2
$\begingroup$

I'm trying to close some outstanding gaps in my understanding of a technique for proving an angle is irrational. While the rest of the details of the proof aren't important for my question, the short version of the proof is that if $\theta$ is a rational angle then $2 \cos(\theta)$ is an algebraic integer.

My gap in understanding has to do with proving something is an algebraic integer. Suppose $2 \cos(\theta) = \sqrt{2} - (1/2)$. If I start out with $x = \sqrt{2} - (1/2)$ and manipulate it to eliminate the $\sqrt{2}$, I wind up with $4x^2 +4x - 7 = 0$. This also happens to be the same polynomial that Wolfram Alpha tells me is the minimal polynomial for $\sqrt{2} - (1/2)$.

Now I look at $4x^2 +4x - 7$ and see the leading $4$ and conclude that it isn't monic so $\sqrt{2} - (1/2)$ isn't an algebraic integer.

But when I check Wikipedia and many other sources I see the definition of the minimal polynomial is:

The minimal polynomial of $\alpha$ is the monic polynomial of least degree among all polynomials in $F[x]$ having $\alpha$ as a root

This seems to imply that for any value $\alpha$ you can find a monic polynomial.

My hunch is that the difference has to do with allowing coefficients to be in the rationals $\mathbb{Q}$ instead of the integers $\mathbb{Z}$ but I'm not certain.

What's going on here? Why do I (and Wolfram Alpha) not get a monic minimum polynomial and why does the minimum polynomial definition seem to suggest I should. Is the difference really just polynomials with coefficients in $\mathbb{Q}$ versus $\mathbb{Z}$ or is there something more subtle going on here?

Finally, is the argument I outlined above sufficient to prove that $2 \cos(\theta) = \sqrt{2} - (1/2)$ implies $\theta$ isn't a rational angle?

$\endgroup$
8
  • 1
    $\begingroup$ Surely you meant: A rational multiple of $\pi$ instead of a rational angle. $\endgroup$
    – Jyrki Lahtonen
    Oct 24, 2016 at 5:32
  • $\begingroup$ Yes $\theta = \frac{2 \pi}{n}$ for any integer $n$. I thought the term "rational angle" was unambiguous. $\endgroup$
    – Brandon Enright
    Oct 24, 2016 at 5:34
  • $\begingroup$ And, yes, the minimal polynomial of $\sqrt2-1/2$ is $m(x)=x^2+x-7/4$. You can always clear the denominators, and get a polynomial with integer coefficients. But because $m(x)\notin\Bbb{Z}[x]$ we conclude that $\sqrt2-1/2$ is not an algebraic integer. It is an algebraic number though.. $\endgroup$
    – Jyrki Lahtonen
    Oct 24, 2016 at 5:35
  • 1
    $\begingroup$ And if $\theta=\pi m/n$ for some integers $m,n$, $\gcd(m,n)=1$, then $$2\cos\theta=\zeta+\overline{\zeta}$$ with $\zeta=e^{2\pi i m/n}$. Here $\zeta$ and $\overline{\zeta}$ are zeros of $x^n-1$, so they are algebraic integers. Therefore so is their sum $2\cos\theta$. Consequently, whenever you show that the minimal polynomial of $2\cos\theta$ over $\Bbb{Q}$ does not have integer coefficients, then $\theta$ is not a rational multiple of $\pi$. $\endgroup$
    – Jyrki Lahtonen
    Oct 24, 2016 at 5:41
  • 1
    $\begingroup$ Oh, $\overline{\zeta}$ is just the complex conjugate $e^{-2\pi i m/n}$. Yes, both $\zeta$ and $\overline{\zeta}$ are $n$th roots of unity, and the rest follows from the familiar identity $e^{ix}=\cos x+i\sin x$. BTW $x^n-1$ is not the minimal polynomial here, but there is a result stating that if a number is a zero of a monic polynomial with integer coefficients, then its (monic) minimal polynomial also has integer coefficients. $\endgroup$
    – Jyrki Lahtonen
    Oct 24, 2016 at 6:02

2 Answers 2

Reset to default
3
$\begingroup$

Basically, the minimal polynomial for $\alpha$ over $\Bbb Q$ is the polynomial $f(x)$ having rational coefficients and smallest degree such that $f(\alpha)=0$.

Problem is, this does not give a unique polynomial: if $f$ satisfies these requirements then so does any (non-zero) constant times $f$. So, if we want to get a single definite minimal polynomial, we have to standardise it somehow. There are two common ways of doing this.

  • Divide by the leading coefficient so that the polynomial is monic, with rational coefficients. In this case, if the coefficients are actually integers, then $\alpha$ is an algebraic integer.
  • Multiply by a greatest common denominator of the coefficients and then divide by any remaining common factor, so that the polynomial has integer coefficients with no common factor. In this case, if the polynomial is monic, then $\alpha$ is an algebraic integer.

Which do you use? Really doesn't matter. It should not be hard to convince yourself that the two methods are equivalent. Sometimes people use the terminology that the first gives the "minimal polynomial of $\alpha$ over $\Bbb Q$" and the second gives the "minimal polynomial of $\alpha$ over $\Bbb Z$".

It appears that Wikipedia is using the former definition and Wolfram Alpha is using the latter.

$\endgroup$
2
  • $\begingroup$ "Divide by the leading coefficient so that the polynomial is monic, with rational coefficients. In this case, if the coefficients are actually integers, then α is an algebraic integer." So in the OP... $x^2 + x + \frac 74$ is not monic and $\sqrt{2} - 1/2$ is not algebraic integer? So the OP had it right? But $2\cos(\pi \theta) = \sqrt{2} - 1/2$ has a rational theta? So the OP's premise, rational angle => algebraic integer is not true? $\endgroup$
    – fleablood
    Oct 24, 2016 at 5:56
  • $\begingroup$ @fleablood As far as I can see, $\theta$ was not specified in the OP. But since $2\cos(\pi\theta)$ is not an algebraic integer, $\theta$ cannot be rational. $\endgroup$
    – David
    Oct 25, 2016 at 0:05
2
$\begingroup$

" the short version of the proof is that if θ is a rational angle then 2cos(θ) is an algebraic integer."

My first answer was completely wrong. Sorry.

But appearently this is not an if and only if statement.

If $2\cos(\pi\theta)$ is an algebraic integer then $ \theta$ is rational.

But the converse is not true. $\theta $ being rational does not imply $2\cos(\pi \theta)$ is an algebraic integer.

So apparently $2\cos(\pi \theta) = \sqrt{2} - \frac 12$ is a counter example.

Thanks to David in the comments for clearing this up.

And I apologize for originally answer a question that simply was not one that had been asked.

$\endgroup$
5
  • $\begingroup$ The result that $2 \cos(\theta)$ is an algebraic integer when $\theta$ is a rational angle can be found many places including arxiv.org/abs/1006.2938 I think the result is originally due to Ivan Niven in the book "Irrational Numbers". $\endgroup$
    – Brandon Enright
    Oct 24, 2016 at 5:33
  • $\begingroup$ Perhaps I'm using "rational angle" wrong? When I say rational angle I mean a rational multiple of $\pi$ not just the argument to the $\cos()$ function is a rational number. $\endgroup$
    – Brandon Enright
    Oct 24, 2016 at 5:37
  • $\begingroup$ I admit I could be wrong. Is arcsin root(2) -1/2 rational? $\endgroup$
    – fleablood
    Oct 24, 2016 at 5:37
  • $\begingroup$ Okay, I'll admit I misunderstood what the OP meant. I assumed s/he meant the angle was constructible. I may remove this post.... $\endgroup$
    – fleablood
    Oct 24, 2016 at 5:40
  • $\begingroup$ @fleablood thank you very much for taking the time to answer. It's appreciated. Sorry I wasn't precise in my wording. $\endgroup$
    – Brandon Enright
    Oct 24, 2016 at 5:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .