Problem involving irreducible element -Ring theory

[chwala]

Homework Statement

see attached...

Relevant Equations

Ring theory

Hey guys, i need insight on the highlighted part...the steps before this are quite clear:

where is the $\dfrac{3}{4}M$ coming from? and how to show the argument does not work for $n=-3$? i should be able to check on this ...later.

and why is the Norm of say;

$ϒ =a+c\sqrt b$

taken as

$N(ϒ)=a^2-c^2b$

and not

$N(ϒ)=a^2+c^2b$?
cheers

 

[pasmith]

In \mathbb{Q}[X : X^2 \in \mathbb{Q}] \supset \mathbb{Z}[X] we have (a + bX)(a - bX) = a^2 - b^2X^2 \in \mathbb{Q}. If this is zero, then we have either a + bX = 0 or a - bX = 0 and in either case a = b = 0. Hence for a + bX \neq 0 we have <br /> (a +bX)^{-1} = \frac{a - bX}{a^2 - b^2X^2}. Note that if X^2 = - N &lt; 0, as in your case, then <br /> a^2 - b^2 X^2 = a^2 + Nb^2.

 

[chwala]

Homework Statement:: see attached...
Relevant Equations:: Ring theory

Hey guys, i need insight on the highlighted part...the steps before this are quite clear:

View attachment 323675

where is the $\dfrac{3}{4}M$ coming from? and how to show the argument does not work for $n=-3$? i should be able to check on this ...later.

and why is the Norm of say;

$ϒ =a+c\sqrt b$

taken as

$N(ϒ)=a^2-c^2b$

and not

$N(ϒ)=a^2+c^2b$?
cheers

...
I am trying to follow this with a practical example...i still have doubts on the highlighted, let us consider; the gcd $(7,4)$ with $n=-2$ for example, then we shall have (using Euclid algorithm),

$7=1⋅4+3$

...
$\dfrac{7}{4}= \dfrac{7(l-m\sqrt{-2})}{l^2+2m^2}=\dfrac{7l}{l^2+2m^2}-\dfrac{m\sqrt{-2}}{l^2+2m^2}$

that is by following the attached literature... then

$\dfrac{t}{M}=\dfrac{7l}{l^2+2m^2}$

and

$\dfrac{s\sqrt{n}}{M}=\dfrac{m\sqrt{-2}}{l^2+2m^2}$

...let $X, Y$ be the closest integers to the two ratios on the right...not quite understanding this statement...do they mean $X$ an integer value i.e close to $\dfrac{7l}{l^2+2m^2}$ and $Y$ an integer value close to $\dfrac{m\sqrt{-2}}{l^2+2m^2}$?

 

[chwala]

...
I am trying to follow this with a practical example...i still have doubts on the highlighted, let us consider; the gcd $(7,4)$ with $n=-2$ for example, then we shall have (using Euclid algorithm),

$7=1⋅4+3$

...
$\dfrac{7}{4}= \dfrac{7(l-m\sqrt{-2})}{l^2+2m^2}=\dfrac{7l}{l^2+2m^2}-\dfrac{m\sqrt{-2}}{l^2+2m^2}$

that is by following the attached literature... then

$\dfrac{t}{M}=\dfrac{7l}{l^2+2m^2}$

and

$\dfrac{s\sqrt{n}}{M}=\dfrac{m\sqrt{-2}}{l^2+2m^2}$

...let $X, Y$ be the closest integers to the two ratios on the right...not quite understanding this statement...do they mean $X$ an integer value i.e close to $\dfrac{7l}{l^2+2m^2}$ and $Y$ an integer value close to $\dfrac{m\sqrt{-2}}{l^2+2m^2}$?

I think i now understand this part;

It basically means;$|\frac{t}{M} - X|≤\frac{1}{2}$

and
$|\frac{s\sqrt {n}}{M} - Y|≤\frac{1}{2}$

i.e

$-\frac{1}{2}≤\frac{t}{M} - X≤\frac{1}{2}$

and$-\frac{1}{2}≤\frac{s\sqrt{n}}{M} - Y≤\frac{1}{2}$

it is true that, (on adding them)

$0≤\frac{t}{M}+\frac{s\sqrt{n}}{M}- X-Y≤1$

 

[SammyS]

Homework Statement:: see attached...
Relevant Equations:: Ring theory

Hey guys, i need insight on the highlighted part...the steps before this are quite clear:

View attachment 323675

where is the $\dfrac{3}{4}M$ coming from? and how to show the argument does not work for $n=-3$? i should be able to check on this ...later.cheers

The proof uses an arbitrary value for the integer, $n$, through the point where

$\displaystyle N(r) \le M( (1/2)^2-(n)(1/2)^2)$ .

Then substituting $\displaystyle n=-2$ one sees that $\displaystyle M( (1/2)^2-(-2)(1/2)^2)=\dfrac{3}{4}M<N(b)$ .

What do you get for $n=-3$ ?

 

[SammyS]

...
I am trying to follow this with a practical example...i still have doubts on the highlighted, let us consider; the gcd $(7,4)$ with $n=-2$ for example, then we shall have (using Euclid algorithm),

$7=1⋅4+3$

...
$\dfrac{7}{4}= \dfrac{7(l-m\sqrt{-2})}{l^2+2m^2}=\dfrac{7l}{l^2+2m^2}-\dfrac{m\sqrt{-2}}{l^2+2m^2}$

that is by following the attached literature... then

$\dfrac{t}{M}=\dfrac{7l}{l^2+2m^2}$

and

$\dfrac{s\sqrt{n}}{M}=\dfrac{m\sqrt{-2}}{l^2+2m^2}$

...let $X, Y$ be the closest integers to the two ratios on the right...not quite understanding this statement...do they mean $X$ an integer value i.e close to $\dfrac{7l}{l^2+2m^2}$ and $Y$ an integer value close to $\dfrac{m\sqrt{-2}}{l^2+2m^2}$?

This is not a particularly good example of two elements in $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ .

Writing them as elements of $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ , we have $\displaystyle 7= 7+0\sqrt{-2\,}$ , and $\displaystyle 4= 4+0\sqrt{-2\,}$

So, we have $l=4\,,\ m=0\,,\ \text{ and } M=16$ .

This gives $\displaystyle \dfrac{t}{M}=\dfrac{7l}{M}=\dfrac{28}{16}\, ,$ which reduces to $\displaystyle \dfrac 7 4 = 1.75$ . And for $\displaystyle \dfrac{s\sqrt{-2\,}}{M}$ we have $\displaystyle \dfrac{m\sqrt{-2\,}}{M}=0$ .

Being the integer closest to $1.75$, we have $X=2$ . It should be apparent that $Y=0$ . This gives $\displaystyle q= X+Y\sqrt{-2\,}=2+0\,\sqrt{-2\,}=2$

Therefore, $\displaystyle \ r = a-q\,b = 7-2\cdot 4 =-1$.

This is very similar to the standard version of Euclidean Division, except that here the remainder falls in the interval $\displaystyle [-b/2 \,, \ b/2]\,$. (The author should clean that up to make one of the ends of the interval be open.)

The Euclidean Algorithm goes further than this. The next step: Promote $b$ to take the role of $a$ and $r$ to take the role of $b$, i.e. - divide $4$ by $-1$ find a quotient and a remainder.. If the remainder is zero, you are nearly (or half) done. If the new $r\ne 0$, promote $(b,\ r)$ to $(a,\ b)$ and repeat . . . until some $r=0$ . The previous value of $r$ is the $\gcd (a,\ b)$ .

The final step in the Euclidean Algorithm has you backtrack through your previous steps to express the gdc as $\displaystyle \gcd (a,\ b)=ua+vb$ for some $u$ and $v$.

 

[chwala]

This is not a particularly good example of two elements in $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ .

Writing them as elements of $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ , we have $\displaystyle 7= 7+0\sqrt{-2\,}$ , and $\displaystyle 4= 4+0\sqrt{-2\,}$

So, we have $l=4\,,\ m=0\,,\ \text{ and } M=16$ .

This gives $\displaystyle \dfrac{t}{M}=\dfrac{7l}{M}=\dfrac{28}{16}\, ,$ which reduces to $\displaystyle \dfrac 7 4 = 1.75$ . And for $\displaystyle \dfrac{s\sqrt{-2\,}}{M}$ we have $\displaystyle \dfrac{m\sqrt{-2\,}}{M}=0$ .

Being the integer closest to $1.75$, we have $X=2$ . It should be apparent that $Y=0$ . This gives $\displaystyle q= X+Y\sqrt{-2\,}=2+0\,\sqrt{-2\,}=2$

Therefore, $\displaystyle \ r = a-q\,b = 7-2\cdot 4 =-1$.

This is very similar to the standard version of Euclidean Division, except that here the remainder falls in the interval $\displaystyle [-b/2 \,, \ b/2]\,$. (The author should clean that up to make one of the ends of the interval be open.)

The Euclidean Algorithm goes further than this. The next step: Promote $b$ to take the role of $a$ and $r$ to take the role of $b$, i.e. - divide $4$ by $-1$ find a quotient and a remainder.. If the remainder is zero, you are nearly (or half) done. If the new $r\ne 0$, promote $(b,\ r)$ to $(a,\ b)$ and repeat . . . until some $r=0$ . The previous value of $r$ is the $\gcd (a,\ b)$ .

The final step in the Euclidean Algorithm has you backtrack through your previous steps to express the gdc as $\displaystyle \gcd (a,\ b)=ua+vb$ for some $u$ and $v$.

Thanks @SammyS ...i will go through your remarks later...cheers.

 

[chwala]

This is not a particularly good example of two elements in $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ .

Writing them as elements of $\displaystyle \mathbb{Z} [\sqrt{-2\,} \,]$ , we have $\displaystyle 7= 7+0\sqrt{-2\,}$ , and $\displaystyle 4= 4+0\sqrt{-2\,}$

So, we have $l=4\,,\ m=0\,,\ \text{ and } M=16$ .

This gives $\displaystyle \dfrac{t}{M}=\dfrac{7l}{M}=\dfrac{28}{16}\, ,$ which reduces to $\displaystyle \dfrac 7 4 = 1.75$ . And for $\displaystyle \dfrac{s\sqrt{-2\,}}{M}$ we have $\displaystyle \dfrac{m\sqrt{-2\,}}{M}=0$ .

Being the integer closest to $1.75$, we have $X=2$ . It should be apparent that $Y=0$ . This gives $\displaystyle q= X+Y\sqrt{-2\,}=2+0\,\sqrt{-2\,}=2$

Therefore, $\displaystyle \ r = a-q\,b = 7-2\cdot 4 =-1$.

This is very similar to the standard version of Euclidean Division, except that here the remainder falls in the interval $\displaystyle [-b/2 \,, \ b/2]\,$. (The author should clean that up to make one of the ends of the interval be open.)

The Euclidean Algorithm goes further than this. The next step: Promote $b$ to take the role of $a$ and $r$ to take the role of $b$, i.e. - divide $4$ by $-1$ find a quotient and a remainder.. If the remainder is zero, you are nearly (or half) done. If the new $r\ne 0$, promote $(b,\ r)$ to $(a,\ b)$ and repeat . . . until some $r=0$ . The previous value of $r$ is the $\gcd (a,\ b)$ .

The final step in the Euclidean Algorithm has you backtrack through your previous steps to express the gdc as $\displaystyle \gcd (a,\ b)=ua+vb$ for some $u$ and $v$.

Am trying to substitute your values into the final inequality but it seems it does not satisfy unless i am not getting it correctly, since $Y=0$ then,

From this line,

$M\left[\dfrac{t}{M} - X\right]^2 ≤ M[(0.5)^2-n(0.5)^2] ≤ 0.75M$

given that, $\dfrac{t}{M}=1.75$, $M=16$, $X=2$, $n=-1$ and $m=0$.

We shall have

$16(1.75-2)^2 ≤16(0.25+0.25) ≤12$

$1≤8≤12$

Looks fine i had made a mistake on the value of $X$. Next question suppose we had a value for $m$ how would that work out? Cheers.

Note
Author indicated that the argument works for $n=-1,2,3$ but not for $n=-3$. I have verified this in the following step; When $n=-3$ we shall have,

$1≤16≤12$

which is a contradiction.