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Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

  • A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.
  • Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.

Community
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goblin GONE
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    What do you mean by "indistinguishable" in abstract algebra? – Qiaochu Yuan May 21 '13 at 04:10
  • @QiaochuYuan, if I knew, I wouldn't be asking! – goblin GONE May 21 '13 at 04:11
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    I like the variety in the answers, including the general ones! – Asaf Karagila May 21 '13 at 05:11
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    Would you consider $i$ and $-i$ indistinguishable? – Christian Blatter May 21 '13 at 08:32
  • @ChristianBlatter, no. – goblin GONE May 21 '13 at 08:34
  • In continuation to @Christian's question, how do you view $\Bbb C$? Is it a field extension over $\Bbb R$ or over $\Bbb Q$? How do you distinguish between $i$ and $-i$? What tools do you have for this task, i.e. in what context do we consider $i$ and $-i$? – Asaf Karagila May 21 '13 at 09:19
  • @ChristianBlatter, I think it would depend on the operations available. If $+$ is available, well $i+i=2i$ and $i+(-i)=0$ are very different numbers. So I'm guessing $i$ and $-i$ would be distinguishable. On the other hand, if all we had available was a relation $\leq$ with $w \leq z$ iff $\mathrm{Re}(w) \leq \mathrm{Re}(z)$, then I would say that $i$ and $-i$ are indistinguishable. – goblin GONE May 21 '13 at 12:06
  • In my opinion it makes sense to say that two objects are indistinguishable only wrt some fixed structure. For example, when you say that two distinct points are indistinguishable in a topological space, you mean wrt the topology (e.g. using open sets). But since you took distinct points, they can't be equal. Also, I think that ruling out some relations (like '=') as "monsters" isn't the way to go. – A.P. May 22 '13 at 20:34
  • @A.P., instead of ruling out $=$, what would you suggest? Note this is exactly what we do when defining indistinguishability in a preorder. – goblin GONE May 23 '13 at 03:50
  • You could start by defining what kind structure you are considering. For example, AFAIK in any category isomorphic objects behave exactly in the same way, so they are effectively indistinguishable in that framework. Maybe you are dealing with a concrete category and considering the underlying sets you could distinguish your objects, but that would then be outside of the scope of your starting structure. – A.P. May 23 '13 at 07:32

5 Answers5

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Consider the field $\Bbb Q(t,s)$ where $t,s$ are two algebraically independent transcendental numbers.

Then these two numbers are completely inseparable by a first-order formula in the language of fields.

Generally speaking, if $\cal L$ is some first-order language of some structure, then there are at most $\aleph_0\cdot|\cal L|$ definable elements in any given structure. If by "indistinguishable" we mean "inseparable by a first-order formula with limited parameters$^1$", then any sufficiently large structure will invariably contain a lot of indistinguishable elements.

One good place to learn about these things is model theory, and in particular the concept of "type".

Edit: To your last edit, about $(\Bbb N,f)$ note that $0$ is a definable element of the structure with the formula $x=f(x)$. And since we don't have any other symbols in the language it's really impossible to express anything else. Therefore it's very easy to see that over the empty set, every two non-zero elements satisfy the same formulas with one free variable.

(To see that we can't express anything else, at least without parameters, note that if $m,n$ are non-zero then there is an automorphism which exchanges between the two. Therefore every two non-zero elements must satisfy the same formulas [in one free variable].)

Footnotes:

  1. Of course if we allow any parameter then $\varphi(x,y)$ defined as $\lnot(x=y)$ is sufficient to distinguish between any two members. But if, like in the first example, we allow no parameters - or parameters from a small substructure - then if the universe of the structure is large enough, there will be many indistinguishable elements.
Asaf Karagila
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  • Asaf, I was going to write in my question: "Edit: It would be nice to see a definition of 'indistinguishable' for arbitrary first-order structures. Something like: two elements of a first-order structure are indistinguishable iff no relations definable in terms of the data of that structure can tell the difference. This doesn't work, however, because the $=$ relation can always tell the difference!" Is this basically what your answer is talking about, or is your answer different? – goblin GONE May 21 '13 at 05:19
  • @user18921: Almost, but not exactly. Suppose $M$ is a structure for some first-order language. We can say that $B\subseteq M$ cannot distinguish between $x,y$ if for every $\varphi(u,u_1,\ldots,u_n)$ and for every $b_1,\ldots,b_n\in B$ we have $M\models\varphi(x,b_1,\ldots,b_n)\leftrightarrow\varphi(y,b_1,\ldots,b_n)$. Equality doesn't help us here because we have to fix one of the elements. This means that every property that we can describe with parameters from $B$ is not enough to distinguish between $x$ and $y$. – Asaf Karagila May 21 '13 at 06:36
  • Hmmm to be honest, I'm not really getting it. In the case of a structure $M$ that is a preorder, does this notion yield the usual notion of 'antisymmetric'? – goblin GONE May 21 '13 at 06:43
  • @user18921: I'm not clear on your question. But you also mentioned that you don't know exactly what you are looking for as an answer. What exactly do you mean by "yield the usual notion of 'antisymmetric'?" – Asaf Karagila May 21 '13 at 06:45
  • Sorry, I am speaking extremely imprecisely. What I'm trying to say is, okay suppose $M$ is a preorder. We want two elements $x,y \in M$ to be indistinguishable iff $x \leq y$ and $y \leq x$. Does the notion you speak of yield precisely this notion of indistinguishability? (wow, that's a long word.) I'm not sure what to take as $B$. – goblin GONE May 21 '13 at 06:47
  • @user18921: I don't see how exactly. But this is not the form of "distinguishable" to which I refer. Two elements are distinguishable (over $B$) if one satisfies a unary predicate (definable with parameters, perhaps) that the other doesn't. If you want definable binary formulas (with or without parameters) then equality is an obvious counterexample, but if we agree to disallow the equality symbol then you can use $x\leq y\land y\leq x$ to define this sort of relation. – Asaf Karagila May 21 '13 at 07:01
  • Yes, absolutely. Equality messes everything up, so lets disallow it. Then we want a definition of "indistinguishable" that yields $x \leq y \wedge y \leq x$ for preorders, but yields a different notion for a different kind of structure. Basically, we want the appropriate notion for each kind of structure. – goblin GONE May 21 '13 at 07:04
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    But the issue here is that you don't distinguish between $\sqrt2$ and $\sqrt3$ because one is a smaller number -- what if you are in $\Bbb C$ and there is no order to begin with? You distinguish them because one of them is a root of $1+1$ and the other is not. So indistinguishability should be some sort of way for us to define a set which includes one object and not the other; rather than define a relation which is not symmetric with respect to the two elements. – Asaf Karagila May 21 '13 at 07:09
  • Let me edit my question and maybe that will clarify things. – goblin GONE May 21 '13 at 07:11
  • Does my edit clarify? – goblin GONE May 21 '13 at 07:16
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    @user18921: Unfortunately your edit didn't clarify things very much. You really just repeated your comments from this conversation. I would again point out that in model theory the concept of "types" are very important to understand the different models of a theory and how much they vary, and from that you can conclude things about provability too. For monoids you can't really do it because either your language has only $1,\cdot$ and $=$ which you avoid. So you have absolutely no relations to work with, so you can't define anything. Or you allow $=$ and then you can distinguish everything. – Asaf Karagila May 21 '13 at 07:19
  • On the other hand, if you consider types, in monoids, then two elements have the same type over $X$ if they are not in the [sub-]monoid that $X$ generates. This means that the elements of $X$ do not help us to distinguish between two objects. – Asaf Karagila May 21 '13 at 07:21
  • Okay, let me ponder what you're saying for a bit. – goblin GONE May 21 '13 at 07:23
  • By the way, we can remove equality from the language of monoids. We just have a relation $\phi(x,y,z)$ which encodes the statement $xy=z$. – goblin GONE May 21 '13 at 07:24
  • @user18921: It's not clear to me that it doesn't, because nowadays people don't explicitly include equality anymore. They take it for granted as a logical symbol, and its axioms as the axioms of logic. – Asaf Karagila May 21 '13 at 07:25
  • Perhaps this is an instance where that convention is bad. – goblin GONE May 21 '13 at 07:27
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    How would you suggest writing the associative properties of $\cdot$ without $=$, by the way? Or the properties of $1$? – Asaf Karagila May 21 '13 at 07:34
  • We don't have to! After all, we're not trying to generalize monoids. Rather, we're trying to look at monoids from a different point of view, so that we can generalize indistinguishability. – goblin GONE May 21 '13 at 07:37
  • But you still want to recognize that your structure is a monoid. What axioms should the structure satisfy if you can't write both of them? (the closure axiom is implicit in the definition of a function symbol.) – Asaf Karagila May 21 '13 at 07:39
  • Begin with a monoid $(M,\cdot,1)$. Define $\phi$ such that $\phi(x,y,z)$ iff $x\cdot y = z$. Define $\psi$ such that $\psi(x,y)$ iff $x = 1$. Now consider the structure $(M,\phi,\psi)$. This yields a notion of indistinguishable elements $\sim$ on $M$. The relation $\sim$ is what you're looking for. – goblin GONE May 21 '13 at 07:44
  • But you don't have $=$ in the language. How did you define that relation? Remember that you're looking for a first-order definition, this means that the structure should be able to interpret the formula all by itself. If the structure doesn't know what is equality, then nothing that you wrote makes sense. – Asaf Karagila May 21 '13 at 07:46
  • Well $(M,\cdot,1)$ is an object of the underlying set theory. That set theory can distinguish between all distinct elements of $M$. – goblin GONE May 21 '13 at 07:57
  • @user18921: That's not a first-order definition anymore. – Asaf Karagila May 21 '13 at 08:00
  • Okay, but why is that a problem? – goblin GONE May 21 '13 at 08:00
  • Let me quote you, "Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary first-order structures." – Asaf Karagila May 21 '13 at 08:02
  • Umm, yes. I didn't mean that the definition had to be first order. Obviously, this would be preferable. – goblin GONE May 21 '13 at 08:03
  • But every structure is a structure of every logic. So if you want first-order this means that you want to define the property in a first-order language over the structure. – Asaf Karagila May 21 '13 at 08:05
  • Oh. I've read that "structure" is an ambiguous term, that's why I qualified it with "first-order." – goblin GONE May 21 '13 at 08:06
  • We can calculate the truth value for any sentence in any logic over any structure. Whether or not we can write these sentences in a particular logic is a different question. – Asaf Karagila May 21 '13 at 08:07
  • I removed the qualification "first-order" from the question. Does this resolve the ambiguity? I really need to do my assignments. – goblin GONE May 21 '13 at 08:09
  • Well, even less than before! Now you have so many more relations then what your language allowed you to have. In particular you always have equality, even if you don't have an explicit name for it in the language. So you can again distinguish between every two elements. – Asaf Karagila May 21 '13 at 08:18
  • Listen, I need to go do homework now. I'll reply here later if I can see how to resolve the issue. – goblin GONE May 21 '13 at 08:23
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    I should go to sleep... It's almost noon! :-) – Asaf Karagila May 21 '13 at 08:27
  • Respect for holding it together this long! – goblin GONE May 21 '13 at 08:34
  • I am usually only awake during the day when I have to teach, which is Sunday this semester. Otherwise nights are preferable. – Asaf Karagila May 21 '13 at 08:35
  • I used to be like that, but I sorted it out. Send me a message sometime if you're interested in knowing how I did it. Although in my experience, most night-dwellers are rather happy that way. – goblin GONE May 21 '13 at 08:37
  • Nah, I'm fine. Being a fatalist I don't usually try to change things. Change is bad. :-) – Asaf Karagila May 21 '13 at 08:41
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A category usually has distinct but isomorphic objects. This generalizes both of your bulleted examples: a preorder is a category in which there is at most one morphism between any two objects, and a pseudometric space is an enriched category (see Lawvere metric space). Getting rid of this extra ambiguity amounts to taking a skeleton.

Since categories are ubiquitous, this gives a wealth of examples. Here are some which are more algebraic in flavor:

  • A set $X$ together with an action of a group $G$ may be regarded as a category (in fact a groupoid) with a morphism $x \to y$ for every $g \in G$ such that $gx = y$. Two objects are isomorphic iff they are in the same orbit with respect to the group action.
  • Given a field $k$ we can consider the category of algebraic extensions of $k$. This category contains various algebraic closures of $k$, all of which are isomorphic.
Qiaochu Yuan
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A commutative ring $A$ with $1$ can contain nilpotent elements, which form an ideal of $A$, called the nilradical $nil(A)$ of $A$. In some contexts, it makes sense to kill off these nilpotents and pass to $A_{red} := A/nil(A)$, the underlying reduced ring. (Reduced means "all nilpotents are zero".)

Matt E
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  • Indeed, $A$ and $A_{red}$ have indistinguishable spectra, i.e. they have the same set of prime ideals. (I'm writing this for the benefit of the OP, since I think this is what you were hinting at). – A.P. May 21 '13 at 05:12
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    @A.P.: Dear A.P., I was hinting at this, and more than this; e.g. in many situations one might have a moduli or deformation space which is in principal non-reduced, but the objects that one is studying via this space may be known to be reduced for some other reason (i.e. one is only interested in individual points of the moduli space, or the example one wants to study happens to be reduced), and then one can pass to the underlying reduced space and not lose anything. (This comes up in the deformation theory of Galois representations, for example.) Regards, – Matt E May 21 '13 at 18:15
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Yes.

  • add to a ring some null elements $\eta_i$ with $\eta_i x = 0$ for all $x$.

  • add extra coordinates $X_i$ and equations $X_i = 0$ to the presentation of an algebraic variety by equations.

  • for an algebraic structure that has a notion of representation, consider elements that act the same in all representations as indistinguishable.

The quotient implied in the last example is to prevent stupidity like taking a finite ring, adjoining uncountably many null elements, and thinking of that as a "big" structure.

zyx
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  • So $\eta_i = \eta_i 1 = 0$? – Ryan Reich May 21 '13 at 05:22
  • @ryan, there might not be a $1$. – zyx May 21 '13 at 05:27
  • But there might be. And this makes your first example not generally applicable, unlike the other two. – Ryan Reich May 21 '13 at 05:49
  • The first example is for any ring $R$ and any abelian group $N$ of "null" elements: on $R \oplus N$ define the product of $(r,\eta)$ and $(r',\eta ')$ to be $(rr',0)$. The unit of $R$, if there is one, is not a unit for the ring so constructed, and what the calculation of $\eta 1$ shows is that no other element is a unit (there is no "1"), not that the construction is limited if a unit was present. @RyanReich – zyx May 21 '13 at 20:00
  • Well, some of us like our rings to have units. – Ryan Reich May 21 '13 at 23:44
  • The example was deliberately described (in the answer) in words that are compatible with either convention. All that changes is whether the result of the construction is called a "ring". I was less careful in comments, where I did call it a ring, but substituting a different name would not change anything. @RyanReich – zyx May 22 '13 at 03:35
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$i$ and $-i$ are algebraically indistinguishable points of $\mathbb C$ because conjugation is a field automorphism over $\mathbb R$.

lhf
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  • I'm not sure this is a good example. If we "quotient out" in the way you seem to be suggesting, what do we get? – goblin GONE May 21 '13 at 03:56
  • You get the upper half plane (with boundary). Not an algebraic structure though. – lhf May 21 '13 at 03:59
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    The corresponding thing for topological spaces is to consider points which are related by a homeomorphism of the space. I don't think this is what the OP is looking for. – Qiaochu Yuan May 21 '13 at 04:29
  • Yes, thank you @QiaochuYuan, I was searching for the words to say exactly that. – goblin GONE May 21 '13 at 04:30
  • More like indistinguishable points of $\operatorname{Spec}\mathbb{R}[x]$ - kinda. – Aleksei Averchenko May 21 '13 at 05:17
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    @user18921 I was thinking something similar to $\pm i$, namely consider the group ${ 1, a, b}$ of order three where $a$ and $b$ are the generators. How do we distinguish $a$ and $b$? Well, we can see the difference between $aa$ (gives $b$) and $ab$ (gives $1$). Is that distinguishability? In $\mathbb{Q}(t, s)$, are $tt$ and $ts$ "different"? One is a square. – Jeppe Stig Nielsen May 21 '13 at 13:34
  • @JeppeStigNielsen, in the first case, yes I think that's distinguishability. For another example along a similar vein, consider the poset on $P={0,a,b,1}$ with order relation ${00,0a,0b,aa,bb,a1,b1,01,11}.$ The resultant indistinguishability relation $\sim$ defined by $x \sim y$ iff $x \leq y \wedge y \leq x$ is simply equality. So all elements are distinguishable, in my opinion, despite that an automorphism permutes $a$ and $b$. – goblin GONE May 21 '13 at 15:10