Sum of two algebraic integers is also an algebraic integer

How do you show that the sum of two algebraic integers is also an algebraic integer? Also, how do we find the minimal polynomial of the sum of the two algebraic integers given the minimal polynomials of both algebraic integers? For example, let $a=2+\sqrt{3}$ and $b=\sqrt[3]{3}$ The minimal polynomial of $a$ is $x^2-4x+1$ and for $b$ is $x^3-3$ , now how would we find the minimal polynomial of $a+b$ ? Furthermore, how would we prove that $a+b$ is also an algebraic integer?

#Algebra #NumberTheory #Goldbach'sConjurersGroup #algebraicintegers

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

http://www.incertia.net/blog/some-proofs-regarding-symmetric-polynomials-and-algebraic-numbers/

Will Song - 7 years, 6 months ago

It is easier to prove that $a+b$ is an algebraic integer. Let $f(x)$ be a (not necessarily minimal) integer polynomial such that $f(a) = 0$ , with roots $a_i$ . Let $g(x)$ be a (not necessarily minimal) integer polynomial such that $g(b) = 0$ , with roots $b_j$ . Then, consider the polynomial

$h(x) = \prod ( x - a_i - b_j)$

It remains to show that the coefficients are integers. If you expand it, you can use Veita's formulas (though tediously). This is motivated from Galois theory which tells us that the algebraic integers form a ring.

The reason why $h(x)$ need not be the minimal polynomial, is because things might cancel out. For example, if $a = \sqrt{2}$ with minimal polynomial $x^2 - 2$ and $b = \sqrt{2}$ with minimal polynomial $x^2 -2$ , then the above construction gives us $h(x) = x^4 - 4 x^2$ , while $a+b= 0$ which has minimal polynomial of $x$ . I don't think a general solution exists, but I might be wrong.

Chung Kevin - 7 years, 6 months ago

This keeps only proves that $a + b$ is an algebraic number because $h$ is not necessarily an integer polynomial. This is because the symmetric sums of $a_i, b_i$ are rational from Vieta, not integral. If $f, g$ were monic, then this would be entirely correct.

Will Song - 7 years, 6 months ago

The definition of an algebraic integer, is that it is a root of a monic polynomial. I should have mentioned that, as it wasn't stated in the post.

Of course, there are many other ways of understanding algebraic integers, especially if you have a knowledge of field theory / abstract algebra. I wasn't certain how much he knew, and I opted to go for the most basic (and crude / brute force) approach. For example, if you used the definition that $\alpha$ is an algebraic integer if $\mathbb{Z} [\alpha]$ is a finitely generated $\mathbb{Z} -$ module, the result is immediate. However, that wouldn't have been helpful to many people.

Chung Kevin - 7 years, 6 months ago

Thanks for the solution! Can you elaborate on the use of Vieta's formulas to prove that all the coefficients of $h(x)$ are integers?

Justin Stevens - 7 years, 6 months ago

We can use the fundamental theorem of symmetric polynomial with the same polynomial defined by chung repeatedly to prove in general.

Vinayak Jha - 4 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$