Addition Of Algebraic Element

Date: 2023/06/25
Last Updated: 2024-06-05T11:18:36.166Z
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 $n > 2$ 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 $n$ positive integers, $a_1 < a_2 < \dots < a_n$, such that $\sqrt{\frac{a_{j}}{a_{i}}} \notin \mathbb{Q}$ for all $i < j$ and $\sqrt{a_{i}} \notin \mathbb{Q}$ for all $i$. 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 $j$, 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 $g$ is irreducible over $\mathbb{Q}$, $g$ must divides $f$. Thus $f( q + \lambda_{k+1}\sqrt{a_{k+1}} ) = 0$

Thus, there exits $\alpha_{1}, \alpha_{2} \dots \alpha_{n}$ which is either $1$ or $-1$ 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 $i$, 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 $n$ 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 $i$. And $b_{i}b_{j}^{-1} \notin \mathbb{F}$ for all $i \neq j$, and $b_{i} \notin \mathbb{F}$ for all $i$. 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 $i$. :::

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 $n$ 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 $i$. 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 $i$. :::

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 $n$ 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 $i$. 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 $i$. :::

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 $f'(x)$ of $f(x)$ 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 $f(x), g(x) \in \mathbb{F}[x]$, $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$ :::

::: 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 $f(x)$ can be factorized into product of linear factors in $\mathbb{K}$.

Chose a root $\alpha$ of $f(x)$ in $\mathbb{K}$. Then $(x-\alpha)^2$ divides $f(x)$ in $\mathbb{K}$, if and only if $f'(x) = 0$. :::

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 $0$.

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

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

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