case of a field for the obvious reasons. For this reason, I am opposed to using the $R^\times$ notation the way you've used it.

   A note: At least in commutative algebra, the standard notation for the group of units of a ring $R$ is $R^*$. The notation $R^\times$ is meant to denote the entire multiplicative monoid of the ring. These are identical in the case of a field for the obvious reasons.

   For this reason, I am opposed to using the $R^\times$ notation the way you’ve used it.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: TobyBartels
   Format: MarkdownItex>The notation $R^\times$ is meant to denote the entire multiplicative monoid of the ring. I don't undersand what you mean by this. This conflicts with the notation $\mathbf{C}^{\times}$ for the non-zero complex numbers that you championed on the other thread. And it is simply not true that >These are identical in the case of a field for the obvious reasons. Instead, $R^*$ is $R^{\times} \setminus \{0\}$ when $R$ is a field (if $R^{\times}$ is the entire multiplicative monoid of $R$). It seems to me that you want $R^{\times}$ to be the multiplicative monoid of *non-zero* elements of the ring $R$. Except that, unless $R$ is an integral domain, this is not a monoid! However, it has a largest subset which is a monoid: the set of non-zero-divisors. And that is precisely what I want to denote by $R^{\times}$!

   The notation $R^\times$ is meant to denote the entire multiplicative monoid of the ring.

   I don’t undersand what you mean by this. This conflicts with the notation $\mathbf{C}^{\times}$ for the non-zero complex numbers that you championed on the other thread. And it is simply not true that

   These are identical in the case of a field for the obvious reasons.

   Instead, $R^*$ is $R^{\times} \setminus \{0\}$ when $R$ is a field (if $R^{\times}$ is the entire multiplicative monoid of $R$).

   It seems to me that you want $R^{\times}$ to be the multiplicative monoid of non-zero elements of the ring $R$. Except that, unless $R$ is an integral domain, this is not a monoid! However, it has a largest subset which is a monoid: the set of non-zero-divisors. And that is precisely what I want to denote by $R^{\times}$!

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • (edited Jun 29th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexYes, you're right, my mistake. The notation for the group of units is still true at least.

   Yes, you’re right, my mistake. The notation for the group of units is still true at least.

6. • CommentRowNumber6.
   • CommentAuthorTobyBartels
   • CommentTimeJun 29th 2010
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   Author: TobyBartels
   Format: MarkdownItexSo is $R^{\times}$ used in my sense by commutative algebraists? Or do you just withdraw it entirely? I have now introduced $R^*$ for the group of units (although it is just an aside) at [[zero-divisor]].

   So is $R^{\times}$ used in my sense by commutative algebraists? Or do you just withdraw it entirely?

   I have now introduced $R^*$ for the group of units (although it is just an aside) at zero-divisor.

7. • CommentRowNumber7.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 29th 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexYour use of $R^\times$ generalizes the common usage and reduces correctly, which is why I withdrew my objection.

   Your use of $R^\times$ generalizes the common usage and reduces correctly, which is why I withdrew my objection.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeJun 29th 2010
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   Author: Mike Shulman
   Format: MarkdownItexI'm not sure what I think about the notation $M^\div$, but I observe that an element of a ring is a non-zero-divisor iff it is cancellable, i.e. $x y = x z$ implies $y = z$ and oppositely. And cancellability makes sense in any monoid.

   I’m not sure what I think about the notation $M^\div$, but I observe that an element of a ring is a non-zero-divisor iff it is cancellable, i.e. $x y = x z$ implies $y = z$ and oppositely. And cancellability makes sense in any monoid.

9. • CommentRowNumber9.
   • CommentAuthorTobyBartels
   • CommentTimeJun 30th 2010
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   Author: TobyBartels
   Format: MarkdownItex@ Mike Good point! But then we get two different meanings of $R^{\times}$ when $R$ is an arbitrary rig. Probably the cancellability condition is actually the better one in a rig, and my definition at [[zero-divisor]] (or at least the claim that the definition makes sense in any rig) is wrong. So then, yes, $M^{\times}$ makes sense for any monoid $M$.

   @ Mike

   Good point!

   But then we get two different meanings of $R^{\times}$ when $R$ is an arbitrary rig. Probably the cancellability condition is actually the better one in a rig, and my definition at zero-divisor (or at least the claim that the definition makes sense in any rig) is wrong. So then, yes, $M^{\times}$ makes sense for any monoid $M$.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJun 30th 2010
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    Author: Urs
    Format: MarkdownItexFor what it's worth, what Harry was thinking of above was $(R,\times)$.

    For what it’s worth, what Harry was thinking of above was $(R,\times)$.