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When we define something like a ring, we often say that the elements $0$ and $1$ are "distinguished elements". What does this mean? It obviously doesn't mean they are distinct.

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up vote 14 down vote accepted

You are correct: "distinguished" doesn't mean that the points are different from each other. A "distinguished point" in an algebraic structure can be thought of as a point that has a label attached to it: in a ring, "I AM THE ADDITIVE IDENTITY!" for point 0, and "I AM THE MULTIPLICATIVE IDENTITY!" for point 1. In the trivial ring, a point can have both these labels attached to it.

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Thanks. This clears things up. – nigelvr Dec 4 '12 at 19:35
Also there is the (implicit?) assertion of existence: a ring must have at least the element $0$, and if it is said to have "a unit" must have an element $1$. This use of "distinguished" is pretty bad, in my opinion, since it does not connect with either colloquial or substantial mathematical usage, although there does appear to be a pattern of this usage in the last few decades. – paul garrett Dec 5 '12 at 0:37


set-apart-from; notable; noted; eminent; of particular importance; distinct, distinctive.

To disambiguate the different connotations of "distinguished", we need context.

E.g. In algebra, e.g. in a ring or field, $0$ is distinguished by the fact that it represents the additive identity, and $1$ is distinguished by the fact that it is the multiplicative identity. So $0$ and $1$ are certainly distinguished (set-apart-from other elements, of particular importance, which happen also to be distinct, except for the case of the trivial ring).

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This answer rather misses the point: “set apart” is too ambiguous to answer the question at all, and as OP said themself, “distinct” is certainly not what’s meant, even though the elements may happen to be distinct as well. – Peter LeFanu Lumsdaine Dec 4 '12 at 21:23
@Peter: that's why I pointed out that 0 is the additive identity, and 1 is the multiplicative identity: these elements are of particular importance in that, without them, we would not have a ring, or a field. The fact that they also usually happen to be distinct is besides the point! – amWhy Dec 4 '12 at 23:26
Fair enough — yes, I think it’s a lot clearer since the edits. – Peter LeFanu Lumsdaine Dec 4 '12 at 23:38

In my experience, in algebra, "distinguished elements" never implies that the elements are distinct. One usage is to denote named constants (a.k.a. nullary operations) in algebraic structures, such as constants $0$ and $1$ in rings - the additive and multiplicative identity elements, respectively. Similarly one may speak of structures with distinguished maps, e.g. groups with distinguished operators, vector spaces with distinguished linear maps e.g. (when deriving normal forms for matrices), and rings with a distinguished automorphism (e.g. in difference algebra, which studies solutions of difference equations = recurrences).

A quick search of Google Books confirms this usage, e.g. see definitions of rings and fields in Cohn, Skew Fields p. 3; Enderton, Set Theory, p. 122; Jacobson, Basic Algebra, p. 86; Lidl and Niederreiter, Finite Fields, p.12; Rowen, Ring Theory p. 2, etc.

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I'd strongly second @Bill Dubuque's note. That is, "distinguished elements" should not (necessarily) be construed to mean that the things named are different-from-each-other, even if their names are different. Rather, they are "constants" in the "theory of" the objects at hand. The only real "corollaries" of such assertions are things like the point that the underlying set of a group cannot be the empty set, because it contains at least the "identity" element. The "issue" of whether $0=1$ in a ring is not a real issue, and this is not the issue at hand, either. – paul garrett Dec 5 '12 at 0:33

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