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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!

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  • $\begingroup$ 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... $\endgroup$
    – Steven Stadnicki
    Apr 15, 2016 at 21:15
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    $\begingroup$ Have you learned yet the theorem that the sum, difference, or product of algebraic integers is again an algebraic integer? $\endgroup$
    – Barry Smith
    Apr 15, 2016 at 21:15
  • $\begingroup$ @BarrySmith Yes I have. I believe I have more reading to do though before I try to continue on this problem. $\endgroup$
    – mathamphetamines
    Apr 15, 2016 at 21:24
  • $\begingroup$ @AlexWertheim We have not yet discussed nor do I believe we will cover finite extensions. $\endgroup$
    – mathamphetamines
    Apr 15, 2016 at 21:25
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    $\begingroup$ 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. $\endgroup$
    – Lubin
    Apr 15, 2016 at 22:29

3 Answers 3

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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.

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    $\begingroup$ Thank you for your help. A clear and concise explanation in which even my simple mind can comprehend! $\endgroup$
    – mathamphetamines
    Apr 15, 2016 at 21:45
  • $\begingroup$ You are welcome. I'm glad that the input is helpful. $\endgroup$
    – Ennar
    Apr 15, 2016 at 21:46
  • $\begingroup$ Is this the same concept of integral dependence that Atiyah MacDonald discusses in Chapter 5? $\endgroup$
    – mathamphetamines
    Apr 15, 2016 at 21:48
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    $\begingroup$ @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. $\endgroup$
    – Ennar
    Apr 15, 2016 at 21:54
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    $\begingroup$ @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. $\endgroup$
    – Barry Smith
    Apr 15, 2016 at 23:13
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$(7-\sqrt{13})/2$ 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.

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

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  • $\begingroup$ 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. $\endgroup$
    – mathamphetamines
    Apr 15, 2016 at 21:36

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