p-th root of rational numbers 2

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

p-th root of rational numbers 2

G H J

Reveal topic tags

algebra proposed

L

G H BBookmark kLocked kLocked NReply

Source: Iran 3rd round 2012-Algebra exam-P5

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1347 posts

#1 h Sep 20, 2012, 9:34 AM • 4 Y

Y by Amir Hossein, mahanmath, Adventure10, Mango247

Let $p$ be an odd prime number and let $a_1,a_2,...,a_n \in \mathbb Q^+$ be rational numbers. Prove that $\mathbb Q(\sqrt[p]{a_1}+\sqrt[p]{a_2}+...+\sqrt[p]{a_n})=\mathbb Q(\sqrt[p]{a_1},\sqrt[p]{a_2},...,\sqrt[p]{a_n}).$

This post has been edited 1 time. Last edited by goodar2006, Sep 30, 2012, 5:16 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2326 posts

yugrey

#2 h Sep 20, 2012, 8:40 PM • 1 Y

Y by Adventure10

What does it mean to take $Q$ of some number of set of numbers?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2903 posts

#3 h Sep 21, 2012, 1:54 AM • 1 Y

Y by Adventure10

$F(\alpha)$ denotes the minimal ring containing $F, \alpha$ and $\frac{1}{\alpha}$ . $F(x_1,x_2,...,x_n)$ just means $F(x_1)(x_2)...(x_n)$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

93 posts

#4 h Mar 15, 2024, 6:58 PM

Y by

Merge $a_i$ which are the same up to powers of $p$ , and WLOG assume $a_i$ are integers. We generalize to arbitrary coefficients of the $\sqrt[p]{}$ sum.
Let $t = b_1\sqrt[p]{a_1}+b_2\sqrt[p]{a_2}+\dots+b_n\sqrt[p]{a_n}$ has a minimal polynomial with degree $p^p$ .

Claim: No non-trivial linear relation exists between $\sqrt[p]{a_1}, \sqrt[p]{a_2}, \sqrt[p]{a_3}, \dots, \zeta$ where $\zeta$ is the $p$ th primitive root of unity.
Proof. Omitted. $\blacksquare$

Claim: In ${\mathbb Q}(\sqrt[p]{a_1}, \sqrt[p]{a_2}, \sqrt[p]{a_3}, \dots, \zeta)$ , the map $\sigma$ that maps $\sqrt[p]{a_i}$ to $\zeta^{c_i} \cdot \sqrt[p]{a_1}$ is a well defined automorphism.
Proof. Expand it all out to get the result. $\blacksquare$

Claim: $t = b_1\sqrt[p]{a_1}+b_2\sqrt[p]{a_2}+\dots+b_n\sqrt[p]{a_n}$ has a minimal polynomial with degree $p^p$ .
Proof. Apply the $\sigma$ automorphisms which map to giving galois conjugates of $\zeta^{c_1} b_1\sqrt[p]{a_1}+ \zeta^{c_2} b_2\sqrt[p]{a_2} + \dots + \zeta^{c_n} b_n\sqrt[p]{a_n}$ . $\blacksquare$
Now, note that ${\mathbb Q}(t) \subset {\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n})$ .
We can compute that $[{\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n}) : {\mathbb Q}] = p^n$ . However, since $t$ has a minimal polynomial of degree of $p^n$ , it follows that $1, t, t^2, \dots, t^{p^n-1}$ is a basis of $[{\mathbb Q}(t) \mid {\mathbb Q}]$ .
It follows by tower law that $[{\mathbb Q}(\sqrt[p]{a_1},\sqrt[p]{a_2},\dots,\sqrt[p]{a_n}) : {\mathbb Q}(t)] = \frac{p^n}{p^n} = 1$ which gives the result.

This post has been edited 1 time. Last edited by TheHazard, Mar 15, 2024, 6:59 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a