Addition Of Algebraic Element

Date: 2023/06/25
Last Updated: 2024-03-07T20:23:34.701Z
Categories: Math
Tags: Math, Galois Theory, Field Theory, Notes
Read Time: 4 minutes

This is some generalization of a question.

Link to the pdf file: Addition Of Algebraic Element

Link to the source code: Addition Of Algebraic Element

::: question Question 1. Proof that 1+2+3ā‹Æ+n\sqrt{1} + \sqrt{2} + \sqrt{3} \dots + \sqrt{n} for n>2n > 2 is irrational. :::

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

::: question Question 2. Given nn positive integers, a1<a2<ā‹Æ<ana_1 < a_2 < \dots < a_n, such that ajaiāˆ‰Q\sqrt{\frac{a_{j}}{a_{i}}} \notin \mathbb{Q} for all i<ji < j and aiāˆ‰Q\sqrt{a_{i}} \notin \mathbb{Q} for all ii. Also given Ī»1,Ī»2ā€¦Ī»nāˆˆQ\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{Q}. Prove that āˆ‘i=1nĪ»iaiāˆˆQ\sum_{i=1}^{n} \lambda_{i} \sqrt{a_{i}} \in \mathbb{Q} if and only if Ī»1=Ī»2=ā‹Æ=Ī»n=0\lambda_{1} = \lambda_{2} = \dots = \lambda_{n} = 0. :::

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

Suppose there exist Ī»1,Ī»2ā€¦Ī»k+1āˆˆQ\lambda_{1}, \lambda_{2} \dots \lambda_{k+1} \in \mathbb{Q} such that āˆ‘i=1k+1Ī»iai=qāˆˆQ\sum_{i=1}^{k+1} \lambda_{i}\sqrt{a_{i}} = q \in \mathbb{Q}

If Ī»j=0\lambda_{j} = 0, then āˆ’Ī»jaj+āˆ‘i=1k+1Ī»iai=qāˆˆQ-\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 Ī»jā‰ 0\lambda_{j} \neq 0 for all jj, then we have āˆ‘i=1kĪ»iai=qāˆ’Ī»k+1ak+1\sum_{i=1}^{k} \lambda_{i}\sqrt{a_{i}} = q - \lambda_{k+1}\sqrt{a_{k+1}}

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

Also, the polynomial f=āˆ(xāˆ’(Ā±Ī»1a1+Ā±Ī»2a2+ā‹Æ+Ā±Ī»nan))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 Q[x]\mathbb{Q}[x] by induction.

Also,

f(qāˆ’Ī»k+1ak+1)=f(āˆ‘i=1k+1Ī»iai)=0\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 gg is irreducible over Q\mathbb{Q}, gg must divides ff. Thus f(q+Ī»k+1ak+1)=0f( q + \lambda_{k+1}\sqrt{a_{k+1}} ) = 0

Thus, there exits Ī±1,Ī±2ā€¦Ī±n\alpha_{1}, \alpha_{2} \dots \alpha_{n} which is either 11 or āˆ’1-1 such that āˆ‘i=1kĪ±iĪ»iai=q+Ī»k+1ak+1\sum_{i=1}^{k} \alpha_{i} \lambda_{i}\sqrt{a_{i}} = q + \lambda_{k+1}\sqrt{a_{k+1}}

Thus

{āˆ‘i=1kĪ±iĪ»iai=q+Ī»k+1ak+1āˆ‘i=1kĪ»iai=qāˆ’Ī»k+1ak+1\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} āˆ‘i=1kĪ±iĪ»iai+āˆ‘i=1kĪ»iai=q+Ī»k+1ak+1+qāˆ’Ī»k+1ak+1āˆ‘i=1k(Ī±i+1)Ī»iai=2qāˆˆQ\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 (Ī±i+1)Ī»i=0(\alpha_{i} + 1)\lambda_{i} = 0 for all iā‰¤ki \le k. As Ī»iā‰ 0\lambda_{i} \neq 0 for all ii, we have Ī±i=āˆ’1\alpha_{i} = -1 for all iā‰¤ki \le k.

Thus,

āˆ‘i=1k(Ī±i+1)Ī»iai=0\begin{aligned} \sum_{i=1}^{k} ( \alpha_{i} + 1 ) \lambda_{i}\sqrt{a_{i} } &= 0 \end{aligned}

which implies q=0q = 0.

Thus,

āˆ‘i=1k+1Ī»iai=0ak+1(āˆ‘i=1k+1Ī»iai)=0āˆ‘i=1kĪ»iak+1ai=āˆ’Ī»k+1ak+1āˆˆQ\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 ak+1ai,iā‰¤ka_{k+1}a_{i}, i \le k satisfy the assumption in the question, and then the induction assumption will apply, which proves that Ī»i=0,iā‰¤k\lambda_{i} = 0, i \le k.

Given any i<ji < j, we have

ajak+1aiak+1=ajaiāˆ‰Qajak+1=ak+1ajak+1āˆ‰Q\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}

ā—» :::

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

::: question Question 3. Given nn distinct element a1,a2,ā€¦,anāˆˆFa_{1}, a_{2}, \dots, a_{n} \in \mathbb{F}, let bib_{i} be a root of x2āˆ’aix^{2} - a_{i} for all ii. And bibjāˆ’1āˆ‰Fb_{i}b_{j}^{-1} \notin \mathbb{F} for all iā‰ ji \neq j, and biāˆ‰Fb_{i} \notin \mathbb{F} for all ii. Then given Ī»1,Ī»2ā€¦Ī»nāˆˆF\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}, āˆ‘i=0nĪ»ibiāˆˆF\sum_{i=0}^{n} \lambda_{i}b_{i} \in \mathbb{F} if and only if Ī»i=0\lambda_{i} = 0 for all ii. :::

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

::: question Question 4. Given nn distinct irreducible quadratic polynomial f1,f2ā€¦fnf_{1}, f_{2} \dots f_{n} in F[x]\mathbb{F}[x].

And given any iā‰ ji \neq j, fjf_{j} does not have root in F[x]/fi\mathbb{F}[x]/f_{i}.

And bib_{i} be all roots of fif_{i} for all ii. Then given Ī»1,Ī»2ā€¦Ī»nāˆˆF\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}, āˆ‘i=0nĪ»ibiāˆˆF\sum_{i=0}^{n} \lambda_{i}b_{i} \in \mathbb{F} if and only if Ī»i=0\lambda_{i} = 0 for all ii. :::

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 nn distinct irreducible polynomial f1,f2ā€¦fnf_{1}, f_{2} \dots f_{n} in F[x]\mathbb{F}[x] with orders greater or equal than 2.

Given any iā‰ ji \neq j. fjf_{j} does not have root in F[x]/fi\mathbb{F}[x]/f_{i}.

And bib_{i} be all roots of fif_{i} for all ii. Then given Ī»1,Ī»2ā€¦Ī»nāˆˆF\lambda_{1}, \lambda_{2} \dots \lambda_{n} \in \mathbb{F}, āˆ‘i=0nĪ»ibi,1āˆˆF\sum_{i=0}^{n} \lambda_{i}b_{i,1} \in \mathbb{F} if and only if Ī»i=0\lambda_{i} = 0 for all ii. :::

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

::: definition Definition 1. Given a polynomial f(x)āˆˆF[x]f(x) \in \mathbb{F}[x], f(x)=xn+a1xnāˆ’1+a2xnāˆ’2+ā‹Æ+anāˆ’1x+anf(x) = x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + \dots + a_{n-1}x + a_{n}

Define the derivative fā€²(x)f'(x) of f(x)f(x) as fā€²(x)=nxnāˆ’1+(nāˆ’1)a1xnāˆ’2+(nāˆ’2)a2xnāˆ’3+ā‹Æ+2anāˆ’2x+anāˆ’1f'(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)āˆˆF[x]f(x), g(x) \in \mathbb{F}[x], (f(x)g(x))ā€²=fā€²(x)g(x)+f(x)gā€²(x)(f(x)g(x))' = f'(x)g(x) + f(x)g'(x) :::

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

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

As fā€²(x)f'(x) is still a polynomial in F[x]\mathbb{F}[x], and we can choose arbitrary K\mathbb{K}. This lemma implies that irreducible polynomial f(x)f(x) have distinct roots if and only if fā€²(x)ā‰ 0f'(x) \neq 0.

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

Let, f(x)=(xāˆ’Ī±)2g(x)f(x) = (x-\alpha)^{2}g(x)

Then, fā€²(x)=2(xāˆ’Ī±)g(x)+(xāˆ’Ī±)2gā€²(x)f'(x) = 2(x-\alpha)g(x) + (x-\alpha)^{2}g'(x)

Then fā€²(Ī±)=0f'(\alpha) = 0. As f(x)f(x) is irreducible, and f(x)f(x) and fā€²(x)f'(x) have common root. So, f(x)f(x) divides fā€²(x)f'(x).

As, the degree of fā€²(x)f'(x) is less than f(x)f(x), then fā€²(x)=0f'(x) = 0.

Conversely, if fā€²(x)=0f'(x) = 0.

Let, f(x)=(xāˆ’Ī±)g(x)f(x) = (x-\alpha)g(x)

Then, fā€²(x)=g(x)+(xāˆ’Ī±)gā€²(x)f'(x) = g(x) + (x-\alpha)g'(x)

Then,

0=fā€²(Ī±)=g(Ī±)+(Ī±āˆ’Ī±)gā€²(Ī±)=g(Ī±)\begin{aligned} 0 & = f'(\alpha) \\ & = g(\alpha) + (\alpha-\alpha)g'(\alpha) \\ & = g(\alpha) \end{aligned}

So, Ī±\alpha is a root of g(x)g(x).

Thus, (xāˆ’Ī±)2(x-\alpha)^2 divides f(x)f(x) in K\mathbb{K}.Ā ā—» :::

::: lemma Lemma 3. Given distinct irreducible polynomials f(x),g(x)āˆˆF[x]f(x), g(x) \in \mathbb{F}[x], where the characteristic of F\mathbb{F} equals 00.

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

Let a1,a2ā€¦ana_{1}, a_{2} \dots a_{n} be all roots of f(x)f(x) in K\mathbb{K}, and b1,b2ā€¦bmb_{1}, b_{2} \dots b_{m} be all roots of g(x)g(x) in K\mathbb{K}.

Then, h(x)=āˆi=1nāˆj=1m(xāˆ’(ai+bj))h(x) = \prod_{i=1}^{n}\prod_{j=1}^{m}(x-(a_{i} + b_{j})) have coefficients in F\mathbb{F}, :::


å¤Ŗ阳你ē…§č€€ęˆ‘

å’Œé£Žä½ å¹é†’ęˆ‘

å¦‚ę­¤ēš„ ē®€å•

åœØčæ™é‡Œ ę„Ÿå—ä½ ēš„ęø©ęš–

ē…§č€€ęˆ‘ēš„天ēœŸ

čæ™ę°øčæœé—Ŗ光ēš„ę¢¦ęƒ³

ē©æčæ‡ęƏäø€äøŖęø…ę™Ø

ęƏäø€äøŖäøēœ å¤œę™š

ę„æꉀ꜉ēš„ę‚²ä¼¤

éƒ½åŒ–ęˆå–œę‚¦ēš„力量

就像你ēˆ±ē€äø–ē•Œ

ä½ ę— å°½ēš„å…‰čŠ’