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 wildrabbitt 2019-07-01 20:00

dividing an algebraic integer by another

Hi,

I've worked out that the algebraic integer $$6+23\sqrt{2}$$ is divisible by
$$2+\sqrt{2}$$. I find finding these factors by looking at norms quite tiring.

Is another way to work out

$$\frac{6+23\sqrt{2}}{2+\sqrt{2}}$$

in it's simplest form?

A division algorithm for example. Please show me how it goes.

 VBCurtis 2019-07-01 20:10

It's really elementary, so I may be misunderstanding your question:
Multiply top and bottom by the conjugate of the bottom. That makes the denominator a real number, so regular primary-school division can proceed.

 CRGreathouse 2019-07-01 20:15

Note that, in general, you don't get an algebraic integer (just like dividing an integer by another integer doesn't give an integer, in general, but a rational number).

 wildrabbitt 2019-07-01 21:05

[QUOTE]
It's really elementary, so I may be misunderstanding your question:
Multiply top and bottom by the conjugate of the bottom. That makes the denominator a real number, so regular primary-school division can proceed. [/QUOTE]Thanks. You weren't misunderstanding my question. I believe that there's something hidden with maths like it's a mystery or I'm sereptticiously tricking myself into making simple things seem hard so I can feel better about how good I am at maths. Perhaps also that I think that the simple form should be smaller than it's numerator and multiplying the numerator by a number greater than 1 wouldn't have occurred to me.

So,

$$\frac{6+23\sqrt{2}}{2+\sqrt{2}}=\frac{(6+23\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} =\frac{12-46+(46-6)\sqrt{2}}{2}=\frac{-34+40\sqrt{2}}{2}=-17+20\sqrt{2}$$

/* editted out the mistakes */

[QUOTE]
Note that, in general, you don't get an algebraic integer (just like dividing an integer by another integer doesn't give an integer, in general, but a rational number).

[/QUOTE]I take your point but isn't it okay to say that in the ring $$Z[\sqrt{2}]$$,$$6+23\sqrt{2}$$ has a factor $$2+\sqrt{2}$$?

Thanks very much to both of you.

 wildrabbitt 2019-07-01 21:42

I guess I prompted CR's post

by writing
[QUOTE]
I've worked out that the algebraic integer [COLOR=inherit][/COLOR][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]23[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT] is divisible by
[COLOR=inherit][/COLOR][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT]. I find finding these factors by looking at norms quite tiring.
[/QUOTE]

I should have said has a factor instead of divisble.

 Dr Sardonicus 2019-07-01 22:32

Using the computational might of Pari-GP,

? f=x^2-2;z=Mod((6+23*x)/(2+x),f)
%1 = Mod(20*x - 17, x^2 - 2)

? charpoly(%)
%2 = x^2 + 34*x - 511

 CRGreathouse 2019-07-02 15:02

[QUOTE=wildrabbitt;520487]I take your point but isn't it okay to say that in the ring $$Z[\sqrt{2}]$$,$$6+23\sqrt{2}$$ has a factor $$2+\sqrt{2}$$?[/QUOTE]

Yes, just like in the ring Z it's right to say that 15 has a factor 3. :smile: But if you want to divide integers in general (apart from denominator 0, of course), you get rational numbers. Similarly, if you want to divide algebraic integers in general (or for a particular ring), you get algebraic numbers (or those in your ring).

 Dr Sardonicus 2019-07-03 16:17

If f is an irreducible polynomial in Q[x], and b is a nonzero polynomial in Q[x] of degree less than the degree of f, then the polmod Mod(b, f) is invertible. Thus, Mod(a/b,f) is defined for any polynomial a in Q[x]. In Pari-GP calculations, f is usually [i]monic[/i] (leading coefficient is 1) with integer coefficients. It is often the defining polynomial of a number field.

In the above example, I took f = x^2 - 2, and I also found the characteristic polynomial of Mod(a/b, f). The point of doing that was that Mod(a/b, f) is an algebraic [i]integer[/i] precisely when its characteristic polynomial is monic and has integer coefficients. In fields of degree greater than 2, there can be cases where algebraic integers have polynomial expressions (mod f) which have fractional coefficients.

 wildrabbitt 2019-07-03 19:04

Right. Well thanks for those posts. I thought my thread had come to it's natural end but I'm glad it hasn't.

Please don't take this to mean I'm not interested in the latest replies, but since I wasn't expecting anymore I was busy looking into some more things.

I hope therefore it can be considered not to be without due interest in matters raised in this thread that I ask the following ;

As I understand it, A Euclidean Domain had a Euclidean Norm and a Euclidean Algorithm for division.
I'm fine with that.
What I'm confused about is that in the same way that Every Euclidean Domain is a UFD, every Field is a Euclidean Domain.

It seems logical to me that every field therefore has a Euclidean Norm and a Euclidean Algorithm so I'm totally puzzled about the fact that in Thomas Hardy's book The Theory of Numbers, a distinction is made between Euclidean Fields and Non-Euclidean fields.

For example, k(sqrt(23)), the real quadratic field is said not to be Euclidean whereas k(sqrt(2)) the real quadratic field associated with root 2, is said to be Euclidean.

 Dr Sardonicus 2019-07-04 13:18

[QUOTE=wildrabbitt;520689]As I understand it, A Euclidean Domain had a Euclidean Norm and a Euclidean Algorithm for division.
I'm fine with that.
What I'm confused about is that in the same way that Every Euclidean Domain is a UFD, every Field is a Euclidean Domain.[/QUOTE]
In a [i]field[/i], every division (by any nonzero element) "comes out even" with a remainder of [i]zero[/i].

You're unclear on the definitions. As the term is used in Hardy and Wright, "Euclidean field" is a number field whose ring of algebraic integers has a Euclidean (division with quotient and remainder) algorithm. The remainder is either 0 or is "smaller" than the divisor. The usual function used to measure the "size" of integers is the absolute value of the norm. You might try reading [url=https://info.maths.ed.ac.uk/assets/files/Vacation%20Projects/Sarah%20Mackie%2016.pdf]The Euclidean Algorithm in Quadratic Number Fields[/url].

 CRGreathouse 2019-07-04 15:42

In this context, fields are boring because all nonzero elements are units.

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