# A sum of irrational numbers is an algebraic integer

I am asked to show that $$\sqrt{2} + \sqrt[3]{5} - \sqrt{17} \Big(\frac{7 - \sqrt{13}}{2} \Big)$$ is an algebraic integer.

$\textbf{Definition:}$ An algebraic integer is the root of a monic polynomial with integer coefficients.

So we start by letting $$x = \sqrt{2} + \sqrt[3]{5} - \sqrt{17} \Big(\frac{7 - \sqrt{13}}{2} \Big)$$ and then raise both sides to some power. I tried squaring both sides, but that didn't seem to help, nor did I think it would get rid of the cube root of five. My next guess is to raise both sides to the $6th$ power. Before I try this by hand, I entered it into Wolfram alpha, and found that the minimal polynomial was a $24$th degree polynomial.

Is this method I am trying correct? What is the trick that I am missing? Any help or advice would be greatly appreciated. Thank you for your help!

• If you need to find a suitable polynomial 'by hand' then you can peel off: $x-\sqrt{2}+\sqrt{17}(\frac12(7-\sqrt{13}))=\sqrt[3]{5}$, so you can cube both sides of this, etc. This gets painful in a hurry, though. Instead, you might consider what operations algebraic integers are closed under... Apr 15, 2016 at 21:15
• Have you learned yet the theorem that the sum, difference, or product of algebraic integers is again an algebraic integer? Apr 15, 2016 at 21:15
• @BarrySmith Yes I have. I believe I have more reading to do though before I try to continue on this problem. Apr 15, 2016 at 21:24
• @AlexWertheim We have not yet discussed nor do I believe we will cover finite extensions. Apr 15, 2016 at 21:25
• It is possible, though hardly instructive, to find a monic polynomial over $\Bbb Z$ with your number as a root. Better by far to use the approach of @Ennar and Robert Israel. Apr 15, 2016 at 22:29

Note that algebraic integers form a ring, so since $\sqrt 2$, $\sqrt[3]5$ and $\sqrt{17}$ are obviously algebraic integers, it is enough to show that $\alpha=\frac{7-\sqrt{13}}2$ is an algebraic integer.

Since we have $7-2\alpha = \sqrt{13}$, by squaring it follows that $4\alpha^2-28\alpha + 36 = 0$. Dividing by $4$ we get that $\alpha$ is a root of monic $x^2 - 7x + 9$, thus an algebraic integer. The claim follows.

• Thank you for your help. A clear and concise explanation in which even my simple mind can comprehend! Apr 15, 2016 at 21:45
• You are welcome. I'm glad that the input is helpful. Apr 15, 2016 at 21:46
• Is this the same concept of integral dependence that Atiyah MacDonald discusses in Chapter 5? Apr 15, 2016 at 21:48
• @mathamphetamines, currently I don't have a copy at my disposal, but the key is that $\alpha$ is algebraic integer if and only if $\mathbb Z[\alpha]$ is finitely generated abelian group. Using this, it's easy to show that algebraic integers do form a ring (i.e. are closed under sum and multiplication) and this is all that is needed for this answer to be formally correct. Apr 15, 2016 at 21:54
• @mathamphetamines This is a special case of integral dependence. Within the complex numbers, a number is integral over $\mathbb{Z}$ (i.e., satisfies an equation of integral dependence over $\mathbb{Z}$) if and only if it is an algebraic integer. Apr 15, 2016 at 23:13

$(7-\sqrt{13})/2$ is a root of the quadratic $x^2 - 7 x + 9$, so it is an algebraic integer. Of course $\sqrt{2}$, $\sqrt[3]{5}$, $\sqrt{17}$ are algebraic integers. The rest is taken care of by the fact that the algebraic integers form a ring.

• Awesome. Thank you very much. That makes more sense, especially since I don't have a clue about finite extensions. Apr 15, 2016 at 21:40

Hint: Observe that you have $\sqrt{2}$, $\sqrt{13}$, $\sqrt{17}$, and $\sqrt[3]{5}$ in the expression. Each of these (individually) gives an extension of degree $2$, $2$, $2$, and $3$, respectively. The product of these is $24$.

Therefore, we only need to look at: $1,x,x^2,\cdots,x^{24}$. If we treat each $x^i$ as a linear combination of products of $\sqrt{2}$, $\sqrt{13}$, $\sqrt{17}$, and $\sqrt[3]{5}$, we can use linear algebra to find the appropriate coefficients to have a nontrivial solution (polynomial).

This is not the most efficient solution, but it will work. A more computational solution would use resultants (but that's more algebraic geometry).

• So finite extensions is the most efficient way to tackle this problem? The linear algebra "trick" you mentioned was shown to us last lecture. Thank you for the hint. Apr 15, 2016 at 21:36