Is $(1 + \sqrt{17})/(2i\sqrt{19})$ an algebraic integer?

Question

I'm learning the basics of algebraic number theory for the first time and I'm confused about a simple question: how can I decide if the number $$\alpha = \frac{1 + \sqrt{17}}{2i\sqrt{19}}$$ is an algebraic integer?

I know $\alpha$ is an algebraic number (since $\overline{\mathbb{Q}}$ is a field), so I thought of using the following criterion:

An algebraic number is an algebraic integer iff its minimum polynomial over $\mathbb{Q}$ has coefficients in $\mathbb{Z}$. (Source: Algebraic number theory and Fermat's last theorem by Stewart and Tall.)

But how can I compute its minimal polynomial? Not knowing how to do this, I though of using another criterion:

If a complex number $\theta$ is a root of a monymonial whose coefficients are algebraic integers, then $\theta$ is an algebraic integer (Stewart and Tall).

By inspection, I found out that $\alpha$ is a root of $t^2 +it/\sqrt{19} + 4/19$, but that does not help since $(9 + i)/38$ is not an algebraic integer. I'm sorry if this is too basic, but I would appreciate any help.

Answer 1

$\alpha$ is not an algebraic integer. As you mentioned, one way to check this would be to show that its minimal polynomial has integer coefficients. This would be the polynomial with roots: $$\frac{1 \pm \sqrt{17}}{\pm 2i\sqrt{19}}$$ since these are the Galois conjugates of $\alpha$. You can expand this polynomial as $x^4 + \tfrac{9}{19}x^2 + \tfrac{16}{361}$. The computation isn't too awful since the polynomial is even (because the negative of a root is also a root).

However, a simpler way to conclude this would be to use the norm of $\mathbb{Q}(\sqrt{17}, \sqrt{-19})/\mathbb{Q}$ directly. Observe that the norm of the denominator is divisible by $19$ while the norm of the numerator is the square of $(1+\sqrt{17})(1-\sqrt{17}) = -16$, which is not divisible by $19$. Thus, the norm of $\alpha$ is not a rational integer, meaning $\alpha$ is not an algebraic integer. (Note that this is detected by the constant term in the minimal polynomial of $\alpha$ being a non-integer)

Answer 2

There are clever approaches, but as always the way we can just compute a problem to death without thinking is with linear alegbra.

It's clear that $\alpha$ lives in the finite extension $\mathbb{Q}(\sqrt{17},\sqrt{19},i)$, so we can write "multiplication by $\alpha$" in terms of the basis for this extension viewed as a $\mathbb{Q}$-vector space -- this is (extremely) tedious, but very doable. Indeed, if we rationalize $\alpha$ as $\frac{i \sqrt{19} + i \sqrt{19}\sqrt{17}}{-38}$ and work with the basis $\{1, \sqrt{17}, \sqrt{19}, i, \sqrt{17}\sqrt{19}, \sqrt{17} i, \sqrt{19} i, \sqrt{17} \sqrt{19} i \}$ we can just do the multiplications and see that multiplication by $\alpha$ corresponds to the action of the following matrix on row vectors:

$$ \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & \frac{-1}{38} & \frac{-1}{38} \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{-17}{38} & \frac{-1}{38} \\ 0 & 0 & 0 & \frac{-1}{2} & 0 & \frac{-1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{38} & 0 & \frac{1}{38} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{-17}{2} & 0 & \frac{-1}{2} & 0 & 0 \\ 0 & 0 & \frac{17}{38} & 0 & \frac{1}{38} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{17}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$

Said another way, the $n$th row of this matrix is $\alpha$ times the $n$th basis vector, written in terms of the basis. But now to compute the minimal polynomial of $\alpha$ is just to compute the minimal polynomial of this matrix, and this is well studied (see, eg, the wikipedia article). If you do this computation, you'll find the minimal polynomial of this matrix is $x^4 + \frac{9}{19} x^2 + \frac{16}{361}$, which is thus also the minimal polynomial of $\alpha$.

These computations should really be done using a computer algebra package like sage. It quickly tells us:

sage: alpha = (I * sqrt(19) + I * sqrt(19) * sqrt(17))/(-38)
sage: M = matrix([[0,0,0,0,0,0,-1/38, -1/38],[0,0,0,0,0,0,-17/38,-1/38],[0,0,0,-1/2,0,-1/2,0,0],[0,0,1/38,0,1/38,0,0,0],[0,0,0,-17/2,0,-1/2,0,0],[0,0,17/38,0,1/38,0,0,0],[1/2,1/2,0,0,0,0,0,0],[17/2,1/2,0,0,0,0,0,0]])
sage: M.minpoly()
x^4 + 9/19*x^2 + 16/361
sage: alpha.minpoly()
x^4 + 9/19*x^2 + 16/361

In particular, the minimal polynomial does not have integer coefficients, so this is not an algebraic integer.

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I hope this helps ^_^

Answer 3

To compute the minimal polynomial, let $\alpha = \frac{1+\sqrt{17}}{2i\sqrt{19}}$, we have

$$\alpha^2 = \frac{18+2\sqrt{17}}{-4\cdot 19}\Rightarrow 76\alpha^2+18=-2\sqrt{17}\Rightarrow (76\alpha^2+18)^2=4\cdot 17 = 68$$

Hence $\frac{(76x^2+18)^2-68}{(76)^2}$ is the candidate minimal polynomial, but to know this for sure we have to show it's irreducible or equivalently $[\mathbb Q[\alpha]:\mathbb Q]=4$ (to be briefly justified at the end).

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Here is a more conceptual approach without much calculation.

Algebraic integers are closed under addition, multiplication and Galois action.

Hence if $\alpha=\frac{1+\sqrt{17}}{2i\sqrt{19}}$ is one, so are $2\alpha = \frac{1+\sqrt{17}}{\sqrt{-19}}$ and its Galois conjugate $\sigma(2\alpha)=\frac{1-\sqrt{17}}{\pm\sqrt{-19}}$, and finally

$$\frac{1+\sqrt{17}}{\sqrt{-19}}\frac{1-\sqrt{17}}{\pm\sqrt{-19}} = \pm\frac{16}{19}\in\mathbb Q$$

But a rational number is an algebraic integer iff it's a (rational) integer.

That is, there must exist a $\sigma$ such that $\sigma(\sqrt{17})=\sqrt{17}$, and though we don't know what $\sigma(\sqrt{-19})$ is, it must be one of $\pm \sqrt{-19}$. And in either case, the above argument works.

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While we don't really need to compute the Galois group, to know e.g. there is a Galois element $\sigma\in \mathrm{Gal}(\mathbb Q(\sqrt{17}, \sqrt{-19})\mid\mathbb Q)$ such that $\begin{cases} \sigma(\sqrt{17})=-\sqrt{17} \\ \sigma(\sqrt{-19}) = \sqrt{-19}\end{cases}$

we still summarize:

• $[\mathbb Q(\sqrt{17}):\mathbb Q]=2$ because $x^2-17$ is irreducible by Eisenstein.

• $[\mathbb Q(\sqrt{17}, \sqrt{-19}):\mathbb Q(\sqrt{17})]=2$, because $\sqrt{-19}\not\in\mathbb R$ hence not in $\mathbb Q(\sqrt{17})$.

• Hence $[\mathbb Q(\sqrt{17}, \sqrt{-19}):\mathbb Q]=[\mathbb Q(\sqrt{17},\sqrt{-19}):\mathbb Q(\sqrt{17})][\mathbb Q(\sqrt{17}):\mathbb Q]=4$

• Now it's clear that $\mathrm{Gal}(\mathbb Q(\sqrt{17}, \sqrt{-19})\mid\mathbb Q)\simeq S_2\times S_2$.

• Check that $\sigma$ fixes $\alpha$ iff it fixes both $\sqrt{17}$ and $\sqrt{-19}$, so $[\mathbb Q(\alpha):\mathbb Q] = 4$.