Addition Of Algebraic Element

Date: 2023/06/25
Last Updated: 2023-11-26T21:30:00.000Z
Categories: Math
Tags: Math, Galois Theory, Field Theory, Notes
Read Time: 4 minutes

This is some generalization of a question.

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::: question Question 1. Proof that $\sqrt{1} + \sqrt{2} + \sqrt{3} \dots + \sqrt{n}$ for is irrational. :::

However, if we consider the linear combination of $\sqrt{1}$ , $\sqrt{2}$ , $\sqrt{3}$ $\dots$ $\sqrt{n}$ over $\mathbb{Q}$ , then it is likely that we will get a similar result as we kick out some special cases.

::: question Question 2. Given positive integers, $a_1 < a_2 < \dots < a_n$ , such that $\sqrt{\frac{a_{j}}{a_{i}}} \notin \mathbb{Q}$ for all and $\sqrt{a_{i}} \notin \mathbb{Q}$ for all . Also given $\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{Q}$ . Prove that $\sum_{i=1}^{n} \lambda_{i} \sqrt{a_{i}} \in \mathbb{Q}$ if and only if $\lambda_{1} = \lambda_{2} = \dots = \lambda_{n} = 0$ . :::

::: proof Proof. Suppose this stand for n=k , then it is suffice to prove n=k+1 .

Suppose there exist $\lambda_{1}, \lambda_{2} \dots \lambda_{k+1} \in \mathbb{Q}$ such that $\sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}} = q \in \mathbb{Q}$

If $\lambda_{j} = 0$ , then $-\lambda_{j}\sqrt{a_{j}} + \sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}} = q \in \mathbb{Q}$ , and we are done according to the assumption.

If $\lambda_{j} \neq 0$ for all , then we have $\sum_{i=1}^{k} \lambda_{i}\sqrt{a_{i}} = q - \lambda_{k+1}\sqrt{a_{k+1}}$

Clearly, $q - \lambda_{k+1}\sqrt{a_{k+1}}$ is a root of a irreducible quadratic polynomial $g \in \mathbb{Q}[x]$ . And the polynomial must have another root which is $q + \lambda_{k+1}a_{k+1}$ .

Also, the polynomial $f = \prod (x - ( \pm \lambda_{1}\sqrt{a_{1}} + \pm \lambda_{2}\sqrt{a_{2}} + \dots + \pm \lambda_{n}\sqrt{a_{n}} ))$ is also in $\mathbb{Q}[x]$ by induction.

Also,

$\begin{aligned} f( q - \lambda_{k+1}\sqrt{a_{k+1}} ) &= f( \sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}} ) \\ &= 0 \end{aligned}$

As is irreducible over $\mathbb{Q}$ , must divides . Thus $f( q + \lambda_{k+1}\sqrt{a_{k+1}} ) = 0$

Thus, there exits $\alpha_{1}, \alpha_{2} \dots \alpha_{n}$ which is either or such that $\sum_{i=1}^{k} \alpha_{i} \lambda_{i}\sqrt{a_{i}} = q + \lambda_{k+1}\sqrt{a_{k+1}}$

Thus

$\begin{cases} \sum_{i=1}^{k} \alpha_{i} \lambda_{i}\sqrt{a_{i}} = q + \lambda_{k+1}\sqrt{a_{k+1}} \\ \sum_{i=1}^{k} \lambda_{i}\sqrt{a_{i}} = q - \lambda_{k+1}\sqrt{a_{k+1}} \end{cases}$ $\begin{aligned} \sum_{i=1}^{k} \alpha_{i} \lambda_{i}\sqrt{a_{i}} + \sum_{i=1}^{k} \lambda_{i}\sqrt{a_{i}} &= q + \lambda_{k+1}\sqrt{a_{k+1} }+ q - \lambda_{k+1}\sqrt{a_{k+1}} \\ \sum_{i=1}^{k} ( \alpha_{i} + 1 ) \lambda_{i}\sqrt{a_{i}} &= 2q \in \mathbb{Q} \\ \end{aligned}$

By the induction assumption, we have $(\alpha_{i} + 1)\lambda_{i} = 0$ for all $i \le k$ . As $\lambda_{i} \neq 0$ for all , we have $\alpha_{i} = -1$ for all $i \le k$ .

Thus,

$\begin{aligned} \sum_{i=1}^{k} ( \alpha_{i} + 1 ) \lambda_{i}\sqrt{a_{i} } &= 0 \end{aligned}$

which implies q = 0 .

Thus,

$\begin{aligned} \sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}} &= 0 \\ \sqrt{a_{k+1}} (\sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}}) &= 0 \\ \sum_{i=1}^{k} \lambda_{i}\sqrt{a_{k+1}a_{i}} &= - \lambda_{k+1} a_{k+1} \in \mathbb{Q}\\ \end{aligned}$

The only thing left to proof is that $a_{k+1}a_{i}, i \le k$ satisfy the assumption in the question, and then the induction assumption will apply, which proves that $\lambda_{i} = 0, i \le k$ .

Given any i < j , we have

$\begin{aligned} \sqrt{\frac{a_{j}a_{k+1}}{a_{i}a_{k+1}}} &= \sqrt{\frac{a_{j}}{a_{i}}} \notin \mathbb{Q} \\ \sqrt{a_{j}a_{k+1}} &= a_{k+1} \sqrt{\frac{a_{j}}{a_{k+1}}} \notin \mathbb{Q} \end{aligned}$

◻ :::

As, we does not use any special property of $\mathbb{Q}$ other then the fact that it is a field. So, the proof should be valid for any field $\mathbb{F}$ that characteristic equals 0, if we rephrase the question in term of $\mathbb{F}$ .

::: question Question 3. Given distinct element $a_{1}, a_{2}, \dots, a_{n} \in \mathbb{F}$ , let $b_{i}$ be a root of $x^{2} - a_{i}$ for all . And $b_{i}b_{j}^{-1} \notin \mathbb{F}$ for all $i \neq j$ , and $b_{i} \notin \mathbb{F}$ for all . Then given $\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}$ , $\sum_{i=0}^{n} \lambda_{i}b_{i} \in \mathbb{F}$ if and only if $\lambda_{i} = 0$ for all . :::

If we try to shift $b_{k}$ to $b_{k} + c_{k}$ such that $c_{k} \in \mathbb{F}$ , then the result clearly still stand. So, the previous question can be generalised a bit.

::: question Question 4. Given distinct irreducible quadratic polynomial $f_{1}, f_{2} \dots f_{n}$ in $\mathbb{F}[x]$ .

And given any $i \neq j$ , $f_{j}$ does not have root in $\mathbb{F}[x]/f_{i}$ .

And $b_{i}$ be all roots of $f_{i}$ for all . Then given $\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}$ , $\sum_{i=0}^{n} \lambda_{i}b_{i} \in \mathbb{F}$ if and only if $\lambda_{i} = 0$ for all . :::

If we are not satisfied with the quadratic polynomials, we can try to generalise the question to any polynomial. However, we may need to have a more strict condition on the roots.

::: question Question 5. Given distinct irreducible polynomial $f_{1}, f_{2} \dots f_{n}$ in $\mathbb{F}[x]$ with orders greater or equal than 2.

Given any $i \neq j$ . $f_{j}$ does not have root in $\mathbb{F}[x]/f_{i}$ .

And $b_{i}$ be all roots of $f_{i}$ for all . Then given $\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}$ , $\sum_{i=0}^{n} \lambda_{i}b_{i,1} \in \mathbb{F}$ if and only if $\lambda_{i} = 0$ for all . :::

To prove the above question, we need another definition and some relating lemma.

::: definition Definition 1. Given a polynomial $f(x) \in \mathbb{F}[x]$ , $f(x) = x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + \dots + a_{n-1}x + a_{n}$

Define the derivative of as $f'(x) = n x^{n-1} + (n-1)a_{1}x^{n-2} + (n-2)a_{2}x^{n-3} + \dots + 2a_{n-2}x + a_{n-1}$ :::

Through some simple calculation, we can prove the following lemma.

::: lemma Lemma 1. Given polynomials $f(x), g(x) \in \mathbb{F}[x]$ , :::

::: lemma Lemma 2. Given a irreducible polynomial $f(x) \in \mathbb{F}[x]$ , and a field $\mathbb{K}$ , such that $\mathbb{F} \subseteq \mathbb{K}$ and can be factorized into product of linear factors in $\mathbb{K}$ .

Chose a root $\alpha$ of in $\mathbb{K}$ . Then $(x-\alpha)^2$ divides in $\mathbb{K}$ , if and only if . :::

As f'(x) is still a polynomial in $\mathbb{F}[x]$ , and we can choose arbitrary $\mathbb{K}$ . This lemma implies that irreducible polynomial f(x) have distinct roots if and only if $f'(x) \neq 0$ .

::: proof Proof. If $(x-\alpha)^2$ divides f(x) in $\mathbb{K}$ .

Let, $f(x) = (x-\alpha)^{2}g(x)$

Then, $f'(x) = 2(x-\alpha)g(x) + (x-\alpha)^{2}g'(x)$

Then $f'(\alpha) = 0$ . As f(x) is irreducible, and f(x) and f'(x) have common root. So, f(x) divides f'(x) .

As, the degree of f'(x) is less than f(x) , then f'(x) = 0 .

Conversely, if f'(x) = 0 .

Let, $f(x) = (x-\alpha)g(x)$

Then, $f'(x) = g(x) + (x-\alpha)g'(x)$

Then,

$\begin{aligned} 0 & = f'(\alpha) \\ & = g(\alpha) + (\alpha-\alpha)g'(\alpha) \\ & = g(\alpha) \end{aligned}$

So, $\alpha$ is a root of g(x) .

Thus, $(x-\alpha)^2$ divides f(x) in $\mathbb{K}$ . ◻ :::

::: lemma Lemma 3. Given distinct irreducible polynomials $f(x), g(x) \in \mathbb{F}[x]$ , where the characteristic of $\mathbb{F}$ equals .

Chose any field $\mathbb{K}$ , such that $\mathbb{F} \subseteq \mathbb{K}$ , and and can be factorized into product of linear factors in $\mathbb{K}$ .

Let $a_{1}, a_{2} \dots a_{n}$ be all roots of in $\mathbb{K}$ , and $b_{1}, b_{2} \dots b_{m}$ be all roots of in $\mathbb{K}$ .

Then, $h(x) = \prod_{i=1}^{n}\prod_{j=1}^{m}(x-(a_{i} + b_{j}))$ have coefficients in $\mathbb{F}$ , :::

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