# Problem involving irreducible element -Ring theory

• chwala
In summary, the conversation discusses the steps before a certain point and then focuses on the highlighted part. The highlighted part is about finding the norm of an element in a ring and why it is taken as a squared difference instead of a squared sum. The conversation also touches on a practical example and the use of the Euclidean algorithm to find the gcd. It concludes with a discussion about expressing the gcd as a linear combination of the two elements.
chwala
Gold Member
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

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 $$(a +bX)^{-1} = \frac{a - bX}{a^2 - b^2X^2}.$$ Note that if $X^2 = - N < 0$, as in your case, then $$a^2 - b^2 X^2 = a^2 + Nb^2.$$

chwala
chwala said:
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 said:
...
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##

chwala said:
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## ?

Last edited:
chwala said:
...
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
SammyS said:
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.

SammyS said:
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##

Last edited:

## 1. What is an irreducible element in ring theory?

An irreducible element in ring theory is an element of a ring that cannot be factored into a product of two non-units. In other words, it is an element that cannot be broken down any further into simpler components.

## 2. How is an irreducible element different from a prime element?

While both irreducible and prime elements cannot be factored into simpler components, the main difference is that a prime element must also be a non-zero divisor. This means that a prime element cannot be multiplied by any other element to equal zero. However, an irreducible element can be a zero divisor in some cases.

## 3. What role do irreducible elements play in ring theory?

Irreducible elements are important in ring theory because they help to classify the structure of a ring. They also play a crucial role in understanding the properties of factorization in rings, which is essential in many areas of mathematics.

## 4. How can irreducible elements be identified in a ring?

One way to identify irreducible elements in a ring is to look for elements that cannot be factored into a product of two non-units. This can be done by using the unique factorization property of rings, which states that every non-zero element can be written as a product of irreducible elements.

## 5. Can an element be both irreducible and prime in a ring?

Yes, in some rings, an element can be both irreducible and prime. These rings are known as unique factorization domains (UFDs). However, not all rings have this property, so it is important to distinguish between irreducible and prime elements in general.

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