Algebraic Elements Are Like Limit Points!

 
 
 
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When you hear the word closure, what do you think of? I think of wholeness - you know, tying loose ends, wrapping things up, filling in the missing parts. This same idea is behind the mathematician's notion of closure, as in the phrase "taking the closure" of a set. Intuitively this just means adding in any missing pieces so that the result is complete, whole. For instance, the circle on the left is open because it's missing its boundary. But when we take its closure and include the boundary, we say the circle is closed.

As another example, consider the set of all real numbers strictly between 0 and 1, i.e. the open interval (0,1). Notice that we can get arbitrarily close to 0, but we can't quite reach it. In some sense, we feel that 0 might as well be included in set, right? I mean, come on, 0.0000000000000000000000000000000000000001 is basically 0, right? So by not considering 0 as an element in our set, we feel like something's missing. The same goes for 1.

 
 

We say an element is a limit point of a given set if that element is "close" to the set,* and we say the set's closure is the set together with its limit points. (So 0 and 1 are both limit points of (0,1) and its closure is [0,1].) It turns out the word closure is also used in algebra, specifically the algebraic closure of a field, but there it has a completely different definition which has to do with roots of polynomials, called algebraic elements. Now why would mathematicians use the same word to describe two seemingly different things? The purpose of today's post is to make the observation that they're not so different after all! This may be somewhat obvious, but it wasn't until after a recent conversation with a friend that I saw the connection:

 

 

algebraic elements of a field

are like

limit points of a sequence!

 

(Note: I'm not claiming any theorems here, this is just a student's simple observation.)

 
 
 
Let's lay down some definitions:
  • Let $F$ be a field and let $E$ be an extension of $F$. An element $\alpha\in E$ is algebraic over $F$ if there is a polynomial $f(x)$ in $F[x]$ such that $f(\alpha)=0$ (i.e. $\alpha\in E$ is an algebraic element if it is the root of some polynomial with coefficients in $F$).
  • Let $F$ be a field and let $E$ be an extension of $F$. If $K\subset E$ is a subset of $E$ which contains all elements which are algebraic over $F$, then $K$ is actually a subfield of $E$ and an algebraic extension of $F$. We call $K$ the algebraic closure of $F$ and denote it by $\overline{F}$. [1]
 
 

It's a fact that an algebraic closure $\overline{F}$ exists for every field $F$ (and is actually unique up to isomorphism). So we can draw a containment picture like the one above.

Those familar with some topology and/or analysis will notice that such a "field tower" is suggestive of a vaguely analogous result: given a topological space $X$ we can always (assuming some conditions about $X$, namely it being locally-compact Hausdorff) stick an open set $V$ and its closure $\overline{V}$ between a certain compact set $K$ and open set $U$: $$K\subset V \subset \overline{V} \subset U.$$

Now, don't buy too much into the analogy. I only mention this topological result to motivate the fact that the closure of a set and the algebraic closure of a field do indeed convey the same concept: wholeness. It seems then that we can view algebraic elements as the mathematical cousins of limit points of sequences of real numbers. Why? Because, topologically speaking, what is the closure of a set? The collection of limit points of that set, right? So in particular, when we let our topological space be $\mathbb{R}$, the set of real numbers (with the usual topology) and consider the subset $\{x_n\}_{n=1}^\infty$ - some sequence of real numbers,

  • we say $x\in\mathbb{R}$ is a limit point of $\{x_n\}$ if for every $\epsilon >0$ there is an $n\in \mathbb{N}$ such that $|x_n-x|<\epsilon$.
Then in light of our comments above, we can make the analogous statement for a subset of polynomials $\{f_n(x)\}\subset F[x]\subset E[x]$:
  • We say $\alpha\in E$ is an algebraic element (over $F$) if there is an $n\in\mathbb{N}$ such that $f_n(\alpha)=0$**.

Notice there's no need for an approximation by $\epsilon$ in the second bullet. Why? Well, imagine placing a "metric" $d$ on $E[x]$ by $d:E[x]\times E[x]\to E$ via***

 
 

(So intuitively, $f(x)$ is far away from $\alpha$ if $\alpha$ is not a root, but if $\alpha$ is a root of $f(x)$, then $f(x)$ and $\alpha$ are just as close as they can be.) In this way, the distance between an algebraic element and its corresponding polynomial is precisely 0. So in this case there's no need to approximate a distance of zero by an arbitrarily small $\epsilon$-ball - we have zero exactly!

And thus we have stumbled upon another insight into one of the main differences between analysis and algebra: you know the adage -

Analysts like inequalies; algebraists like equalities!

 
 
 
 
 

It would be interesting to see if there's something in the language of category theory which allows one to see that closure of an algebraic field and closure of a topological set really are the same. Now I don't know much about categories, but as one of my classmates recently suggested, we might want to look for a functor from the category of fields to the category of topological spaces such that the operation of closure is equivalent in each. In this case, perhaps it's more appropriate to relate an algebraic closure to the completion of a topological space, as opposed to its closure. Admittedly, I'm not sure about all the details, but I think it's worth looking into!

 

 

Footnotes:

* This is actually a bit deceiving. How we measure "closeness" really depends on the topology of the space we're working on. For example, we can place the ray topology on $\mathbb{R}$ so that the open sets are intervals of the form $(a,\infty)$ for $a\in \mathbb{R}$. Then in the strict definition of a limit point we see that -763 is a limit point of the interval $(0,1)$ even though it's "far away"!

** Okay okay... there's no reason to assume an arbitrary collection of polynomials is countable. I really should write $\mathcal{F}$ for some family of polynomials in which case this statement would read "...if there is some $f\in\mathcal{F}$ such that $f(\alpha)=0$." But bear with me for analogy's sake.

*** I put "metric" in quotes here because as defined $d$ is not a metric in the strict sense of the word. Indeed, we don't have the condition $d(f(x),\alpha)=0$ if and only if "$f(x)=\alpha$" since the latter is like comparing apples and oranges! But it would be interesting to see if we could place looser version of a metric on a polynomial ring. For instance, the way I've defined $d$ here, an open ball centered at $\alpha$ would correspond to all polynomials in $E[x]$ which have $\alpha$ as a root! This idea seems to be related to Hilbert's Nullstellensatz and the Zariski topology.

Reference:

[1] Basic Abstract Algebra. Bhattacharya, P.B., Jain, S.K., and Nagpaul, S.R. 1994 (2nd ed), p. 293