Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question is from a very old exercise sheet, I do not know what the notation $\mathbb{Z}_{\mathbb{C}}$ means:

  1. $1\equiv 1 \mod 4 $ in $\mathbb{Z}$. Show that $\frac{1+\sqrt{m}}{2} \in \mathbb{Z}_{\mathbb{C}}$ 2. Let $a= (19)^{1/3}$. Show that $\frac{1+a+a^{2}}{3} \in \mathbb{Z}_{\mathbb{C}}$

1. It will now be shown that $\frac{1+\sqrt{m}}{2}$ is an algebraic integer, that means that it fulfills a polynomial with leading coefficient 1 (monic) and rest integral. Wikipedia article on algebraic integers says that it suffises the monic polynomial: $x^{2}-x+\frac{1-m}{4}$ This is true because: $(x-\frac{1}{2}(1+\sqrt{m})) (x-\frac{1}{2}(1-\sqrt{m})) = x^{2}-x+\frac{1-m}{4}$

2. Let $b = 19$, then $\frac{1+b+b^{2}}{3} = \frac{1+19+19^{2}}{3}$

Then one examines : $(x-(\frac{1+19+19^{2}}{3}))(x-\frac{-1-19-19^{2}}{3}) = (x-127)(x+127) = x^{2}- 16129$

now subtituting $x$ with $x^{3}$ gives: $x^{6}-16129$, and this is the monic polynomial for $a= (19)^{1/3}$

Is this done correctly? Please do tell

share|improve this question
2  
From the first part, my guess is that $\mathbf Z_\mathbf{C}$ is the set of complex numbers which satisfy a monic polynomial with integer coefficients. –  Dylan Moreland Nov 24 '11 at 1:21
2  
@Dylan: Also known as algebraic integers. –  joriki Nov 24 '11 at 1:27
    
Thanks at both of you. –  VVV Nov 24 '11 at 1:30
    
@joriki Indeed! I should have mentioned those words. Touch typing makes me a little too economical at times. I was also somewhat worried because while I've heard "algebraic numbers" to refer to the algebraic closure of $\mathbf Q$ in $\mathbf C$, I hadn't seen a corresponding definition for integral elements. –  Dylan Moreland Nov 24 '11 at 1:31
    
Just how old is this "very old" problem sheet? –  KCd Nov 24 '11 at 20:55

1 Answer 1

up vote 5 down vote accepted

Your approach seems too complicated in both cases. Try these simpler approaches:

  1. Let $\alpha =\frac{1+\sqrt{m}}{2}%$. Then $2 \alpha -1 = \sqrt{m}$. Can you now find a monic equation for $\alpha$? Make sure you use the hypothesis on $m \bmod 4$.

  2. Let $\beta=\frac{1+a+a^{2}}{3}$. Note that $\beta = \frac{a^3-1}{3(a-1)}=\frac{6}{a-1}$. Then $a=\frac{6}{\beta}+1$. Now use that $a^3=19$ to find a monic equation for $\beta$.

share|improve this answer
    
I made an edit, are my proofs right? Thanks. –  VVV Nov 24 '11 at 14:14
    
@VVV, see my edited answer. –  lhf Nov 24 '11 at 20:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.