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I have been doing some exercises about finite fields lately and I think I've obtained some understanding of what they are. What seems to be missing though is some kind of picture. Learning to work with complex numbers seems pretty much equivalent to learning what the field operations do on the complex plane geometrically. At least, it makes things much, much easier. Is there a way to imagine finite fields and their algebraic closure geometrically?

I was thinking that perhaps I should look at $F_p$ as the set $\mathbb C_p$ of $p$-th roots of $1$ in $\mathbb C.$ But it doesn't seem to work very well. The additive structure of $F_p$ translates well (via an isomorphism) to the multiplicative structure of $\mathbb C_p.$ But transporting the multiplicative structure $F_p$ to $\mathbb C_p$ via the same function gives something I can't really visualize. It's the operation $\star$ given by the following formula:

$$\left(\cos\frac{2k\pi}{p}+i\sin\frac{2k\pi}{p}\right)\star \left(\cos\frac{2l\pi}{p}+i\sin\frac{2l\pi}{p}\right)=\left(\cos\frac{2kl\pi}{p}+i\sin\frac{2kl\pi}{p}\right).$$

But even if it's the right way to do it, it still doesn't immediately give a way of seeing $F_{p^n}$ there too, let alone the algebraic closure $\overline{F_p}.$ And of course, it would be good to have an interpretation of $\overline{F_p}$ in which all $F_{p^n}$ for $n\in \mathbb N$ could be easily distinguished.

So my question is

Are there any useful ways of interpreting the finite fields and their algebraic closures geometrically? If so what are they?

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en.wikipedia.org/wiki/… –  Qiaochu Yuan Mar 26 '12 at 15:31
    
@QiaochuYuan Thank you for the link. I've only heard of projective spaces in the context of complex analysis but I didn't really understand it. Unfortunately, I don't understand whether (and how exactly) what's on that Wikipedia page answers my question. I see that what they call the Fano plane is somehow connected to the field $F_2.$ They say it can be seen as the set of non-zero vectors in the three dimensional vector space over $F_2.$ Can I see $F_8$ there? –  user23211 Mar 26 '12 at 15:59
    
You can define projective spaces over any finite field, but of course they'll be bigger. In projective spaces over a field one can do projective geometry, and as it turns out this geometry is enough to recover the addition, multiplication, and division on the underlying field. I don't know a good reference for this, though. –  Qiaochu Yuan Mar 26 '12 at 16:05
    
@QiaochuYuan I see. So this must be the right way do look at this then. If it's imossible to explain it in full generality here, I will accept an answer giving a practical way of using these spaces to visualize a finite field, without the underlying theory. I think I might have trouble absorbing this much theory anyway... –  user23211 Mar 26 '12 at 16:17
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@ymar: See M. K. Bennett's book Affine and Projective Geometry, especially Chapter 4. –  KCd Mar 26 '12 at 22:15

2 Answers 2

up vote 3 down vote accepted

The short answer is that the visualizations of finite fields are different from what you seem to want. I will elaborate a bit, though. It is not clear that anything I suggest is really a "visualization" or a "geometric point of view". But they do provide a way of getting your hands on a finite field starting from structures inside the complex numbers.

Let's first look at the prime field $F_p=\mathbf{Z}/p\mathbf{Z}$. As you observed, we can faithfully map the additive structure of $F_p$ to the unit circle via the (obviously well-defined) map $$ e:m+p\mathbf{Z}\mapsto e^{2m\pi i/p}. $$ This is a homomorphism from the additive group of $F_p$ to the multiplicative group $\mathbf{C}^*$ of non-zero complex numbers, or $$ e(x+y)=e(x)e(y)$$ for all $x,y\in F_p$. In other words $e$ is what is called an additive character of $F_p$. As you observed, it doesn't really reflect the multiplicative structure of $F_p$ at all. This was to be expected, as we used the multiplicative structure of $\mathbf{C}^*$ to represent the additive structure of $F_p$, so naturality was broken down. Characters and character sums do play a role in the study of number theoretic properties of finite fields.

If we want to take a geometric viewpoint of the prime field $F_p$, $p$ a prime, then I would stick to the number line $\mathbf{R}$. Except that I really need restrict to the integers $\mathbf{Z}$, and then go to the quotient ring, or equivalently identify the integers within a coset of $p\mathbf{Z}$.

You can identify some finite fields by similarly identifying cosets of the ideal $p\mathbf{Z}[i]$, where $$\mathbf{Z}[i]=\{a+bi\mid a,b\in\mathbf{Z}\}.$$ Geometrically you identify here points $a+bi$ and $a'+b'i$, iff $p\mid (a-a')$ and $p\mid (b-b')$, or instead of "wrapping the line around a circle" you "wrap the plane around a torus". An immediate word of warning. This only works, if $p\equiv-1\pmod4$. So you can construct $F_9$ in this wasy, but you cannot construct $F_4$ or $F_{25}$. One of the reasons why this construction does not work in those cases is that while the resulting quotient ring has the desired number of elements, it is not a field because it has zero divisors. For example, $(1+i)^2=2i\equiv 0 \pmod{2\mathbf{Z}[i]}$ in the ring $\mathbf{Z}[i]/2\mathbf{Z}[i]$, and $(2+i)(2-i)=5\equiv 0$ in the ring $\mathbf{Z}[i]/5\mathbf{Z}[i]$ (I plead guilty to the standard abuse of identifying a number with the coset it represents, so I really meant to write that the square of the coset $(1+i)+2\mathbf{Z}[i]$ is equal to zero in the first example above.

To get fields like $F_4$ and $F_{25}$ we need quotient rings of other lattices of complex numbers. Let $\omega=e^{2\pi i/3}=(-1+i\sqrt3)/2$. Then $\mathbf{Z}[\omega]=\{a+b\omega\mid a,b\in\mathbf{Z}\}$ is a ring (a subring of the complex numbers), and we get (requires some checking) $$ F_4=\mathbf{Z}[\omega]/2\mathbf{Z}[\omega]\qquad\text{and}\qquad F_{25}=\mathbf{Z}[\omega]/5\mathbf{Z}[\omega]. $$ But a construction for all the finite fields (let alone the algebraic closure) in this way is a bit trickier. The field $F_{p^n}$ is an $n$-dimensional space over $F_p$. A lattice in the complex plane is naturally (at most) a two-dimensional thing, so when $n>2$ it gets dirty. No more wrapping this nicely. It can be done by using certain subrings of $\mathbf{C}$ namely rings of integers of an algebraic number field, but some algebraic number theory is needed. See this question for a description.

We can think of a finite field using roots of unity. But the (non-zero) elements of $F_q$, $q=p^n$, $p$ a prime, are roots of unity of order that is a factor of $q-1$. So, if we want to represent ther multiplicative structure, we need to use those roots of unity. For example, we can also view $F_7$ as the quotient ring $\mathbf{Z}[\omega]/(3+\omega)\mathbf{Z}[\omega]$. When we do that the non-zero elements of $F_7$ are represented by the cosets of the sixth roots of unity, i.e. the cosets $(-\omega)^j+(3+\omega)\mathbf{Z}[\omega]$. This is similar in spirit to the discussion in the linked question.

Things become more uniform, when instead of using the complex numbers we use the $p$-adic integers $\mathbf{Z}_p$ (not to be confused with the residue class ring $\mathbf{Z}/p\mathbf{Z}$. If $q=p^n$, and $\zeta$ is a root of unity of order $q-1$, then we have an isomorphism $$ F_q=\mathbf{Z}_p[\zeta]/p\mathbf{Z}_p[\zeta] $$ between the finite field $F_q$ and the quotient ring of an extension ring of the $p$-adic integers. That's a rather different animal, and this last paragraph may be meaningless to you unless you have the right background. The $p$-adic integeres themselves cannot be readily visualized geometrically, because the metric there is very weird.

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I'm afraid this does not help at all in seeing $F_p$ inside $F_{p^2}$ (much less the other inclusions), and the algebraic closure has really dropped out of sight. –  Jyrki Lahtonen Mar 26 '12 at 16:32
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Concerning the last sentence: the $p$-adic integers definitely can be visualized geometrically, as a subset of C with the induced topology (but its ring structure doesn't align with that in C at all). Take a look at Al Cuoco's article "Visualizing the p-adic integers," Amer. Math. Monthly, 98 (1991), 355-364. –  KCd Mar 26 '12 at 22:21
    
Thank you Jyrki and @KCd –  user23211 Mar 28 '12 at 1:11

My own idiosyncratic way of looking at these things: Remember that the multiplicative group of ${\mathbb{F}}_{p^n}$ is cyclic, so you could, if you wished, model this also as a group of roots of unity in $\mathbb{C}$. This helps tremendously in doing hand computations in the finite field, by the way, since once you find a generator (“primitive root”) and write down all its powers, you have a nice log table for quick multiplication. At any rate, these finite cyclic groups all fit together very nicely, so that the multiplicative group of the algebraic closure of your finite field of characteristic $p$ is (noncanonically) isomorphic to the group of all $m$-th roots of unity in $\mathbb{C}$ for $m$ ranging through the integers prime to $p$.

Edit: I originally said that the group in question was also isomorphic to ${\mathbb{Q}}_p/{\mathbb{Z}}_p$, and this was quite wrong.

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Thank you for your answer. What does noncanonically isomorphic mean? –  user23211 Mar 26 '12 at 21:32
    
“Noncanonical” in this case means that among the (†) uncountably many isomorphisms that exist, there is nothing to prefer one over another. (†) That there is even one depends on a weak kind of Axiom of Choice; but once you have one, you see that the set of all is a free ${\mathbb{Z}}_p$-module of rank one. –  Lubin Mar 29 '12 at 4:50
    
May I ask for a reference? I would like to see a proof of this, because there are still things that are new to me in what you said. If I had a longer text I could work with it. –  user23211 Mar 29 '12 at 7:27
    
For a reference, I would always give Lang's Algebra. Send me an e-mail, and we can discuss your question much more fully. But to go into detail in comments here seems inefficient. –  Lubin Apr 1 '12 at 23:51
    
I see that I blundered in trying to describe the set of isomorphisms from the multiplicative group in question to ${\mathbb{C}}^*$. It’s far more complicated than what I said in my first comment. (Apologies!) –  Lubin Apr 27 '12 at 5:55

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