# Can an algebraic structure have indistinguishable elements?

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.

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What do you mean by "indistinguishable" in abstract algebra? – Qiaochu Yuan May 21 '13 at 4:10
@QiaochuYuan, if I knew, I wouldn't be asking! – goblin May 21 '13 at 4:11
I like the variety in the answers, including the general ones! – Asaf Karagila May 21 '13 at 5:11
Would you consider $i$ and $-i$ indistinguishable? – Christian Blatter May 21 '13 at 8:32
@ChristianBlatter, no. – goblin May 21 '13 at 8:34

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.
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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 May 21 '13 at 5:19
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 7:09
@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 7:19
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 7:34
I should go to sleep... It's almost noon! :-) – Asaf Karagila May 21 '13 at 8:27

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.
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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".)

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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 5:12
@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

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.

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So $\eta_i = \eta_i 1 = 0$? – Ryan Reich May 21 '13 at 5:22
@ryan, there might not be a $1$. – zyx May 21 '13 at 5:27
But there might be. And this makes your first example not generally applicable, unlike the other two. – Ryan Reich May 21 '13 at 5:49
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

$i$ and $-i$ are algebraically indistinguishable points of $\mathbb C$ because conjugation is a field automorphism over $\mathbb R$.

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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 May 21 '13 at 3:56
You get the upper half plane (with boundary). Not an algebraic structure though. – lhf May 21 '13 at 3:59
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 4:29
Yes, thank you @QiaochuYuan, I was searching for the words to say exactly that. – goblin May 21 '13 at 4:30
@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