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I am willing to give a general audience lecture about prime factorization, and opening towards the lack of unique factorization in the case of e.g. $\mathbb{Q}(\sqrt{5})$. However, I have two issues :

  • how to introduce $\mathbb{Q}(\sqrt{5})$ naturally (for general audience or high-schoolers)? They don't have naturally complex numbers, but maybe I can stay murky about it, or say it is the least we can do to solve $x^2-5=0$ (why is this equation important, though?)
  • are there nice, yet accessible, applications (e.g. geometrically, or in cryptography, etc.) of the unique factorization in such fields?
TheStudent
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  • $\mathbb{Q}(\sqrt{5})$ is a field and hence a UFD :) – tetra Oct 27 '22 at 09:50
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    If forced to talk $K=\Bbb{Q}(\sqrt5)$ specifically, I would make it a story about the golden ration $\phi=(1+\sqrt5)/2$ and think of $K=\Bbb{Q}(\phi)$ instead. Then you still have unique factorization in $\Bbb{Z}[\phi]$. But the units of the ring are $\pm\phi^n,n\in\Bbb{Z}$. Applications? After you have given the audience their fill of the golden ratio, you can mention how it is used in WLAN standard. – Jyrki Lahtonen Oct 27 '22 at 12:25
  • This may be a better source for Golden code. Emanuele chose to denote it by $\theta$ instead :-) – Jyrki Lahtonen Oct 27 '22 at 12:27
  • Note that ${\bf Z}[\sqrt5]$ is not the ring of integers of ${\bf Q}[\sqrt5]$. If you want an example where ${\bf Z}[\sqrt d]$ is the ring of integers of ${\bf Q}[\sqrt d]$, then you might want to use $d=2$ (with unique factorization), or $d=6$ (without unique factorization). – Gerry Myerson Oct 28 '22 at 09:31

2 Answers2

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This is how I would do it.

  • Explain what it meant to be prime or irreducible.

  • Tell them, preferably with a justification, that there are numbers, such as $\sqrt2, \sqrt3, \cdots$, that are not rational. But be cautious not to go beyond positive square roots.

  • That gives you a way to expand the horizon by adding these extra numbers, one at a time. Like $\mathbb{Z}[\sqrt2], \mathbb{Q}[\sqrt2],$
    Also, $\mathbb{Z}[\sqrt6]\subset\mathbb{Z}[\sqrt2][\sqrt3]$ while $\mathbb{Q}[\sqrt6]=\mathbb{Q}[\sqrt2][\sqrt3]$

  • Tell them, while this gives you some extra freedom (I would show them that $\mathbb{Q}[\Delta]$ is where the solutions of a rational quadratic equation with discriminant $\Delta$ lives or casually mention something like this), sometimes you lose some nice algebraic properties. For example $\mathbb{Z}[\sqrt5]$ is not a unique factorization domain. $$(3-\sqrt5)(3+\sqrt5)=2^2$$

I wouldn't talk about complex numbers or Gaussian integers at all. Later when they see complex numbers, they will remember your talk.

Bumblebee
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  • ${\bf Z}[\sqrt2][\sqrt3]$ is not ${\bf Z}[\sqrt6]$. – Gerry Myerson Oct 28 '22 at 09:27
  • @GerryMyerson: Thanks for the corrections. I shouldn't have done that careless edit. – Bumblebee Oct 28 '22 at 09:46
  • Nice points! However, what about, for example, solving actual equations? For example, to find all the integer solutions of some equation, you interpret the equation over $\mathbb Q[\sqrt 5]$ instead of $\mathbb Q$ and use some properties? I mean, isn't that the reason we invent field extensions after all? – Sarvesh Ravichandran Iyer Oct 28 '22 at 09:48
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Why are you choosing this specific topic of $\mathbf Q(\sqrt{5})$, which really sounds like it is about $\mathbf Z[\sqrt{5}]$? Students are going to find the notation intimidating.

And is the audience high school students or a true general audience? If it's a general audience, then your topic is too sophisticated. Find something simpler, like puzzles (15-puzzle, Rubik's cube).

If the audience is high school students who have already taken algebra and trigonometry, and if the goal is to discuss a setting where there is failure of unique factorization, then use an example they already know but had not thought much about: the system of trigonometric polynomials, which are things like the function $$ \sin^2(x)\cos^3(x) - 4\sin(x)\cos^2(x) + 2\cos(x) - 9. $$ It's a polynomial in $\sin x$ and $\cos x$. You can add and multiply such functions, and thanks to $\sin^2(x) + \cos^2(x) = 1$ you can write the same expression in multiple ways. In fancier language, these are precisely the finite Fourier series (linear combinations of $\sin(mx)\cos(nx)$ for $m, n \geq 0$), but you don't have to mention that.

Anyway, it turns out that $\sin x$ and $\cos x$ are irreducible: they're not the product of two other nonconstant trigonometric polynomials. And also $1 \pm \sin x$ and $1 \pm \cos x$ are irreducible. This implies that the familiar relation from trigonometry $$ \sin^2 x = 1 - \cos^2 x = (1+\cos x)(1 - \cos x) $$ is an example of nonunique irreducible factorization: $1 + \cos x$ and $1 - \cos x$ are not scale multiples of $\sin x$ (why? neither of the functions $1 + \cos x$ and $1 - \cos x$ vanish at all the values where $\sin x = 0$, so $1\pm \cos x$ isn't a scalar multiple of $\sin x$). Here is Hale Trotter's short paper about this example: http://alpha.math.uga.edu/~pete/trotter.pdf.

KCd
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